Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics

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Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics

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Samacheer Kalvi 12th Physics Optics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Optics Multiple Choice Questions 

12th Physics Chapter 6 Book Back Answers Question 1.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

12th Physics Wave Optics Book Back Answers Question 2.
A rod of length 10 cm lies along the principal axis of a concav e mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is, (AIPMT Main 2012)
(a) 2.5 cm
(b) 5cm
(c) 10 cm
(d) 15cm
Answer:
(b) 5cm
Hint:
By mirror formula, image distance of A
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\) ; \(\frac { { 1 } }{ { v }_{ A } } \) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
\(\frac { { 1 } }{ { v }_{ A } } \) + \(\frac { 1 }{ u }\) + \(\frac { 1 }{ (-30) }\) = \(\frac { 1 }{ (-10) }\)
12th Physics Chapter 6 Book Back Answers Optics Samacheer Kalvi
∴ vA = – 15 cm
Image distance of C, vc = – 20 cm
The length of image = |vA – vc|
= |-15 + 20| = 5 cm

Samacheer Kalvi Guru 12th Physics Question 3.
An object is placed in front of a convex mirror of focal length of/and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified. (IEE Main 2009)]
(a) 2ƒ and c
(b) c and ∞
(c) ƒ and O
(d) None of these
Answer:
(d) None of these
Hint:
There is no maximum minimum object distance for convex mirror to form real and inverted image.

Samacheer Kalvi Physics Question 4.
For light incident from air onto a slab of refractive index 2. Maximum possible angle of refraction is,
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(a) 30°
Hint:
From Snell’s law, µ = \(\frac { sin i }{ sin r }\)
Now consider an angle of incident is 90°
\(\frac { sin 90° }{ 2 }\)
sin r = sin-1 (0.5)
r = 30°

Samacheerkalvi.Guru 12th Physics Question 5.
If the velocity and wavelength of light in air is Va and λa and that in water is Va and λw, then the refractive index of water is,
(a) \(\frac { { V }_{ w } }{ { V }_{ a } } \)
(b) \(\frac { { V }_{ a } }{ { V }_{ w } } \)
(c) \(\frac { { λ }_{ w } }{ { λ }_{ a } } \)
(d) \(\frac { { V }_{ a }{ \lambda } }{ { V }_{ w }{ \lambda }_{ w } } \)
Answer:
(b) \(\frac { { V }_{ a } }{ { V }_{ w } } \)

Samacheer Kalvi.Guru 12th Physics Question 6.
Stars twinkle due to
(a) reflection
(b) total internal reflection
(c) refraction
(d) polarisation
Answer:
(c) refraction

Physics Samacheer Kalvi Question 7.
When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index,
(a) less than one
(b) less than that of glass
(c) greater than that of glass
(d) equal to that of glass
Answer:
(d) equal to that of glass
Hint:
According to len’s maker formula,
\(\frac { 1 }{ ƒ } \) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Biconvex lens dipped in a liquid, acts as a plane sheet of glass, ƒ = ∞; \(\frac { 1 }{ ƒ } \) = 0
\(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) -1 = 0;  n2 = n1

Samacheer Kalvi Guru 12 Physics Question 8.
The radius of curvature of curved surface at a thin planoconvex lens is 10 cm and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be,
(a) 5 cm
(b) 10 cm
(c) 15 cm
(d) 20 cm
Answer:
(b) 10 cm
Hint:
According to len’s maker formula,
\(\frac { 1 }{ ƒ }\) = (n – 1) \(\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
R1 = ∞; R2 = -6
∴ ƒ = \(\frac { R }{ \left( n – 1 \right) } \) (R= 10 cm; n= 1.5
ƒ = \(\frac { 10}{ \left( 1.5 – 1 \right) } \) =20 cm

Question 9.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm
Hint:
Let d1 = 5 cm and d2 = 3 cm ; n = 1.5
Actual width is the sum of real depth from 2 sides
Thickness of slab = d1n + d2 n
= (5 x 1.5) +(3 x 1.5)= 12 cm

Question 10.
A ray of light travelling in a transparent medium of refractive index n falls, on a surface separating the medium from air at an angle of incidents of 45°. The ray can undergo total internal reflection for the following n,
(a) n= 1.25
(b) n = 1.33
(c) n= 1.4
(d) n= 1.5
Answer:
(d) n= 1.5
Hint:
For total internal reflection
sin i > sin c            where, i – angle of incidence
But, sin c = 1/n      c – critical angle
sin i > 1/n
n > \(\frac { 1 }{ sin i }\)
n > \(\frac { 1 }{ sin 45 }\) ; n > √2 ; n > 1.414

Question 11.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
(a) red
(b) yellow
(c) green
(d) violet
Answer:
(d) violet
Hint:
Letters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.

Question 12.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1 m
(b) 5 m
(c) 3 m
(d) 6m
Answer:
(b) 5 m
Hint:
Resolution limit sin θ = \(\frac { Y }{ d }\) = \(\frac { 1.22c }{ d }\)
12th Physics Wave Optics Book Back Answers Chapter 6 Samacheer Kalvi

Question 13.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,
(a) 2D
(b) \(\frac { D }{ 2 }\)
(c) √2D
(d) \(\frac { D }{ √2 }\)
Answer:
(a) 2D
Hint:
Young’s double -slite experiment is
β = \(\frac { λD }{ d}\) ; β’ = \(\frac { λD’ }{ d’}\) ; d’ = 2d
Same fringe space, β = β’
⇒ \(\frac { λD }{ d}\) = \(\frac { λD’ }{ d’}\) ; D’ = 2D

Question 14.
Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are [IIT-JEE 1988]
(a) 5I and I
(b) 5I and 3I
(c) 9I and I
(d) 9I and 3I
Answer:
(c) 9I and I
Hint:
Samacheer Kalvi Guru 12th Physics Solutions Chapter 6 Optics

Question 15.
When light is incident on a soap film of thickness 5 x 10-5
cm, the wavelength of light reflected maximum in the visible region is 5320 A. Refractive index of the film will be,
(a) 1.22
(b) 1.33
(c) 1.51
(cl) 1.83
Answer:
(b) 1.33
Hint.
The condition for constructive interference, (for reflection)
2µ tcos r = (2n +1) \(\frac { λ }{ 2 }\) [∴ cos r = 1]
µ = \(\frac { \left( 2n+1 \right) \lambda }{ 4t } \)
For visible region, n = 2
Samacheer Kalvi Physics 12th Solutions Chapter 6 Optics

Question 16.
First diffraction minimum due to a single slit of width 1.0 x 10-5 cm is at 30°. Then wavelength of light used is,
(a) 400 Å
(b) 500 Å
(c) 600 Å
Answer:
(b) 500 Å
Hint.
For diffraction minima, d sin θ = nλ
λ = \(\frac { dsin\theta }{ n } =\frac { 1\times { 10 }^{ -5 }\times { 10 }^{ -2 }\times sin30° }{ 1 } \) = 0.5 x 10-7
λ = 500 Å

Question 17.
A ray of light strikes a glass plate at an angle 60°. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
(a) √3
(b) \(\frac { 3 }{ 2 }\)
(c) \(\sqrt { \frac { 3 }{ 2 } } \)
(d) 2
Answer:
(a) √3
Hint.
Angle of refraction r = 60° ; Angle of incident i = 30°
sin i =n x sin r
n = \(\frac {sin 30°}{ sin 60°} \) = √3

Question 18.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will,
Samacheerkalvi.Guru 12th Physics Solutions Chapter 6 Optics
(a) get shifted downwards
(b) get shifted upwards
(c) will remain the same
(d) data insufficient to conclude
Answer:
(b) get shifted upwards

Question 19.
Light transmitted by Nicol prism is,
(a) partiallypolarised
(b) unpolarised
(c) plane polarised
(d) elliptically polarised
Answer:
(c) plane polarised

Question 20.
The transverse nature of light is shown in,
(a) interference
(b) diffraction
(c) scattering
(d) polarisation
Answer:
(d) polarisation

Samacheer Kalvi 12th Physics Optics Short Answer Questions

Question 1.
State the laws of reflection.
Answer:
(a) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(b) The angle of incidence i is equal to the angle of reflection r. i = r

Question 2.
What is angle of deviation due to reflection?
Answer:
The angle between the incident and deviated light ray is called angle of deviation of the light ray. It is written as, d= 180 – (i + r).
As, i = r in reflection, we can write angle of deviation ‘ in reflection at plane surface as, d = 180 – 2i

Question 3.
Give the characteristics of image formed by a plane mirror.
Answer:

  1. The image formed by a plane mirror is virtual, erect, and laterally inverted.
  2. The size of the image is equal to the size of the object.
  3. The image distance far behind the mirror is equal to the object distance in front of it.
  4.  If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as, n = \(\left( \frac { 360 }{ \theta } -1 \right) \)

Question 4.
Derive the relation between ƒ and R for a spherical mirror.
Let C be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. The geometry of reflection of the incident ray is shown in figure. The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection. If MP is the perpendicular from M on the principal axis, then from the geometry, The angles ∠MCP = i and ∠MFP = 2i From right angle triangles ∆MCP and ∆MFP,
Samacheer Kalvi.Guru 12th Physics Solutions Chapter 6 Optics
\(\frac { 1 }{ f }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
tan i = \(\frac { PM }{ PC }\) and tan 2 i = \(\frac { PM }{ PF }\)
As the angles are small, tan i ≈ i = \(\frac { PM }{ PC }\) and tan 2 i = \(\frac { PM }{ PF }\)
Simplifying further, 2 = \(\frac { PM }{ PC }\) and tan = \(\frac { PM }{ PF }\) ;2PF = PC
PF is focal length/and PC is the radius of curvature R.
2 ƒ= R (or) ƒ = \(\frac { R }{ 2 }\)
ƒ = \(\frac { R }{ 2 }\) is the relation between ƒ and R.

Question 5.
What are the Cartesian sign conventions for a spherical mirror?
Answer:

  1. The Incident light is taken from left to right (i.e. object on the left of mirror).
  2. All the distances are measured from the pole of the mirror (pole is taken as origin).
  3. The distances measured to the right of pole along the principal axis are taken as positive.
  4. The distances measured to the left of pole along the principal axis are taken as negative.
  5. Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
  6. Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.

Question 6.
What is optical path? Obtain the equation for optical path of a medium of thickness d and refractive index n.
Answer:
Optical path of a medium is defined as the distance d’ light travels in vacuum in the same time it travels a distance d in the medium.
Let us consider a medium of refractive index n and thickness d. Light travels with a speed v through the medium in a time t. Then we can write,
Physics Samacheer Kalvi 12th Solutions Chapter 6 Optics
v = \(\frac { d }{ t }\); rewritten as, t = \(\frac { d }{ v }\)
In the same time, light can cover a greater distance d’
in vacuum as it travels with greater speed c in vacuum.
Then we have,
c = v = \(\frac { d’ }{ t }\); rewritten as, t = \(\frac { d’ }{ c }\)
As the time taken in both the cases is the same, we can equate the time t as,
\(\frac { d’ }{ c }\) = \(\frac { d }{ v }\)
rewritten for the optical path d’ as d’ = \(\frac { c }{ v }\)d
As, \(\frac { c }{ v }\) = n ; The optical path d’ is, d’ = nd
As n is always greater than 1, the optical path d’ of the medium is always greater than d.

Question 7.
State the laws of refraction.
Answer:
Law of refraction is called Snell’s law.
Snell’s law states that,
(a) The incident ray, refracted ray and normal to the refracting surface are all coplanar (i.e. lie in the same plane).
(b) The ratio of angle of incident i in the first medium to the angle of reflection r in the second medium is equal to the ratio of refractive index of the second medium n2 to that of the refractive index of the first medium n1.
\(\frac { sin i }{ sin r }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \).

Question 8.
What is angle of deviation due to refraction?
Answer:
The angle between the incident and deviated light is called angle of deviation. When light travels from rarer to denser medium it deviates towards normal. The angle of deviation in this case is, d = i – r

Question 9.
What is principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 10.
What is relative refractive index?
Answer:
Snell’s law, the term \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \) is called relative refractive index of second medium with respect to the first medium which is denoted as (n21). n21 = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)

Question 11.
Obtain the equation for apparent depth.
Answer:
Light from the object O at the bottom of the tank passes from denser medium (water) to rarer medium (air) to reach our eyes. It deviates away from the normal in the rarer medium at the point of incidence B. The refractive index of the denser medium is n1 and rarer medium is n2. Here, n1 > n2. The angle of incidence in the denser medium is i and the angle of refraction in the rarer medium is r. The lines NN’ and OD are parallel. Thus angle ∠DIB is also r. The angles i and r are very small as the diverging light from O entering the eye is very narrow. The Snell’s law in product form for this refraction is,
n1 sin i = n2 sin r
As the angles i and r are small, we can approximate, sin i ≈ tan i;
n1 tan i = n2 tan i
In triangles ∆DOB and ∆DIB,
tan(i) = \(\frac { DB }{ DO }\) and tan(r) = \(\frac { DB }{ DI }\)
n1 = \(\frac { DB }{ DO }\) n2 = \(\frac { DB }{ DI }\)
DB is cancelled on both sides, DO is the actual depth d and DI is the apparent depth d’.
Rearranging the above equation for the apparent depth d’,
d’ = \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \)d
As the rarer medium is air and its refractive index n2 can be taken as 1, (n2 = 1). And the refractive index n1 of denser medium could then be taken as n, (n1 = n).
In that case, the equation for apparent depth becomes,
d = \(\frac { d }{ n }\)

Question 12.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 13.
What is critical angle and total internal reflection?
Answer:
The angle of incidence in the denser medium for which the refracted ray graces the boundary is called critical angle ic.
The entire light is reflected back into the denser medium itself. This phenomenon is called total internal reflection.

Question 14.
Obtain the equation for critical angle.
Answer:
Snell’s law in the product form, equation for critical angle incidence becomes,
n1 sini ic = n2 sin 90°
n1 sini ic = n2 (∵ sin 90° = 1)
sini ic = \(\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } \right) \)
Here, n1 > n2
If the rarer medium is air, then its refractive index is 1 and can be taken as n itself, i.e. (n2 = 1) and (n1= n).
sini ic = \(\frac { 1 }{ n }\) (or) ic = sin-1 \(\left( \frac { 1 }{ n } \right) \)

Question 15.
Explain the reason for glittering of diamond.
Answer:
Diamond appears dazzling because the total internal reflection of light happens inside the diamond. The refractive index of only diamond is about 2.417. It is much larger than that for ordinary glass which is about only 1.5. The critical angle of diamond is about 24.4°. It is much less than that of glass. A skilled diamond cutter makes use of this larger range of angle of incidence (24.4° to 90° inside the diamond), to ensure that light entering the diamond is total internally reflected from the many cut faces before getting out. This gives a sparkling effect for diamond.

Question 16.
What are mirage and looming?
Answer:
Mirage: Mirage takes place in hot regions. The light from distant objects appears to be reflected from ground. For mirage to form refractive index goes on increasing as we go up. Looming: Looming takes place in cold regions. The light from distant objects appears to be flying. For looming to form refractive index goes on decreasing.

Question 17.
Write a short notes on the prisms making use of total internal reflection.
Answer:
Prisms can be designed to reflect light by 90° or by 180° by making use of total internal reflection. The critical angle ic for the material of the prism must be less than 45°. This is true for both crown glass and flint glass. Prisms are also used to invert images without changing their size.

Question 18.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle ic. The restricted illuminated circular area is called Snell’s window.

Question 19.
Write a note on optical fibre.
Answer:
Transmitting signals through optical fibres is possible due to the phenomenon of total internal reflection. Optical fibres consists of inner part called core and outer part called cladding (or) sleeving. The refractive index of the material of the core must be higher than that of the cladding for total internal reflection to happen. Signal in the form of light is made to incident inside the core-cladding boundary at an angle greater than the critical angle. Hence, it undergoes repeated total internal reflections along the length of the fibre without undergoing any refraction.

Question 20.
Explain the working of an endoscope.
Answer:
An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient’s body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body. Even operations could be carried out with the endoscope cable which has the necessary instruments attached at their ends.

Question 21.
What are primary focus and secondary focus of convex lens?
Answer:
Primary focus: The primary focus F1 is defined as a point where an object should be placed to give parallel emergent rays to the principal axis.
Secondary focus: The secondary focus F2 is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis.

Question 22.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).
(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 23.
Arrive at lens equation from lens maker’s formula.
Answer:
From refraction through a double convex lens, the relation between the object distance u, image distance v1 and radius of curvature R1 as
\(\frac { { \mu }_{ 2 } }{ { v }_{ 1 } } -\frac { { \mu }_{ 1 } }{ u } =\frac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ { R }_{ 1 } } \) …… (1)
The relation between the object distance image distance v1 and radius of curvature R2 can be
\(\frac { { \mu }_{ 1 } }{ { v } } -\frac { { { \mu }_{ 2 } } }{ v_{ 1 } } =\frac { { \mu }_{ 1 }-{ \mu }_{ 2 } }{ { R }_{ 2 } } \) …… (2)
Adding equation (1) and (2)
Samacheer Kalvi Guru 12 Physics Solutions Chapter 6 Optics
If the object is placed at infinity (µ = ∞), the image will be formed at the focus. i.e.v = ƒ
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-9
This is len’s maker’s formula. When the lens is placed in air µ1 = 1 and µ2 = µ.
Equation (4) becomes,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-10
From equation (3) and (4), we have \(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)
=\(\frac { 1 }{ ƒ }\)
This is the len’s equation.

Question 24.
Obtain the equation for lateral magnification for thin lens.
Answer:
Lateral magnification in terms of u and ƒ.
The thin lens formula is
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)=\(\frac { 1 }{ ƒ }\)
Multiplying on both sides by ‘u’
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-11
In terms of v and ƒ multiplying by v, we get
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-12
Hence, lateral magnification for thin lens,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-13

Question 25.
What is power of a lens?
Answer:
The power of a lens P is defined as the reciprocal of its focal length.
p = \(\frac { 1 }{ ƒ }\)
The unit of power is diopter D.

Question 26.
Derive the equation for effective focal length for lenses in contact.
Answer:
Consider a two thin lenses in contact. In the absence of second lens L2, the first lens L1 will form a real image I’. Using thin lens formula.
\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v’ }\)–\(\frac { 1 }{ u }\) ….. (1)
The image I’ acts as a virtual object (u = v’) for the second lens L2 which finally forms its real image I at distance v. Thus
\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v }\)–\(\frac { 1 }{ v’ }\) ….. (2)
Adding equation (1) and (2) we get,
\(\frac { 1 }{ { f }_{ 1 } } \)+\(\frac { 1 }{ { f }_{ 2 } } \)=\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) ….. (3)
For the combination of thin lenses in contact, if ‘f’ is the equivalent focal length, then
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\)=\(\frac { 1 }{ ƒ }\) ….. (4)
From equations (3) and (4), the effective focal length for lenses in contact.
\(\frac { 1 }{ ƒ }\)=\(\frac { 1 }{ { f }_{ 1 } } \)+\(\frac { 1 }{ { f }_{ 2 } } \)

Question 27.
What is angle of minimum deviation?
Answer:
The minimum value of the angle of deviation suffered by a ray on passing through a prism is called the angle of minimum deviation.

Question 28.
What is dispersion?
Answer:
The phenomenon of spliting of white light into its component colours on passing through a refracting medium is called dispersion.

Question 29.
How are rainbows formed?
Answer:
Rainbow is formed by dispersion of sunlight into its constituent colours by raindrops which disperse sunlight by refraction and deviate the colours by total internal reflection.

Question 30.
What is Rayleigh’s scattering?
Answer:
The scattering of light by particles in a medium without a change in wavelength is called as Rayleigh’s scattering.

Question 31.
Why does sky appear blue?
Answer:
Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, intensity of scattered light, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \) So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appears blue.

Question 32.
What is the reason for reddish appearance of sky during sunset and sunrise?
Answer:
During sunrise or sunset, the sun is near the horizon. Sunlight has to travel a greater distance. So shorter waves of blue region are scattered away by the atmosphere. Red waves of longer wavelength are least scattered and reach the observer. So the sun appears red.

Question 33.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 34.
What are the salient features of corpuscular theory of light?
Answer:

  • According this theory, light is emitted as tiny, massless (negligibly small mass) and perfectly elastic particles called corpuscles.
  • As the corpuscles are very small, the source of light does not suffer appreciable loss of mass even if it emits light for a long time.
  • On account of high speed, they are unaffected by the force of gravity and their path is a straight line in a medium of uniform refractive index.
  • The energy of light is the kinetic energy of these corpuscles. When these corpuscles impinge on the retina of the eye, the vision is produced.
  • The different size of the corpuscles is the reason for different colours of light.
  • When the corpuscles approach a surface between two media, they are either attracted or repelled.
  • The reflection of light is due to the repulsion of the corpuscles by the medium and refraction of light is due to the attraction of the corpuscles by the medium.

Question 35.
What is wave theory of light?
Answer:
Light is a disturbance from a source that travels as longitudinal mechanical waves through the either medium that was presumed to pervade all space as mechanical wave requires medium for its propagation. The wave theory could successfully explain phenomena of reflection, refraction, interference and diffraction of light.

Question 36.
What is electromagnetic wave theory of light?
Answer:
Electromagnetic wave theory:
Maxwell (1864) proved that light is an electromagnetic wave which is transverse in nature carrying electromagnetic energy. He could also show that no medium is necessary for the propagation of electromagnetic waves. All the phenomenon of light could be successfully explained by this theory.

Question 37.
Write a short note on quantum theory of light.
Answer:
Albert Einstein (1905), endorsing the views of Max Plank (1900), was able to explain photoelectric effect in which light interacts with matter as photons to eject the electrons. A photon is a discrete packet of energy. Each photon has energy E of,
E = hv
Where, h is Plank’s constant (h = 6.625 x 10-34J s) and v is frequency of electromagnetic wave. As light has both wave as well as particle nature it is said to have dual nature. Thus, it is concluded that light propagates as a wave and interacts with matter as a particle.

Question 38.
What is a wave front?
Answer:
A wavefront is the locus of points which are in the same state or phase of vibration.

Question 39.
What is Huygens’ principle?
Answer:
According to Huygens principle, each point of the wavefront is the source of secondary wavelets emanating from these points spreading out in all directions with the speed of the wave. These are called as secondary wavelets.

Question 40.
What is interference of light?
Answer:
The phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and decrease in intensity at some other points is called interference of light.

Question 41.
What is phase of a wave?
Answer:
Phase is a particular point in time on the cycle of a waveform, measured as an angle in degrees.

Question 42.
Obtain the relation between phase difference and path difference.
Answer:
Phase difference (Φ):
It is the difference expressed in degrees or radians between two waves having same frequency and referenced to same point in time.
Path difference (δ):
It is the difference between the lengths of two paths of the two different having same frequency and travelling at same velocity. δ =\(\frac { \lambda }{ 2\pi } \) Φ

Question 43.
What are coherent sources?
Answer:
Two light sources are said to be coherent if they produce waves which have same phase or constant phase difference, same frequency or wavelength (monochromatic), same waveform and preferably same amplitude.

Question 44.
What is intensity division?
Answer:
Intensity’ or amplitude division: If we allow light to pass through a partially silvered mirror (beam splitter), both reflection and refraction take place simultaneously. As the two light beams are obtained from the same light source, the two divided light beams will be coherent beams. They will be either in-phase or at constant phase difference.

Question 45.
How does wavefront division provide coherent sources?
Answer:
Wavefront division is the most commonly used method for producing two coherent sources. A point source produces spherical wavefronts. All the points on the wavefront are at the same phase. If two points are chosen on the wavefront by using a double slit, the two points will act as coherent sources.

Question 46.
How do source and images behave as coherent sources?
Answer:
Source and images: In this method a source and its image will act as a set of coherent source, because the source and its image will have waves in-phase or constant phase difference. The Instrument, Fresnel’s biprism uses two virtual sources as two coherent sources and the instrument, Lloyd’s mirror uses a source and its virtual image as two coherent sources.

Question 47.
What is bandwidth of interference pattern?
Answer:
The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.

Question 48.
What is diffraction?
Answer:
Diffraction is bending of waves around sharp edges into the geometrically shadowed region.

Question 49.
Differentiate between Fresnel and Fraunhofer diffraction.
Answer:

S.No. Fresnel diffraction Fraunhofer diffraction
1. Spherical or cylindrical wavefront undergoes diffraction Plane wavefront undergoes diffraction
2. Light wave is from a source at finite distance Light wave is from a source at infinity
3. For laboratory conditions, convex lenses need not be used In laboratory conditions, convex lenses are to be used
4. difficult to observe and analyse Easy to observe and analyse

Question 50.
Discuss the special cases on first minimum in Fraunhofer diffraction.
Let us consider the condition for first minimum with (n = 1). a sin θ = λ
The first minimum has an angular spread of, sin θ = \(\frac { \lambda }{ a } \). Special cases to discuss on the condition.
1. When a < λ, the diffraction is not possible, because sin 0 can never be greater than 1.
2. When a ≥ λ, the diffraction is possible.

  • For a = λ, sin θ = 1 i.e, θ = 90°. That means the first minimum is at 90°. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90°.
  • For a >> λ, sin θ << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all.

3. When a > λ and also comparable, say a = 2λ, sin θ = \(\frac { \lambda }{ a } \) = \(\frac { \lambda }{ { 2\lambda } } \) =\(\frac { 1 }{ 2 }\); then θ = 30°. These are practical cases where diffraction could be observed effectively.

Question 51.
What is Fresnel’s distance? Obtain the equation for Fresnel’s distance.
Answer:
Fresnel’s distance is the distance up to which the ray optics is valid in terms of rectilinear propagation of light.
Fresnel’s distance z as, z = \(\frac { { a }^{ 2 } }{ 2\lambda } \).

Question 52.
Mention the differences between interference and diffraction.
Answer:

S.No. Interference Diffraction
1. Superposition of two waves Bending of waves around edges
2. Superposition of waves from two coherent sources. Superposition wavefronts emitted from various points of the same wavefront.
3. Equally spaced fringes. Unequally spaced fringes
4. Intensity of all the bright fringes is almost same Intensity falls rapidly for higher orders
5. Large number of fringes are obtained Less number of fringes are obtained

Question 53.
What is a diffraction grating?
Answer:
A diffraction grating is an optical component with a periodic structure that splits and diffracts light into several beams travelling in different directions.

Question 54.
What are resolution and resolving power?
Answer:
Optical resolution describes the ability of an imaging system to resolve detail in the object that is being imaged. Resolving power is the ability of an optical instrument to resolve or separate the image of two nearby point objects so that they can be distinctly seen. It is equal to the reciprocal of the limit of resolution of the optical instrument.

Question 55.
What is Rayleigh’s criterion?
Answer:
The images of two point objects are just resolved when the central maximum of the diffraction pattern of one falls over the first minimum of the diffraction pattern of the other.

Question 56.
What is polarisation?
Answer:
The phenomenon of restricting the vibrations of light (electric or magnetic field vector) to a particular direction perpendicular to the direction of wave propagation motion is called polarization of light.

Question 57.
Differentiate between polarised and unpolarised light.
Answer:

S.No. Polarised light Unpolarised light
1. Consists of waves having their electric field vibrations in a single plane normal to the direction of ray. Consists of waves having their electric field vibrations equally distributed in all directions normal to the direction of ray.
2. Asymmetrical about the ray direction Symmetrical about the ray direction
3. It is obtained from unpolarised light with the help of polarisers Produced by conventional light sources.

Question 58.
Discuss polarisation by selective absorption.
Answer:
Selective absorption is the property of a material which transmits w’aves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves. The polaroids or polarisers are thin commercial sheets which make use of the property of selective absorption to produce an intense beam of plane polarised light. Selective absorption is also called as dichroism.

Question 59.
What are polariser and analyser?
Answer:
Polariser:
The Polaroid which plane polarises the unpolarised light passing through it is called a polariser.
Analyser:
The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.

Question 60.
What are plane polarised, unpolarised and partially polarised light?
Answer:
Plane polarised:
If the vibrations of a wave are present in only one direction in a plane perpendicular to the direction of propagation of the wave is said to be polarised or plane polarised light.

Unpolarised:
A transverse wave which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light.

Partially polarised light:
If the intensity of light varies between maximum and minimum for every’ rotation of 90° of the analyser, the light is said to be partially polarised light.

Question 61.
State and obtain Malus’ law.
Answer:
When a beam of plane polarised light of intensity I0 is incident on an analyser, the light transmitted of intensity I from the analyser varies directly as the square of the cosine of the angle 0 between the transmission axis of polariser and analyser. This is known as Malus’ law. I = I0 cos2θ

Question 62.
List the uses of polaroids.
Answer:
Uses of polaroids:

  1. Polaroids are used in goggles and cameras to avoid glare of light.
  2. Polaroids are useful in three dimensional motion pictures i.e., in holography.
  3. Polaroids are used to improve contrast in old oil paintings.
  4. Polaroids are used in optical stress analysis.
  5. Polaroids are used as window glasses to control the intensity of incoming light.

Question 63.
State Brewster’s law.
Answer:
The law states that the tangent of the polarising angle for a transparent medium is equal to its refractive index, tan i = n. This relation is known as Brewster’s law.

Question 64.
What is angle of polarisation and obtain the equation for angle of polarisation.
Answer:
The angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of plane polarised light is called polarising angle or Brewster’s angle. It is denoted by ip
ip = 90° – Rp

Question 65.
Discuss about pile of plates.
Answer:
The phenomenon of polarisation by reflection is used in the construction of pile of plates. It consists of a number of glass plates placed one over the other. The plates are inclined at an angle of 33.7° (90° – 56.3°) to the axis of the tube. A beam of unpolarised light is allowed to fall on the pile of plates along the axis of the tube. So, the angle of incidence of light will be at 56.3° which is the polarising angle for glass.

The vibrations perpendicular to the plane of incidence are reflected at each surface and those parallel to it are transmitted. The larger the number of surfaces, the greater is the intensity of the reflected plane polarised light. The pile of plates is used as a polarizer and also as an analyser.

Question 66.
What is double refraction?
Answer:
When a ray of unpolarised light is incident on a calcite crystal, two refracted rays are produced. Hence, two images of a single object are formed. This phenomenon is called double refraction.

Question 67.
Mention the types of optically active crystals with example.
Answer:
Types of optically active crystals:
Uniaxial crystals:
Crystals like calcite, quartz, tourmaline and ice having only one optic axis are called uniaxial crystals.

Biaxial crystals:
Crystals like mica, topaz, selenite and aragonite having two optic axes are called biaxial crystals.

Question 68.
Discuss about Nicol prism.
Answer:
Nicol prism is an optical device incorporated in optical instruments both for producing and analysing plane polarised light. The construction of a Nicol prism is based on the phenomenon of Double refraction. One of the most common forms of the Nicol prism is made by taking a calcite crystal which is a double refracting crystal with its length three times its breadth.

It is cut into two halves along the diagonal so that their face angles are 72° and 108°. The two halves are joined i together by a layer of Canada balsam, a transparent cement.

Question 69.
How is polarisation of light obtained by scattering of light?
Answer:
The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is because of sunlight, which has changed its I direction (having been scattered) on encountering the molecules of the earth’s atmosphere. The electric field of light interact with the electrons present in the air molecules.

Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have an observer looking at 90° to the direction of the sun. Clearly, charges accelerating parallel do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore polarized perpendicular to the plane.

Question 70.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.

  1. Magnification in near point focusing m = 1 + \(\frac { D }{ f }\)
  2. Magnification in normal focusing (angular magnification), m = \(\frac { D }{ f }\)

Question 71.
What are near point and normal focusing?
Answer:

  • Near point focusing:
    The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
  • Normal focusing:
    The image is formed at infinity. In this position the eye is most relaxed to view the image.

Question 72.
Why is oil immersed objective preferred in a microscope?
Answer:
It is best to use an oil-immersed objective at high magnification as the oil compensates for short focal lengths associated with larger magnifications.

Question 73.
What are the advantages and disadvantages of using a reflecting telescope?
Answer:
Advantages:

  • The main advantage is reflector telescope can escape from chromatic aberration because wavelength does not effect reflection.
  • The primary mirror is very stable because it is located at the back of the telescope and can be support in the back.
  • More cost effective than refractor of similar size.
  • Easier to make a high quality mirror than lens because mirror need to only concern with one side of the curvature.

Disadvantages:

  • Optical misalignment can occur quite easily.
  • Require frequent cleaning because the inside is expose to the atmosphere.
  • Secondary mirror can cause diffraction of original incoming light rays causing the “Christmas star effect” where a bright object have spikes.

Question 74.
What is the use of an erecting lens in a terrestrial telescope?
Answer:
A terrestrial telescope has an additional erecting lens to make the final image erect.

Question 75.
What is the use of collimator?
Answer:
The collimator is an arrangement to produce a parallel beam of light.

Question 76.
What are the uses of spectrometer?
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials.

Question 77.
What is myopia? What is its remedy?
Answer:
Myopia (or) short sightedness:
It is a vision defect in which a person can see nearby objects clearly but cannot see the distant objects clearly beyond a certain point.

Remedy (correction):
A myopia eye is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.

Question 78.
What is hypermetropia? What is its remedy?
Answer:
Hypermetropia (or) Long sightedness: It is a vision defect in which a person can see the distant objects clearly but cannot see the nearby objects clearly.
Remedy (correction): A hypermetropic eye is corrected by using a convex lens of suitable focal length.

Question 79.
What is presbyopia?
Answer:
This defect is similar to hypermetropia i.e., a person having this defect cannot see nearby objects distinctly, but can see distant objects without any difficulty. This defect occurs in elderly persons (aged persons).

Question 80.
What is astigmatism?
Answer:
Astigmatism is the defect arising due to different curvatures along different planes in the eye lens. Astigmatic person cannot see all the directions equally well. The defect due to astigmatism is more serious than myopia and hyperopia.

Samacheer Kalvi 12th Physics Optics Long Answer Questions

Question 1.
Derive the mirror equation and the equation for lateral magnification.
Answer:
The mirror equation:
1. The mirror equation establishes a relation among object distance u, image distance v and focal length/for a spherical mirror. An object AB is considered on the principal axis of a concave mirror beyond the center of curvature C’.
2. Let us consider three paraxial rays from point B on the object.
3. The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at D, close to the pole P. After reflection the ray passes through the focus F. The second paraxial ray BP incident at the pole P is reflected along PBThe third paraxial ray BC passing through centre of curvature C, falls normally on the mirror at E is reflected back along the same path.
4. The three reflected rays intersect at the point B’. A perpendicular drawn as A’ B’ to the principal axis is the real, inverted image of the object AB.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-14
As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B’PA’. The triangles ∆BPA and ∆B’PA’ are similar. Thus, from the rule of similar triangles,
\(\frac { A’B’ }{ AB }\) = \(\frac { PA’ }{ PA }\) …….. (1)
The other set of similar triangles are, ADPF and A BA.’ F. (PD is almost a straight vertical line)
\(\frac { A’B’ }{ PD }\) = \(\frac { AF’ }{ PF }\)
As, the distances PD = AB the above equation becomes,
\(\frac { A’B’ }{ AB }\) = \(\frac { AF’ }{ PF }\) ……… (2)
From equations (1) and (2) we can write,
\(\frac { PA’ }{ PA }\) = \(\frac { AF’ }{ PF }\)
As, A’ F = PA’ – PF, the above equation becomes,
\(\frac {PA’ }{ PA }\) = \(\frac { PA’-PF’P }{ PF }\) ……….. (3)
We can apply the sign conventions for the various distances in the above equation.
PA = – u, PA’ = -v, PF = -f
All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
\(\frac { -v }{ -u }\) = \(\frac { -v\left( -f \right) }{ -f } \)
On further simplification,
\(\frac { -v }{ -u }\) = \(\frac { v-f }{ f }\); \(\frac { v }{ u }\)=\(\frac { v }{ f }\)=1
Dividing either side with v,
\(\frac { 1 }{ u }\)=\(\frac { 1 }{ f }\)=\(\frac { 1 }{ v }\)
After rearranging,
\(\frac { 1 }{ v }\)+\(\frac { 1 }{ u }\)=\(\frac { 1 }{ f }\) ……… (4)
The above equation (4) is called mirror equation.

Lateral magnification in spherical mirrors:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics q1
m=\(\frac { h’ }{ h }\) ……..(5)
Applying proper sign conventions for equation (1),
\(\frac { A’B’ }{ AB }\) = \(\frac { PA’ }{ PA }\)
A’B’ = -h, AB = h, PA’ = -v, PA = -u
\(\frac { -h’ }{ h }\)=\(\frac { -v }{ -u }\)
On simplifying we get,
m=\(\frac { h’ }{ h }\)=-\(\frac { v }{ u }\) ………. (6)
Using mirror equation, we can further write the magnification as,
m=\(\frac { h’ }{ h }\)–\(\frac { f-v }{ f }\)=\(\frac { f }{ f-u }\) ………… (7)

Question 2.
Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus:
The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism. The light passing through one cut in the wheel will get reflected by a mirror M kept at a long distance d, about 8 km from the toothed w’heel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working:
The angular speed of rotation of the toothed wheel was increased from zero to a value to until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light:
The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
v = \(\frac { 2d }{ t }\) c ….. (1)
The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed co of the toothed wheel.
The angular speed ω of the toothed wheel when the light disappeared for the first time is,
ω = \(\frac { θ }{ t }\) ……. (2)
Here, 0 is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-15
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics q2
θ = \(\frac { 2π }{ 2N }\) = \(\frac { π }{ N }\)
Substituting for 0 in the equation (2) for ,
ω = \(\frac { \pi /n }{ 2 } \) = \(\frac { π }{ Nt }\)
Rewriting the above equation for t,
t = \(\frac { π }{ Nω }\) ……. (3)
Substituting t from equation (3) in equation (1), v = \(\frac { 2d }{ \pi /N\omega } \)
After rearranging,
v = \(\frac { 2dNω }{ π }\) …….. (4)
Fizeau had some difficulty to visually estimate the minimum intensity of the light when blocked by the adjacent tooth, and his value for speed of light was very close to the actual value.

Question 3.
Obtain the equation for radius of illumination (or) Snell’s window.
Answer:
The radius of Snell’s window can be deduced with the illustration as shown in figure. Light is seen from a point A at a depth d. The Snell’s law in product form, equation n2 sin i = n2 sin r for the refraction happening at the point B on the boundary between the two media is,
n1 sin ic = n2 sin 90° ……. (1)
n1 sin ic= n2 ∵ sin 90° = 1
sin ic = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …….. (2)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-16
From the right angle triangle ∆ABC,
sin ic = \(\frac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \) ……. (3)
Equating the above two equation (3) and equation (2).
\(\frac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-17
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-18
If the rarer medium outside is air, then, n2 = 1, and we can take n1 = n
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-19

Question 4.
Derive the equation for acceptance angle and numerical aperture, of optical fiber. Acceptance angle in optical fibre:
Answer:
To ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core n1, cladding n2 and the outer medium n3. Assume the light is incident at an angle called acceptance angle i at the outer medium and core boundary at A.
The Snell’s law in the product form, equation for this refraction at the point A.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-20
n3 sin ia = n1 sin ra …(1)
To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be atleast critical angle ic. Snell’s law in the product form, equation for the refraction at point B is,
n1 sin ic = n2 sin 90° …(2)
n1 sin ic= n2 ∵ sin 90° =1
∴sin ic= \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …(3)
From the right angle triangle ∆ABC,
ic = 90°-ra
Now, equation (3) becomes, sin (90° – ra) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Using trigonometry’, cos ra = ra = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \) …….. (4)
sin ra = \(\sqrt { 1-{ cos }^{ 2 }{ r }_{ a } } \)
Substituting for cos ra
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-21
If outer medium is air, then n3 = 1. The acceptance angle ia becomes,
ra=sin-1\(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (9)
Light can have any angle of incidence from 0 to ia with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n3 sin ia) is called numerical aperture NA of the optical fibre.
NA =n3 sin ia \(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (10)
If outer medium is air, then n3 = 1. The numerical aperture NA becomes,
NA = sin ia = \(\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) …….. (11)

Question 5.
Obtain the equation for lateral displacement of light passing through a glass slab.
Answer:
Consider a glass slab of thickness t and refractive index n is kept in air medium. The path of the light is ABCD and the refractions occur at two points B and C in the glass slab. The angles of incidence i and refraction r are measured with respect to the normal N1and N2 at the two points B and C respectively. The lateral displacement L is the perpendicular distance CE drawn between the path of light and the undeviated path of light at point C. In the right angle triangle ∆BCE,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-22
sin(i – r) = \(\frac { L }{ BC }\); BC= \(\frac { L }{ sin(i-r) } \) …… (1)
In the right angle triangle ∆BCF,
cos(r) = \(\frac { t }{ BC }\) ; BC= \(\frac { t }{ cos(r) } \) …… (2)
Equating equations (1) and (2)
\(\frac { L }{ sin(i-r) } \) = \(\frac { t }{ cos(r) } \)
After rearranging,
L = t\(\left( \frac { sin(i-r) }{ cos(r) } \right) \)
Lateral displacement depends upon the thickness of the slab. Thicker the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement.

Question 6.
Derive the equation for refraction at single spherical surface.
Answer:
Equation for refraction at single spherical surface:
Let us consider two transparent media having refractive indices n1and n2 are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n1. The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-23
Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C. As n2 > n1, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed.
Snell’s law in product form for the refraction at the point N could be written as,
n1 sin i = n2 sin r …(1)
As the angles are small, sin of the angle could be approximated to the angle itself.
n1 i = n2 r …(2)
Let the angles,
∠NOP = α, ∠NCP = β, ∠NIP = γ
tan α = \(\frac { PN }{ PO }\);tan β = \(\frac { PN }{ PC }\);tan γ = \(\frac { PN }{ PI }\)
As these angles are small, tan of the angle could be approximated to the angle itself.
α = \(\frac { PN }{ PO }\); β = \(\frac { PN }{ PC }\); γ = \(\frac { PN }{ PI }\) …….. (3)
For the triangle, ∆ONC,
I = α + β …….. (4)
For the triangle, ∆INC,
β = r + γ (or) r = β – γ ……… (5)
Substituting for I and r from equations (4) and (5) in the equation (2).
n1 (α + β) = n2 (β – γ)
Rearranging,
n1 α + n2γ = (n2 – n1
Substituting for α, β and γ from equation (3)
n1 = (\(\frac { PN }{ PO }\)) + n2 = (\(\frac { PN }{ PI }\)) = (n2-n1) (\(\frac { PN }{ PC }\))
Further simplifying by cancelling PN,
\(\frac { { n }_{ 1 } }{ PO } \) + \(\frac { { n }_{ 2 } }{ PI } \) = \(\frac { { n }_{ 2 }-{ n }_{ 1 } }{ PC } \)
Following sign conventions, PO = -u, PI = +v and PC = +R in equation (6),
\(\frac { { n }_{ 2 } }{ -v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
After rearranging, finally we get,
\(\frac { { n }_{ 2 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \) ……… (7)
Equation (7) gives the relation among the object distance n, image distance v, refractive indices of the two media (n1 and n2) and the radius of curvature R of the spherical surface. It holds for any spherical surface.
If the first medium is air then, n1 = 1 and the second medium is taken just as n2 = n, then the equation is reduced to,
\(\frac { n }{ v}\)–\(\frac { 1 }{ u}\) = \(\frac { \left( { n }-1 \right) }{ R } \) ………(8)

Question 7.
Obtain lens maker’s formula and mention its significance.
Answer:
Lens maker’s formula and lens equation:
Let us consider a thin lens made up of a medium of refractive index n2, is placed in a medium of refractive index n1. Let R1 and R2 be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole. Consider a point object O on the principal axis. The ray which falls very close to P, after refraction at the surface (1) forms image at I’. Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given by
\(\frac { { n }_{ 2 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
For the refracting surface (1), the light goes from n1ton2.
\(\frac { { n }_{ 2 } }{ v’ } \) – \(\frac { { n }_{ 1 } }{ u } \) = \(\frac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R }_{ 1 } \) …….. (1)
For the refracting surface (2), the light goes from medium n2ton1.
\(\frac { { n }_{ 1 } }{ v } \) – \(\frac { { n }_{ 2 } }{ v’ } \) = \(\frac { \left( { n }_{ 1 }-{ n }_{ 2 } \right) }{ R }_{ 2 } \) …….. (2)
Adding the above two equation (1) and (2)
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\(\frac { { n }_{ 1 } }{ v } \) – \(\frac { { n }_{ 1 } }{ u } \) = (n2 – n1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
Furhter simplifying and rearranging,
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) = \(\left(\frac{n_{2}-n_{1}}{n_{1}}\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ……….. (3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = ∞, v= f. Then the equation becomes.
\(\frac { 1 }{ f }\)–\(\frac { 1 }{ ∞ }\) = \(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
\(\frac { 1 }{ f }\)=\(\left(\frac{n_{2}}{n_{1}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ………… (4)
If the refractive index of the lens is and it is placed in air, then n2 = n and n1 = 1 equation (4) becomes,
\(\frac { 1 }{ f }\) = (n-1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) ………. (5)
The above equation is called the lens maker’s formula, because it tells the lens manufactures what curvature is needed to make a lens of desired focal length with a material of particular refractive index. This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write,
\(\frac { 1 }{ v }\)–\(\frac { 1 }{ u }\) =\(\frac { 1 }{ f }\) …… (6)
This equation is known as lens equation which relates the object distance u and image distance v with the focal length f of the lens. This formula holds good for a any type of lens.

Question 8.
Derive the equation for thin lens and obtain its magnification.
Answer:
Lateral magnification in thin lens:
Let us consider an object 00′ of height h1 placed on the principal axis with its height perpendicular to the principal axis. The ray Op passing through the pole of the lens goes undeviated. The inverted real image II’ formed has a height h2.
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The lateral or transverse magnification m is defined as the ratio of the height of the image to that of the object.
m = \(\frac { II’ }{ OO’ }\) …….. (1)
From the two similar triangles ∆ POO’ and ∆ PII’, we can write,
\(\frac { H’ }{ OO’ }\) = \(\frac { PI }{ PO }\) ……..(2)
Applying sign convention,
\(\frac{-h_{2}}{h_{1}}\) = \(\frac { v }{ -u}\)
Substituting this in the equation (1) for magnification,
m = \(\frac{-h_{2}}{h_{1}}\) = \(\frac { v }{ -u}\)
After rearranging,
m = \(\frac{h_{2}}{h_{1}}\) = \(\frac { v }{ u}\) …….. (3)
The magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and less than one. We can also have the equations for magnification by combining the lens equation with the formula for magnification
as,
m = \(\frac{h_{2}}{h_{1}}=\frac{f}{f+u}\) (or) m = \(\frac{h_{2}}{h_{1}}=\frac{f-v}{f}\) ……… (4)

Question 9.
Derive the equation for effective focal length for lenses in out of contact.
Answer:
Consider a two lenses of focal length f1 and f2 arranged coaxially but separated by a distance d can be considered. For a parallel ray that falls on the arrangement, the two lenses produce deviations δ1 and δ2 respectively and The net deviation δ is,
δ = δ1 + δ2 ……. (1)
From Angle of deviation in lens equation, δ = \(\frac { h }{ f }\)
δ1 = \(\frac{h_{1}}{f_{1}}\); δ2 = \(\frac{h_{2}}{f_{2}}\) δ = \(\frac{h_{1}}{f}\) ….. (2)
The equation (1) becomes,
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\(\frac{h_{1}}{f}\) = \(\frac{h_{1}}{f_{1}}\) + \(\frac{h_{2}}{f_{2}}\) ……. (3)
From the geometry,
h2 – h1 = P2C = CG
h2 – h1 = BG tan δ1 ≈ BG δ1
h2 – h1 = d\(\frac{h_{1}}{f_{1}}\)
h2 = h1 + d\(\frac{h_{1}}{f_{1}}\) …….. (4)
Substituting the above equation in Equation (3)
\(\frac{h_{1}}{f}=\frac{h_{1}}{f_{1}}+\frac{h_{1}}{f_{2}}+\frac{h_{1} d}{f_{1} f_{2}}\)
On further simplification,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{d}{f_{1} f_{2}}\) …….. (5)
The above equation could be used to find the equivalent focal length.

Question 10.
Derive the equation for angle of deviation produced by a prism and thus obtain the equation for refractive index of material of the prism.
Answer:
Angie of deviation produced by prism:
Let light ray PQ is incident on one of the refracting faces of the prism. The angles of incidence and refraction at the first face AB are i1 and r1. The path of the light inside the prism is QR. The angle of incidence and refraction at the second face AC is r2 and i2 respectively. RS is the ray emerging from p the second face. Angle i2 is also called angle of emergence. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
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The two normals drawn at the point of incidence Q and emergence R are QN and RN. They meet at point N. The incident ray and the emergent ray meet at a point M.
The deviation d1 at the surface AB is,
angle ∠RQM = d1 = i1 – r1 …(1)
The deviation d2 at the surface AC is,
angle ∠QRM = d2 = i2 – r2 …(2)
Total angle of deviation d produced is,
d = d1 + d2 …(3)
Substituting for d1 and d2,
d=(i1 – r1) + (i2 – r2)
After rearranging,
d = (i1 – r1) + (i2 – r2) …(4)
In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
∠A + ∠QNR = 180° …(5)
From the triangle ∆QNR,
r1+ r2 ∠QNR = 180° …(6)
Comparing these two equations (5) and (6) we get,
r1+ r2 = A …(7)
Substituting this in equation (4) for angle of deviation,
d= i1+ i2 -A …(8)
Thus, the angle of deviation depends on the angle of incidence angle of emergence and the angle for the prism.
Refractive index of the material of the prism:
At minimum deviation,
i1 = i2 = i and r1 = r2 = r
Now, the equation (8) becomes,
D = i1 + i2-A (or) i = \(\frac { \left( A+D \right) }{ 2 } \)
The equation (7) becomes,
r1 + r2 = A ⇒ 2r = A (or) r = \(\frac { A }{ 2 }\)
Substituting i and r in Snell’s law,
n = \(\frac { sin i }{ sin r }\)
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
The above equation is used to find the refractive index of the material of the prism.

Question 11.
What is dispersion? Obtain the equation for dispersive power of a medium.
Answer:
Dispersion. Dispersion is splitting of white light into its constituent colours.
Dispersive Power:
Consider a beam of white light passes through a prism; It gets dispersed into its constituent colours. Let δV, δR are the angles of deviation for violet and red light. Let nV and nR are the refractive indices for the violet and red light respectively.
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The refractive index of the material of a prism is given by the equation
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
Here A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10°, the prism is said to be a small angle prism. When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a small angle prism and 5 the angle of deviation then the prism formula becomes.
n = \(\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}\) ……… (1)
For small angles of A and δm,
\(\sin \left(\frac{\mathrm{A}+\delta}{2}\right) \approx\left(\frac{\mathrm{A}+\delta}{2}\right)\) ……. (2)
\(\sin \left(\frac{\mathrm{A}}{2}\right) \approx\left(\frac{\mathrm{A}}{2}\right)\) …… (3)
∴ n = \(\frac{\left(\frac{A+\delta}{2}\right)}{\left(\frac{A}{2}\right)}=\frac{A+\delta}{A}=1+\frac{\delta}{A}\)
Further simplifying, \(\frac { δ }{ A }\) = n – 1
δ = (n – 1) A ……. (4)
When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours.
For Violet light, δV = (nV – 1)A …(5)
For Red light, δR = (nR – 1) …(6)
As, angle of deviation for violet colour δV is greater than angle of deviation for red colour δR, the refractive index for violet colour nV is greater than the refractive index for red colour nR. Subtracting δV from δR we get,
δV – δR = (nV – nR)A ….. (7)
The term (δV – δR) is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. Clearly, the angular dispersion produced by a prism depends upon.

  1. Angle of the prism
  2. Nature of the material of the prism.

If we take 8 is the angle of deviation for any middly ray (green or yellow) and n the corresponding refractive index. Then,
8 = (n – 1) A … (8)
Dispersive power (ω) is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour. Dispersive power (ω),
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Substituting for (δVR)and (δ),
ω = \(\frac{\left(n_{\mathrm{V}}-n_{\mathrm{R}}\right)}{(n-1)}\) ……. (10)
Dispersive power is a dimensionless quality. It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.

Question 12.
Prove laws of reflection using Huygens’ principle.
Answer:
Proof for laws of reflection using Huygens’ Principle:
Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY. The incident wavefront is AB and the reflected wavefront is A’B’ in the same medium. These wavefronts are perpendicular to the incident rays L, M and reflected rays L’, M’ respectively.

By the time point A of the incident wavefront touches the reflecting surface, the point B is yet to travel a distance BB’ to touch the reflecting surface a B’. When the point B falls on the reflecting surface at B’ , the point A would have reached A’. This is applicable to all the points on the wavefront.

Thus, the reflected wavefront A’B’ emanates as a plane wavefront. The two normals N and N’ are considered at the points where the rays L and M fall on the reflecting surface. As reflection happens in the same medium, the speed of light is same before and after the reflection. Hence, the time taken for the ray to travel from B to B’ is the same as the time taken for the ray to travel from A to A’. Thus, the distance BB’ is equal to the distance AA’; (AA’= BB’).
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(i) The incident rays, the reflected rays and the normal are in the same plane.
(ii) Angle of incidence,
∠i = ∠NAL = 90° – ∠NAB = ∠BAB’
Angle of reflection,
∠r = ∠N’B’M’ = 90° – ∠N’B’A’ = ∠A’B’A
For the two right angle triangles, A ABB’ and A B’A’A, the right angles, ∠B and ∠A’ are equal, (∠B and ∠A’ = 90°); the two sides, AA’ and BB’ are equal, (AA’ = BB’); the side AB’ is the common. Thus, the two triangles are congruent. As per the property of congruency, the two angles, ∠BAB’ and ∠A’B’A must also be equal. i = r
Hence, the laws of reflection are proved.

Question 13.
Prove laws of refraction using Huygens’ principle.
Answer:
Let us consider a parallel beam of light is incident on a refracting plane surface XY such as a glass surface. The incident wavefront AB is in rarer medium (1) and the refracted wavefront A’B’ is in denser medium (2). These wavefronts are perpendicular to the incident rays L, M and refracted rays L’, M’ respectively. By the time the point A of the incident wavefront touches the refracting surface, the point B is yet to travel a distance BB’ to touch the refracting surface at B’
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When the point B falls on the refracting surface at B’, the point A’ would have reached A in the other medium. This is applicable to all the points on the wavefront. Thus, the refracted wavefront A’B’ emanates as a plane wavefront.
The two normals N and N’ are considered at the points where the rays L and M fall on the refracting surface. As refraction happens from rarer medium (1) to denser medium (2), the speed of light is v1 and v2 before and after refraction and v1 is greater than v2 (v1 > v2). But, the time taken t for the ray to travel from B to B’ is the same as the time taken for the ray to
travel from A to A’.
t = \(\frac { BB’ }{ { v }_{ 1 } } \) = \(\frac { AA’ }{ { v }_{ 2 } } \) (or) \(\frac { BB’ }{ AA’ }\) = \(\frac { { v }_{ 1 } }{ { v }_{ 2 } } \)
(i) The incident rays, the refracted rays and the normal are in the same plane.
(ii) Angle of incidence,
i = ∠NAL = 90° – ∠NAB = ∠BAB’
Angle of refraction,
r =∠N’B’M’ = 90° – ∠N’B’A’ = ∠A’B’A’
For the two right angle triangles ∆ ABB’ and ∆ B’A’A,
\(\frac { sin i }{ sin r }\) = \(\frac{B B^{\prime} / A B^{\prime}}{A A^{\prime} / A B^{\prime}}\) =\(\frac { BB’ }{ AA’ }\) =\(\frac{v_{1}}{v_{2}}=\frac{c / v_{2}}{c / v_{1}}\)
Here, c is speed of light in vacuum. The ratio c/v is the constant, called refractive index of the medium. The refractive index of medium (1) is, c/v1 = n1 and that of medium (2) is, c/v2 = n2.
\(\frac { sin i }{ sin r }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \)
In product form,
n1 sin i = n2 sin r
Hence, the laws of refraction are proved.

Question 14.
Obtain the equation for resultant intensify’ due to interference of light.
Answer:
Let us consider two light waves from the two sources S1 and S2 meeting at a point P. The wave from S1 at an instant t at P is,
y1 = a1 sin ωt … (1)
The wave form S2 at an instant t at P is,
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y2 = a2 sin (ωt + Φ) … (2)
The two waves have different amplitudes a1 and a2, same angular frequency ω, and a phase difference of Φ between them. The resultant displacement will be given by,
y = y1 + y2 = a1 sin ωt + a1 sin2 (ωt + Φ) … (3)
The simplification of the above equation by using trigonometric identities gives the equation,
y = A sin (ωt + θ) … (4)
Where, A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}\) …. (5)
θ = tan-1 \(\frac{a_{2} \sin \phi}{a_{1}+a_{2} \cos \phi}\) …. (6)
The resultant amplitude is maximum,
Amax = \(\sqrt{\left(a_{1}+a_{2}\right)^{2}}\) ; when Φ = 0, ±2π, ±4π …., …… (7)
The resultant amplitude is minimum,
Amax = \(\sqrt{\left(a_{1}-a_{2}\right)^{2}}\) ; when Φ = ±π, ±3,π ±5π …., …… (8)
The intensity of light is proportional to square of amplitude,
I ∝ A2 ….. (9)
Now, equation (5) becomes,
I ∝ I1+I 2+2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) cos Φ …… (10)
In equation (10) if the phase difference, Φ = 0, ±2π, ±4π …., it corresponds to the condition for maximum intensity of light called as constructive interference. The resultant maximum intensity is,
Imax ∝ ( a1 + a2)2 ∝ I1 + I2 +2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) …… (11)
In equation (10) if the phase difference, Φ = ±π, ±3,π ±5π …., it corresponds to the condition for minimum intensity of light called destructive interference. The resultant minimum intensity is,
Imax ∝( a1 – a2)2 ∝ I1 + I2 -2 \(\sqrt { { I }_{ 1 }{ I }_{ 2 } } \) ……. (12)

Question 15.
Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
I Experimental setup:

1. Wavefronts from S1 and S2 spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter s from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
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2. Using an eyepiece the fringes can be seen directly. At the center point O on the screen, waves from S1 and S2 travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

3. The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

II Equation for path difference:

1. Let d be the distance between the double slits S1 and S2 which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S1 and S2 is C and the midpoint of the screen O is equidistant from S1 and S2. P is any point at a distance y from O.
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2. The waves from S1 and S2 meet at P either inphase or out-of-phase depending upon the path difference between the two waves.
3. The path difference δ between the light waves from S1 and S2 to the point P is,
δ = S2 P – S1
4. A perpendicular is dropped from the point S1 to the line S2 P at M to find the path difference more precisely.
δ = S2 P – MP = S2 M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S2 S1M are equal. ∠OCP = ∠S2 S1 M = θ
In right angle triangle ∆S1 S2 M, the path difference, S2 M = d sin θ
δ = d sin θ
If the angle 0 is small, sin θ ≈ tan θ ≈ θ. From the right angle triangle ∆OCP, tan θ = \(\frac { y }{ D }\)
The path difference, δ = \(\frac { dy }{ D }\)

Question 16.
Obtain the equation for bandwidth in Young’s double slit experiment.
Answer:
Condition for bright fringe (or) maxima
The condition for the constructive interference or the point P to be have a bright fringe is,
Path difference, δ = nλ
where, n = 0, 1, 2,. . .
∴ \(\frac { dy }{ D }\) = nλ
y = n \(\frac { λD }{ d }\) (or) yn = n \(\frac { λD }{ d }\)
This is the condition for the point P to be a bright fringe. The distance is the distance of the nth bright fringe from the point O.
Condition for dark fringe (or) minima
The condition for the destructive interference or the point P to be have a dark fringe is,
Path difference, δ = (2n – 1) \(\frac { λ }{ 2 }\)
where, n = 1, 2, 3 …
∴ \(\frac { dy }{ D }\) = (2n – 1) \(\frac { λ }{ 2 }\)
y = \(\frac { (2n-1) }{ 2 }\) \(\frac { λD }{ d }\) (or) yn = \(\frac { (2n-1) }{ 2 }\) \(\frac { λD }{ d }\)
Equation for bandwidth
The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.
The distance between (n +1)th and nth consecutive bright fringes from O is given by,
β = y(n +1) – yn = \(\left((n+1) \frac{\lambda \mathrm{D}}{d}\right)-\left(n \frac{\lambda \mathrm{D}}{d}\right)\)
β = \(\frac { λD }{ d }\)
Similarly, the distance between (n +1)th and nth consecutive dark fringes from O is given by,
β = y(n+1) – yn = \(\left(\frac{(2(n+1)-1)}{2} \frac{\lambda D}{d}\right)-\left(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\right)\)
β = \(\frac { λD }{ d }\)
The above equation show that the bright and dark fringes are of same width equally spaced on either side of central bright fringe.

Question 17.
Obtain the equations for constructive and destructive interference for transmitted and reflected waves in thin films.
Answer:
Interference in thin films:
Let us consider a thin film of transparent material of refractive index p (not to confuse with order of fringe n) and thickness d. A parallel beam of light is incident on the film at an angle i. The wave is divided into two parts at the upper surface, one is reflected and the other is refracted.

The refracted part, which enters into the film, again gets divided at the lower surface into two parts; one is transmitted out of the film and the other is reflected back in to the film. Reflected as well as refracted waves are sent by the film as multiple reflections take place inside the film. The interference is produced by both the reflected and transmitted light.
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For transmitted light:
The light transmitted may interfere to produce a resultant intensity. Let us consider the path difference between the two light waves transmitted from B and D. The two waves moved together and remained in phase up to B where splitting occured. The extra path travelled by the wave transmitted from D is the path inside the film, BC + CD.
If we approximate the incidence to be nearly normal (i = 0), then the points B and D are very close to each other. The extra distance travelled by the wave is approximately twice thickness of the film, BC + CD = 2d. As this extra path is traversed in a medium of refractive index µ, the optical path difference is, δ = 2µd.
The condition for constructive interference in transmitted ray is,
2µd = nλ
Similarly, the condition for destructive interference in transmitted ray is,
2µd = (2n – 1) \(\frac { λ }{ 2 }\)
For reflected light:
It is experimentally and theoretically proved that a wave while travelling in a rarer medium and getting reflected by a denser medium, undergoes a phase change of π. Hence, an additional path difference of \(\frac { λ }{ 2 }\) should be considered.

Let us consider the path difference between the light waves reflected by the upper surface at A and the other wave coming out at C after passing through the film. The additional path travelled by wave coming out from C is the path inside the film, AB + BC. For nearly normal incidence this distance could be approximated as, AB + BC = 2d. As this extra path is travelled in the medium of refractive index p, the optical path difference is, δ = 2µd.
The condition for constructive interference for reflected ray is,
2µd + \(\frac { λ }{ 2 }\) = nλ (or) 2µd = (2n – 1) \(\frac { λ }{ 2 }\)
The additional path difference \(\frac { λ }{ 2 }\) is due to the phase change of n in rarer to denser reflection taking place at A. The condition for destructive interference for reflected ray is,
2µd + \(\frac { λ }{ 2 }\) = (2n + 1) \(\frac { λ }{ 2 }\) (or) 2µd = nλ

Question 18.
Discuss diffraction at single slit and obtain the condition for nth minimum.
Answer:
Diffraction at single slit:
Let a parallel beam of light fall normally on a single slit AB of width. The diffracted beam falls on a screen kept at a distance. The center of the slit is C. A straight line through C perpendicular to the plane of slit meets the center of the screen at O. We would like to find the intensity at any point P on the screen. The lines joining P to the different points on the slit can be treated as parallel lines, making an angle 9 with the normal CO.

All the waves start parallel to each other from different points of the slit and interfere at point P and other points to give the resultant intensities. The point P is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction. We need to give the condition for the point P to be of various minima.

The basic idea is to divide the slit into much smaller even number of parts. Then, add their contributions at P with the proper path difference to show that destructive interference takes place at that point to make it minimum. To explain maximum, the slit is divided into odd number of parts.
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Condition for P to be first minimum:
Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). We have different points on the slit which are separated by the same width (here a/2) called corresponding points.
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The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum. The path difference 8 between waves from these corresponding points
is, δ = \(\frac { a }{ 2 }\) sin θ
The condition for P to be first minimum, \(\frac { a }{ 2 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = λ (first minimum) ….. (1)
Condition for P to be first minimum:
Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). We have different points on the slit which are separated by the same width (here a/2) called corresponding points.
The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum.
The path difference δ between waves from these corresponding points δ = \(\frac { a }{ 4 }\) sin θ.
The condition for P to be first minimum, \(\frac { a }{ 4 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = 2λ (second minimum) …(2)
Condition for P to be third order minimum:
The same way the slit is divided into six equal parts to explain the condition for P to be third
minimum is, \(\frac { a }{ 6 }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = 3λ (third minimum) …(3)
Condition for P to be nth order minimum:
Dividing the slit into 2n number of (even number of) equal parts makes the light produced by one of the corresponding points to be cancelled by its counterpart. Thus, the condition for nth
order minimum is, \(\frac { a }{ 2n }\) sin θ = \(\frac { λ }{ 2 }\)
a sin θ = nλ (nth minimum)

Question 19.
Discuss the diffraction at a grating and obtain the condition for the mth maximum.
Answer:
A plane transmission grating is represented by AB. Let a plane wavefront of monochromatic light with wavelength λ be incident normally on the grating. As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.
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A diffraction pattern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens. Let us consider a point P at an angle 0 with the normal drawn from the center of the grating to the screen. The path difference 5 between the diffracted waves from one pair of corresponding points is,
δ = (a + b) sin θ
This path difference is the same for any pair of corresponding points. The point P will be bright, when
δ = m λ where m = 0, 1 , 2, 3

Combining the above two equations, we get,
(a + b) sin θ = m λ
Here, m is called order of diffraction.

Condition for zero order maximum, m = 0
For (a + b) sin θ = 0, the position, θ = 0. sin θ = 0 and m = 0. This is called zero order diffraction or central maximum.

Condition for first order maximum, m = 1
If (a + b) sin θ1 = λ, the diffracted light meet at an angle θ1 to the incident direction and the first order maximum is obtained.

Condition for second order maximum, m = 2
Similarly, (a + b) sin θ2 = 2λ, forms the second order maximum at the angular position θ2.

Condition for higher order maximum
On either side of central maxima different higher orders of diffraction maxima are formed at different angular positions.
If we take, N = \(\frac { 1 }{ a+b }\)

Then, N gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number N is specified on the grating itself. Now, the equation becomes,
\(\frac { 1 }{ N }\) sin θ = mλ, (or) sin θ = Nmλ

Question 20.
Discuss the experiment to determine the wavelength of monochromatic light using diffraction grating.
Answer:
Experiment to determine the wavelength of monochromatic light:
The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating and a spectrometer. Initially all the preliminary adjustments of the spectrometer are made. The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined. The telescope is brought in line with collimator to view the image of the slit. The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator.

The telescope is turned to one side until the first order diffraction image of the slit coincides with the vertical cross wire of the eye piece. The reading of the position of the telescope is noted. Similarly the first order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted. The difference between two positions gives 2θ. Half of its value gives θ, the diffraction angle for first order maximum. The wavelength of light is calculated from the equation,
λ = \(\frac { sin θ }{ N m}\)
Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Question 21.
Discuss the experiment to determine the wavelength of different colours using diffraction grating.
Answer:
Determination of wavelength of different colours:
When white light is used, the diffraction pattern consists of a white central maximum and on both sides continuous coloured diffraction patterns are formed. The central maximum is white as all the colours meet here constructively with no path difference.

As θ increases, the path difference, (a + b) sin θ, passes through condition for maxima of diffraction of different orders for all colours from violet to red. It produces a spectrum of diffraction pattern from violet to red on either side of central maximum. By measuring the angle at which these colours appear for various orders of diffraction, the wavelength of different colours could be calculated using the formula,
λ = \(\frac { sin θ }{ N m}\)
Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Question 22.
Obtain the equation for resolving power of optical instrument.
Answer:
Resolution:
The effect of diffraction has an adverse impact in the image formation by the optical instruments such as microscope and telescope. For a single rectangular slit, the half angle θ subtended by the spread of central maximum (or position of first minimum) is given by the ‘ relation,
a sin θ = λ … (1)
Similar to a rectangular slit, when a circular aperture or opening (like a lens or the iris of our eye) forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that becomes fainter while moving away from the center. These are known as Airy’s discs. The circle of central maximum has the half angular spread given by the equation,
a sin θ = 1.22λ … (2)
Here, the numerical value 1.22 comes for central maximum formed by circular apertures. This involves higher level mathematics which is avoided in this discussion.
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For small angles, sin θ ≈ θ
a θ = 1.22λ … (3)
Rewriting further, θ = \(\frac { 1.22λ }{ a}\) and \(\frac { { r }_{ 0 } }{ f } \) = \(\frac { 1.22λ }{ a}\)
r0 = \(\frac { 1.22λf }{ a}\) …. (4)

When two point sources close to each another form image on the screen, the diffraction pattern of one point source can overlap with another and produce a blurred image. To obtain a good image of the two sources, the two point sources must be resolved i.e., the point sources must be imaged in such a way that their images are sufficiently far apart that their diffraction patterns do not overlap.

According to Rayleigh’s criterion, for two point objects to be just resolved, the minimum distance between their diffraction images must be in such a way that the central maximum of one coincides with the first minimum of the other and vice versa. Such an image is said to be just resolved image of the object. The Rayleigh’s criterion is said to be limit of resolution.

According to Rayleigh’s criterion the two point sources are said to be just resolved when the distance between the two maxima is at least r0. The angular resolution has a unit in radian (rad) and it is given by the equation.
θ = \(\frac { 1.22λf}{ a}\)
It shows that the first order diffraction angle must be as small as possible for greater resolution. This further shows that for better resolution, the wavelength of light used must be as small as possible and the size of the aperture of the instrument used must be as large as possible. The equation (4) is used to calculate spacial resolution. The inverse of resolution is called resolving power. This implies, smaller the resolution, greater is the resolving power of the instrument. The ability of an optical instrument to • separate or distinguish small or closely adjacent objects through the image formation is said to be resolving power of the instrument.

Question 23.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
Simple microscope:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.

  • Near point focusing :
    The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
  • Normal focusing :
    The image is formed at infinity. In this position the eye is most relaxed to view the image.

Magnification in near point focusing:
Object distance u is less than f. The image distance is the near point D. The magnification m is given by the relation,
m = \(\frac { v}{ u}\)
With the help of lens equation, \(\frac { 1}{ v}\) – \(\frac { 1}{ u}\) \(\frac { 1}{ f}\) the magnification can further be writen as,
m = 1- \(\frac { v}{ f}\)
Substituting for v with sign convention, v = -Derivem
m = 1 + \(\frac { D}{ f}\)
This is the magnification for near point focusing
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Magnification in normal focusing (angular magnification):
We will now find the magnification for the image formed at infinity. If we take the ratio of height of image to height of object \(\left(m=\frac{h^{\prime}}{h}\right)\) to find the magnification, we will not get a practical relation as the image will also be of infinite size when the image is formed at infinity. Hence, we can practically use the angular magnification. The angular magnification is defined as the ratio of angle 0j subtended by the image with aided eye to the angle 90 subtended by the object with unaided eye.
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m = \(\frac{\theta_{i}}{\theta_{0}}\)
For unaided eye, tan θ0 ≈ θ0 = \(\frac { h }{ D }\)
For aided eye, tan θi ≈ θi = \(\frac { h }{ f }\)
The angular magnification is, m = \(\frac{\theta_{i}}{\theta_{0}}\) = \(\frac { h/f }{ h/d }\)
m = \(\frac { d }{ f }\)
This is the magnification for normal focusing. The magnification for normal focusing is one less than that for near point focusing.

Question 24.
Explain about compound microscope and obtain the equation for magnification. Compound microscope:
Answer:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.
Magnification of compound microscope :
From the ray diagram, the linear magnification due to the objective is,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-41
m0 = \(\frac { h’ }{ f}\)
From the figure, tan β = \(\frac { h }{ { f }_{ 0 } } \) =\(\frac { h’ }{ L’}\),then
\(\frac { h’ }{ h}\) = \(\frac { L }{ { f }_{ 0 } } \); m0 =\(\frac { L }{ { f }_{ 0 } } \)
Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length
L of the microscope as f0 and fe are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification me of the eyepiece is,
me = 1 + \(\frac { D }{ { f }_{ e } } \)
The total magnification m in near point focusing is,
m = m0me = \(\left(\frac{L}{f_{o}}\right)\left(1+\frac{D}{f_{e}}\right)\)
If the final image is formed at infinity (normal focusing), the magnification me of the eyepiece is,
me = \(\frac { D }{ { f }_{ e } } \)
The total magnification m in normal focusing is,
m = m0me = \(\left(\frac{L}{f_{o}}\right)\left(\frac{D}{f_{e}}\right)\)

Question 25.
Obtain the equation for resolving power of microscope.
Answer:
Resolving power of microscope:
The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance dmin.
Smaller the value of dmin better will be the resolving power of the microscope.
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The radius of central maxima is, r0 = \(\frac { 1.22λv }{ a }\)
In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is dmin, then the magnification m is, m = \(\frac{r_{o}}{d_{\min }}\)
dmin = \(\frac{r_{o}}{m}\) =\(\frac{1.22 \lambda v}{a m}\) = \(\frac{1.22 \lambda v}{a(v / u)}\) = \(\frac{1.22 \lambda u}{a}\) [∴m = v/u] [∴ [u ≈ ƒ]
On the object side, 2tan β ≈ 2sin β = \(\frac { a }{ f }\)
dmin = \(\frac{1.22 \lambda}{2 sin β }\) ∴ [a = ƒ2 sinβ]
To further reduce the value of dmin the optical path of the light is increased by immersing the objective of the microscope in to a bath containing oil of refractive index n.
dmin = \(\frac{1.22 \lambda}{2n sin β }\)
Such an objective is called oil immersed objective. The term n sin p is called numerical aperture NA.
dmin = \(\frac{1.22 \lambda}{2(NA) }\)

Question 26.
Discuss about astronomical telescope.
Answer:
Astronomical telescope:
An astronomical telescope is used to get the magnification of distant astronomical objects like stars, planets, moon etc. the image formed by astronomical telescope will be inverted. It has an objective of long focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image.
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Magnification of astronomical telescope:
The magnification m is the ratio of the angle β subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye
m = \(\frac { β }{ α }\)
From the diagram, m = \(\frac{h / f_{e}}{h / f_{0}}\)
m = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \)
The length of the telescope is approximately, L = f0 + fe

Question 27.
Mention different parts of spectrometer and explain the preliminary adjustments. Spectrometer:
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are (i) collimator, (ii) prism table and (iii) Telescope.
Adjustments of the spectrometer:
The following adjustments must be made before doing the experiment using spectrometer.

(a) Adjustment of the eyepiece:
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.

(b) Adjustment of the telescope:
The telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.

(c) Adjustment of the collimator:
The telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for parallel rays, a well-defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel.

(d) Levelling the prism table:
The prism table is adjusted or levelled to be in horizontal position by means of levelling screws and a spirit level.

Question 28.
Explain the experimental determination of material of the prism using spectrometer. Determination of refractive index of material of the prism:
Answer:
The preliminary adjustments of the telescope, collimator and the prism table of the spectrometer are made. The refractive index of the prism can be determined by knowing the angle of the prism and the angle of minimum deviation.

(i) Angle of the prism (A):
The prism is placed on the prism table with its refracting edge facing the collimator. The slit is illuminated by a sodium light (monochromotic light). The parallel rays coming from the collimator fall on the two faces AB and AC. The telescope is rotated to the position T1 until the image of the slit formed by the reflection at the face AB is made to coincide with the vertical cross wire of the telescope.
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The readings of the verniers are noted. The telescope is then rotated to the position T2 where the image of the slit formed by the reflection at the face AC coincides with the vertical cross wire. The readings are again noted.
The difference between these two readings gives the angle rotated by the telescope, which is twice the angle of the prism. Half of this value gives the angle of the prism A.

(ii) Angle of minimum deviation (D):
The prism is placed on the prism table so that the light from the collimator falls on a refracting face, and the refracted image is observed through the telescope. The prism table is now rotated so that the angle of deviation decreases. A stage comes when the image stops for a moment and if we rotate the prism table further in the same direction, the image is seen to recede and the angle of deviation increases. The vertical cross wire of the telescope is made to coincide with the image of the slit where it turns back. This gives the minimum deviation position.
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The readings of the verniers are noted. Now, the prism is removed and the telescope is turned to receive the direct ray and the vertical cross wire is made to coincide with the image. The readings of the verniers are noted. The difference between the two readings gives the angle of minimum deviation D. The refractive index of the material of the prism n is calculated using the formula,
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
The refractive index of a liquid may be determined in the same way using a hollow glass prism filled with the given liquid.

Samacheer Kalvi 12th Physics Optics Conceptual Questions

Question 1.
Why are dish antennas curved?
Answer:
Dish antenna is curved so as it can receive parallel signal rays coming from same direction. These parallel signal rays reflect from parabolic dish, and gathered at main antenna part. This increases directivity of antenna, and gives sufficient amplitude signal.

Question 2.
What type of lens is formed by a bubble inside water?
Answer:
Air bubble has spherical surface and is surrounded by medium (water) of higher refractive index. When light passes from water to air it gets diverged. So air bubble behaves as a concave lens.

Question 3.
It is possible for two lenses to produce zero power?
Answer:
Yes. It is possible for two lenses to produce zero power. Both the surfaces of lenses are equally curved, i.e. R1 = R2 and hence
Power (P) = (µ – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=0\)

Question 4.
Why does sky look blue and clouds look white?
Answer:
Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, intensity of scattered light, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \). So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appear blue. Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 5.
Why is yellow light preferred to during fog?
Answer:
Yellow light has longer wavelength than green, blue or violet components of white lights. As scattered intensity, I ∝ \(\frac { 1 }{ { \lambda }^{ 4 } } \). so yellow colour is least scattered and produces sufficient illumination.

Question 6.
Two independent monochromatic sources cannot act as coherent sources, why?
Answer:
Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms. When they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase.

Question 7.
Does diffraction take place at the Young’s double slit?
Answer:
Both diffraction and interference in the double slit experiment. The wavefront is diffracted as it passes through each of the slits. The diffraction causes the wavefronts to spread out as if they were coming from light sources located at the slits. These two wavefronts overlap, and interference occurs.

Question 8.
Is there any difference between colored light obtained from prism and colours of soap bubble?
Answer:
Yes. there is a difference between colored light obtained from the prism is the phenomenon of‘dispersion of light’ and colored light obtained from the soap bubble is the phenomenon of ‘interference of light’.

Question 9.
A small disc is placed in the path of the light from distance source. Will the center of the shadow be bright or dark?
Answer:
When a tiny circular small disc is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the disc because, wave diffracted from the edge of the circular disc interface constructively at the centre of the shadow, which produces bright spot.

Question 10.
When a wave undergoes reflection at a denser medium, what happens to its phase?
Answer:
When a wave undergoes a reflection at a denser medium then it’s crest reflected as trough and vice versa. So, its phase changes at 180°.

Samacheer Kalvi 12th Physics Optics Numerical Problems

Question 1.
An object is placed at a certain distance from a convex lens of focal length 20 cm. Find the distance of the object if the image obtained is magnified 4 times.
Solution:
ƒ = – 20 cm
v = – 4u
According to lens formula
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ (-20) }\) = \(\frac { 1 }{ (-4u) }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ (-20) }\) = \(\left[-\frac{1}{4}+1\right]\)
= \(\frac { 1 }{ u }\) \(\frac { 3 }{ 4 }\)
u = \(\frac { 3 × 20 }{ 4 }\) = -15 cm.

Question 2.
A compound microscope has a magnification of 30. The focal length of eye piece is 5 cm. Assuming the final image to be at least distance of distinct vision, find the magnification produced by the objective.
Solution:
Magnification of compound microscope, M = 30
Focal length, f = 5 cm
Least distance of distinct vision, D = 25 cm
Now, M = M0 x Me
= M0 x \(\left[1+\frac{\mathrm{D}}{f_{e}}\right]\)
30 = M0 x \(\left[1+\frac{25}{5}\right]\)
M0 = \(\frac { 30 }{ 6 }\) = 5

Question 3.
An object is placed in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror.
Solution:
m = + 3 and m = – 3; f = – 20 cm
When, m = 3
m = \(\frac { v }{ u }\) = 3
v = – 3 u
From mirror equation,
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 1 }{ -3u }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ u }\) \(\left[-\frac{1}{3}+1\right]\)
\(\frac { 1 }{ -20 }\) = \(\frac { 2 }{ 3u }\)
u = –\(\frac { 40 }{ 3 }\) cm
When, m = – 3 ⇒ v = 3u
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 1 }{ 3u }\) + \(\frac { 1 }{ u }\)
\(\frac { 1 }{ -20 }\) = \(\frac { 4 }{ 3u }\)
u = \(\frac { 4×20 }{ 3 }\)
u = –\(\frac { 80 }{ 3 }\) cm

Question 4.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Solution:
Actual depth of the bulb in water
d1 = 80 cm = 0.8 m
Refractive index of water, p = 1.33
The given situation is shown in the figure.
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Where, i = Angle of incidence
r = Angle of refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle AT
R= \(\frac { AC }{ 2 }\) = AO = OB
Using snell’s law, the relation for the refractive index of water is sin
µ = \(\frac { sin r }{ sin i }\)
i = sin-1 (0.75) = 48.75°
Using the given figure, we have the relation
tan i = \(\frac { OC }{ BC }\) = \(\frac { R }{ { d }_{ 1 } } \)
R = tan i x d1 = tan 48.75° x 0.8
R = 0.91 m
Area of the surface of water = πR²
= 3.14 x (0.91)² = 2.61 m²
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m².

Question 5.
A thin converging glass lens made of glass with refractive index 1.5 has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index n, it acts as a divergent lens of focal length 100 cm. What must be the value of n?
Solution:
According to Lens maker’s formula, the focal length for a convex lens placed in air can be obtained as
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-47
\(\frac { 0.5 }{ 5 }\) = \(\frac { 1.5 }{ n }\) -1
0.9 = \(\frac { 1.5 }{ n }\)
n = \(\frac { 1.5 }{ 0.9 }\) = \(\frac { 5 }{ 3 }\) -1

Question 6.
If the distance D between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called conjugate foci method. If d is the distance between the two positions of the lens, obtain the equation for focal length of the convex lens.
Solution:
Let us fix the position of object and place the screen to get the enlarged image first. Also, let us fix the position of screen where we get the enlarged image.
Let D be the distance between object and screen. Let us mark the position of lens dv Then let us move the lens away from the object to get a diminished image. Let this position of lens be d2. Let d be the distance between the lens position d1 and d2. Let V be the distance b/w image and lens. Let ‘u’ be the distance between object and lens.
From mirror equation,
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-48
\(\frac { 1 }{ v }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
Let us replace v by substituting v = D – u
\(\frac { 1 }{ D-u }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ f }\)
we get the equation, u² -Du + fD = 0
the quadratic equation for above equation,
u = \(\frac{\mathrm{D} \pm \sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\)
When D = 4f, we get only position of lens to get image.
This corresponds to placing the object at 2f and getting the image at 2f on the otherside. Hence, for displacement method we need D > 4 f. When this condition is satisfied we get
u1 = \(\frac{\mathrm{D}-\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\) ; corresponding v1 – D – u2 = \(\frac{\mathrm{D}+\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\) after changing the location
u1 =\(\frac{\mathrm{D}+\sqrt{\mathrm{D}^{2}-4f \mathrm{D}}}{2}\) ; corresponding v2 – D – u2 = \(\frac{\mathrm{D}-\sqrt{\mathrm{D}^{2}-4 f \mathrm{D}}}{2}\)
now the displacement d = v1 – u1 = u1 – v1 = \(\sqrt{D^{2}-4 f D}\)
Hence we get focal length, ƒ = \(\frac{D^{2}-d^{2}}{4 D}\)

Question 7.
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe?
Solution:
For first minimum (n = 1) on either side of central maximum.
sin θ = \(\frac { λ }{ a }\)
Where a is width of the slit Since 0 is very small, (sin θ ≈ θ)
θ = \(\frac { λ }{ a }\) …… (1)
sin θ ≈ θ = \(\frac { x }{ 2D }\) …… (2)
Where, x is distance of first dark fringe from central maximum.
D is distance between slit and screen.
From equation (1) and (2)
\(\frac { x }{ 2D }\) = \(\frac { λ }{ a }\)
x = \(\frac { 2Dλ }{ a }\) = \(\frac { 2\times 2\times 600\times { 10 }^{ -9 } }{ 1\times { 10 }^{ -3 } } \)
= 2400 x 10-6 = 2.4 x 10-3m
x = 2.4 mm

Question 8.
In Young’s double slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two wavelength λ0 = 750 nm and λ = 900 nm. What is the minimum distance from the common central bright fringe on a screen 2 m from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other?
Solution:
Now from the question we can infer that
D = 2; d = 2
n1 λ1 = n2 λ2
\(\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{1}}\) =\(\frac { 900 }{ 750 }\) \(\frac { 1 }{ 2 }\)
Thus, we have
\(\frac{n_{1}}{n_{1}}=\frac{\lambda_{2}}{\lambda_{1}}\) =\(\frac { 6 }{ 5 }\)
5th and 6th fringes will coincide respectively. The minimum distance is given as
Xmin = \(\frac{n_{2} \lambda_{2} \mathrm{D}}{d}\) = \(\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}}\)
= 4500 x 10-6 = 4.5 x 10-3m
Xmin = 4.5 mm

Question 9.
In Young’s double slit experiment, 62 fringes are seen in visible region for sodium light of wavelength 5893 Å. If violet light of wavelength 4359 Å is used in place of sodium light, then what is the number of fringes seen?
Solution:
From young’s double slit experiment,
λ1 = 5893 Å ; λ2 = 4359 Å
\(\frac{n_{1} \lambda_{1} D}{d}\) = \(\frac{n_{2} \lambda_{2} \mathrm{D}}{d}\)
The above condition is total extent of fringes is constant for both wavelengths.
\(\frac{62 \times 5893 \times 10^{-10} \times D}{d}\) = \(\frac{n_{2} \times 4359 \times 10^{-10} \times D}{d}\)
n2 = \(\frac{62 \times 5893}{4359}\) = \(\frac{365366}{4359}\) = 83.8
n2 = 84

Question 10.
A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece.
Solution:
Magnifying Power, m = 100
Focal length of the objective,f0 = 0.5 cm
Tube length, l = 6.5 cm
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece.
v0e= 6.5 cm ….. (1)
The magnifying power for normal adjustment is given by
m = \(\left(\frac{v_{o}}{u_{o}}\right) \times \frac{\mathrm{D}}{f_{e}}\)
= – \(\left[1-\frac{v_{o}}{f_{\mathrm{o}}}\right] \frac{\mathrm{D}}{f_{e}}\)
100 = – \(\left[1-\frac{v_{o}}{0.5}\right] \times \frac{25}{f_{e}}\)
2v0 -4ƒe= 1
On solving equations (1) and (2), we get
v0 = 4.5 cm and ƒe = 2 cm
Thus, the focal length of the eyepiece is 2 cm.

Samacheer Kalvi 12th Physics Optics Additional Questions

Samacheer Kalvi 12th Physics Optics Multiple Choice Questions

Question 1.
When a ray of light enters a glass slab from air
(b) its wavelength increases
(c) its frequency increases
(d) neither its wavelength nor its frequency changes
Answer:
(a) its wavelength decreases
Hint:
Wavelength, λ = \(\frac { Velocity }{ Frequency }\) = \(\frac { u }{ v }\)
When light travels from air to glass, frequency v remains unchanged, velocity u decreases and hence wavelength X also decreases.

Question 2.
A source emits sound of frequency 600 Hz inside water. The frequency heard in air (velocity of sound in water = 1500 m/s, velocity of sound in air = 300 m/s) will be
(a) 300 Hz
(b) 120 Hz
(c) 600 Hz
(d) 6000 Hz
Answer:
(c) 600 Hz
Hint:
Frequency does not change when sound travels from one medium to another
∴ Frequency of sound in air = Frequency of sound in water = 600 Hz

Question 3.
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be
(a) 30° for both the colours
(b) greater for the violet colour
(c) greater for the violet colour
(d) equal but not 30° for both the colours
Answer:
(a) 30° for both the colours
Hint:
For any prism, r1 = r2 = A
In the position of minimum deviation for any wavelength,
r1 = r2 = \(\frac { A }{ 2 }\) = \(\frac { 60° }{ θ }\) = 30°

Question 4.
To get three images of a single object, one should have two plane mirrors at an angle of
(a) 60°
(b) 90°
(c) 120°
(d) 30°
Answer:
(b) 90°
Hint:
The number of images formed,
n = \(\frac { 360° }{ θ }\) – 1 or 3 = \(\frac { 360° }{ θ }\) -1 or θ = 90°

Question 5.
Which of the following is used in optical fibres?
(a) Total internal reflection
(b) Diffraction
(c) Refraction
(d) Scattering
Answer:
(a) Total internal reflection
Hint:
The working of optical fibres is based on total internal reflection.

Question 6.
Two lenses of power – 15 D and +15D are in contact with each other. The focal length of the combination is
(a) + 10 cm
(b) -20 cm
(c) – 10 cm
(d) + 20cm
Answer:
(c) – 10 cm
Hint P = P1 + P2 = -15 + 5= -10 D
F = \(\frac { 1 }{ P }\) = \(\frac { 1 }{ -10 }\) m = -10 cm

Question 7.
The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let δ1 and δ2 be angles of minimum deviation for red and blue light respectively in a prism of this glass, then
(a) δ1, can be less than or greater than δ2 depending upon the values of δ1 and δ2
(b) δ1 > δ2
(c) δ1 < δ2
(d) δ1 = δ2
Answer:
(c) δ1 < δ2
Hint:
δ1 = (μR – 1)A, δ2 = (μB – 1)A
As, μR < μB ∴ δ1 < δ2

Question 8.
Time image formed by an objective of a compound microscope is
(a) virtual and diminished
(b) real and diminished
(c) real and enlarged
(d) virtual and enlarged
Answer:
(c) real and enlarged
Hint
The image formed by the objective of a compound microscope is real and enlarged.

Question 9.
An astronomical telescope has a large aperture to,
(a) reduce spherical aberration
(b) have high resolution
(c) increase span of observation
(d) have low dispersion
Answer:
(b) have high resolution

Question 10.
Two plane mirrors are inclined to each other at an angle of 60°. A point object is placed in between them. The total number of images produced by both the mirror is
(a) 2
(b) 4
(c) 5
( d) 6
Answer:
(c) 5
Hint:
Number of images formed, θ = \(\frac { 360° }{ θ }\) – 1 = \(\frac { 360 }{ 60 }\) = 1 = 5.

Question 11.
A boy 1.5 m tall with his eye level at 1.38 m stands before a mirror fixed on a wall. The minimum length of mirror required to view the complete image of boy is
(a) 0.75 m
(b) 0.06 m
(c) 0.69 m
(d) 0.12 m
Answer:
(a) 0.75 m
Hint
Minimum length of mirror required \(\frac { 1 }{ 2 }\) x Height of boy = \(\frac { 1 }{ 2 }\) x 1.5 = 0.75 m.

Question 12.
A pencil of light rays falls on a plane mirror and forms a real image, so the incident rays are
(a) parallel
(b) diverging
(c) converging
(d) statement is false
Answer:
(c) converging
Hint:
When converging rays fall on a plane mirror, they get reflected to a point d in front of the mirror forming a real image.

Question 13.
For a real object, which of the following can produce a real image?
(a) plane mirror
(b) concave lens
(c) convex lens
(d) concave mirror
Answer:
(d) concave mirror
Hint:
Only concave mirror produces real image provided the object is not placed between its focus and pole.

Question 14.
Which mirror is to be used to obtain a parallel beam of light from a small lamp?
(a) Plane mirror
(b) Convex mirror
(c) Concave mirror
(d) None of the above
Answer:
(c) Concave mirror
Hint:
When the small lamp is placed at the focus of the concave mirror, the reflected light is a parallel beam.

Question 15.
When a plane electromagnetic wave enters a glass slab, then which of the following will not change?
(a) Wavelength
(b) Frequency
(c) Speed
(d) Amplitude
Answer:
(b) Frequency
Only the frequency of the electromagnetic wave remains unchanged.

Question 16.
If wavelength of light in air is 2400 x 10-10 m, then what will be the wavelength of light in glass (μ = 1.5)?
(a) 1600 Å
(b) 7200 Å
(c) 1080 Å
(d) None of these
Answer:
(a) 1600 Å
Hint:
μ = \(\frac { { \lambda }_{ a } }{ { \lambda }_{ g } } \) ⇒λλg = \(\frac { { \lambda }_{ a } }{ μ } \) = \(\frac{2400 \times 10^{-10} \mathrm{m}}{1.5}\) = 1600 Å

Question 17.
Why is refractive index in a transparent medium greater than one?
(a) Because the speed of light in vacuum is always less than speed in a transparent medium.
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
(c) Frequency of wave changes when it Gasses medium.
(d) None of the above.
Answer:
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-49
As c > v, μ > 1

Question 18.
The wavelength of sodium light in air is 5890 Å. The velocity of light in air is 3 x 108 ms-1. The wavelength of light in a glass of refractive index 1.6 would be close to
(a) 5890 Å
(b) 3681 Å
(c) 9424 Å
(d) 15078 Å
Answer:
(b) 3681 Å
Hint:
μ = \(\frac { { \lambda }_{ a } }{ { \lambda }_{ g } } \) ⇒λg= \(\frac { { \lambda }_{ a } }{ μ } \) = \(\frac { 5890 Å }{ 1.6 }\) = 3681 Å

Question 19.
A glass slab (μ = 1.5) of thickness 6 cm is placed over a paper. Qhat is the shift in the letters?
(a) 4 cm
(b) 2 cm
(c) 1 cm
(d) None of these
Answer:
(b) 2 cm
Hint:
Normal shift, x = \(t\left(1-\frac{1}{\mu}\right)=6\left(1-\frac{1}{1.5}\right)\) cm = 2cm

Question 20.
Light traveling from a transparent medium to air undergoes total internal reflection at an angle of incident of 45°. Then refractive index of the medium may be
(a) 1.5
(b) 1.3
(c) 1.1
(d) \(\frac { 1 }{ √2 }\)
Answer:
(a) 1.5
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac{1}{\sin 45^{\circ}}\) = √2 = 1.414 ≈ 1.5.

Question 21.
A point source of light is placed 4 m below the surface of water of refractive index 5/3. The minimum diameter of a disc which should be placed over the source, on the surface of water to cut-off all light coming out of water is
(a) infinite
(b) 6 cm
(c) 4 cm
(d) 3 cm
Answer:
(b) 6 cm
Hint:
r = \(\frac{h}{\sqrt{\mu^{2}-1}}\) = \(\frac{4}{\sqrt{\left(\frac{5}{3}\right)^{2}-1}}\) = 3 cm

Question 22.
In optical fibres, propagation of light is due to
(a) diffraction
(b) total internal reflection
(c) reflection
(d) refraction
Answer:
(b) total internal reflection
Hint:
In optical fiberes, light propagates due to tatal internal reflection.

Question 23.
Sparkling of diamond is due to
(a) reflection
(b) dispersion
(c) total internal reflection
(d) high refractive index of diamond
Answer:
(c) total internal reflection

Question 24.
For a given lens, the magnification was found to be twice as large as when the object was 0.15m distant from it as when the distance was 0.2 m. The focal length of the lens is
(a) 1.5 m
(b) 0.20 m
(c) 0.10 m
(d) 0.05 m
Asnwer:
(c) 0.10 m
Hint:
Here m1 = 2m2
\(\frac { f }{ f-0.15 }\) = 2 \(\frac { f }{ f-0.20 }\)
2f – 0.30 = f- 0.20 ; f = 0.10 m

Question 25.
Two lenses of focal lengths f1 and f2 are kept in contact coaxially. The resultant power of combination will be
(a) \(\frac { { f }_{ 1 }{ f }_{ 2 } }{ { { f }_{ 1 }-f }_{ 2 } } \)
(b) \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \)
(c) \({ f }_{ 1 }{ +f }_{ 2 }\)
(d) \(\frac { { f }_{ 1 } }{ { f }_{ 2 } } +\frac { { f }_{ 2 } }{ { f }_{ 1 } } \)
Answer:
(b) \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \)
Hint:
P = \(\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { { f }_{ 1 }{ +f }_{ 2 } }{ { { f }_{ 1 }f }_{ 2 } } \).

Question 26.
Two lenses of power 3D and -ID are kept in contact. What is focal length and nature of combined lens?
(a) 50 cm, convex
(b) 200 cm, convex
(c) 50 cm, concave
(d) 200 cm, concave
Answer:
(a) 50 cm, convex
Hint:
P = P1 + P2 = 3 – 1 = 2D
F= \(\frac { 1 }{ P }\) = \(\frac { 1 }{ 2 }\) m = 50 cm

Question 27.
If two thin lenses are kept coaxially together, then their power is proportional (R1,R2) being the radii of curved surfaces) to
(a) R1 + R2
(b) \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
(c) \(\left[\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1} \mathrm{R}_{2}}\right]\)
(d) None of these
Answer:
(b) \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
Hint:
P = \(\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{2}{\mathrm{R}_{1}}+\frac{2}{\mathrm{R}_{2}}\) = 2\(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)
or p ∝ \(\left[\frac{\mathbf{R}_{1}+\mathbf{R}_{2}}{\mathbf{R}_{1} \mathbf{R}_{2}}\right]\)

Question 28.
A ray incident at 15° on one refracting surface of a prism of angle 60°, suffers a deviation of 55°. What is the angle of emergance?
(a) 95°
(b) 45°
(c) 30°
(d) none of these
Answer:
(d) none of these
Hint:
A + δ = i + each
60° + 55° = 15° + each e = 115 -15 =100°

Question 29.
Dispersion of light is caused due to
(a) Wavelength
(b) intensity of light
(c) density of medium
(d) none of these
Answer:
(a) Wavelength
Hint:
Dispersion is due to the dependence of the speed of a wave on its wavelength in any medium.

Question 30.
White light is incident on one of the refracting surfaces of a prism of angle 50°. If the refractive indices for red and blue colours are 1.641 and 1.659 respectively, the angular separation between these two colours when they emerge out of the prism is
(a) 0.9°
(b) 0.09°
(c) 1.8°
(d) 1.2°
Answer:
(b) 0.09°
Hint:
Angular dispersion,
δB – δR = (µB – µR) A
= (1.659 – 1.641) x 5° = 0.09°

Question 31.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 32.
A setting sun appears to be at an altitude higher than it really is. This is because of
(a) absorption of light
(b) reflection of light
(c) refraction of light
(d) dispersion of light
Answer:
(c) refraction of light
Hint:
This is due to refraction of light by the earth’s atmosphere.

Question 33.
The reddish appearance of rising and setting sun is due to
(a) reflection of light
(b) diffraction of light
(c) scattering of light
(d) interference of light
Answer:
(c) scattering of light
Hint:
The reddish appearance of the rising and the setting sun is due to scattering of light.

Question 34.
In the formation of a rainbow, the light from the sun on water droplets undergoes
(a) dispersion only
(b) only total internal reflection
(c) dispersion and total internal reflection
(d) none of the above
Answer:
(c) dispersion and total internal reflection
Hint:
Rainow is formed due to dispersion of sunlight by raindrops which also deviate the colours by total internal reflection.

Question 35.
The angular magnification of a simple microscope can be increased by increasing
(a) focal length of lens
(b) size of object
(c) aperture of lens
(d) power of lens
Answer:
(b) size of object

Question 36.
For compound microscope f0 = 1 cm, fe = 2.5 cm. An object is placed at distance 1.2 cm from objective lens. What should be length of microscope for normal adjustment?
(a) 8.5 cm
(b) 8.3 cm
(c) 6.5 cm
(d) 6.3 cm
Answer:
(a) 8.5 cm
Hint:
In the normal adjustment of a compound microscope, L = υ0 + υe = \(\frac{υ_{0} f_{e}}{υ_{0}+f_{e}}\) + fe = \(\frac { 1.2×1 }{ -1.2+1 }\) + 2.5 = 6 + 25 = 8.5 cm

Question 37.
Magnifying power of an astronomical telescope for normal vision with usual notation is
(a) -f0 / fe
(b) -f0 x fe
(c) -f0 / f0
(d) -f0 + fe
Answer:
(a) -f0 / fe
Hint:
In normal adjustment of the telescope, m = -f0 / fe

Question 38.
F1 and F2 are focal length of objective and eyepiece respectively of the telescope. The angular magnification for the given telescope is equal to
(a) \(\frac {F_{1}}{F_{2}}\)
(b) \(\frac { { F }_{ 2 } }{ { F }_{ 1 } } \)
(c) \(\frac { { F }_{ 1 }{ F }_{ 2 } }{ { F }_{ 1 }+{ F }_{ 2 } } \)
(d) \(\frac { { F }_{ 1 }+{ F }_{ 2 } }{ { F }_{ 1 }{ F }_{ 2 } } \)
Answer:
(a) \(\frac { { F }_{ 1 } }{ { F }_{ 2 } } \)
Hint:
In normal adjustment of the telescope, \(\left| m \right| \) = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \) = \(\frac { { F }_{ 1 } }{ { F }_{ 2 } } \)

Question 39.
Focal length of objective and eyepiece of telescope are 200 cm and 4 cm respectively. What is length of telescope for normal adjustment?
(a) 196 cm
(b) 204 cm
(c) 250 cm
(d) 225 cm
Answer:
(b) 204 cm
Hint L = \(-{ f }_{ 0 }+{ f }_{ e }\) = 200 + 4 = 204 cm

Question 40.
For normal vision, what is minimum distance of object from eye?
(a) 30 cm
(b) 25 cm
(c) Infinite
(d) 40 cm
Answer:
(b) 25 cm
Hint:
For normal eye, the least distance of distinct vision is 25 cm.

Question 41.
The focal length of the objective and eyepiece of a telescope are respectively 100 cm and 2 cm. The moon subtends angle of 0.5° ; the angle subtended by the moon’s image will be
(a) 10°
(b) 250
(e) 100°
(d) 75°
Answer:
b) 250
Hint ;
m = \(\frac { β }{ α }\) β = \(\frac { { f }_{ 0 } }{ { f }_{ e } } \) ; α = \(\frac { 100 }{ 2 }\) x 0.5° = 25°

Question 42.
A person cannot clearly see distance more than 40 cm. He is advised to use lens of power,
(a) – 2.5 D
(b) 2.5 D
(c) – 6.25 D
(d) 1.5 D
Answer:
(a) – 2.5 D
Hint;
For the remedial lens, u = ∞,
v = – 40 cm = – 0.40 m
∴ \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ -0.40 }\) – \(\frac { 1 }{ ∞ }\) = -2.5 ⇒ P = 2.5D

Question 43.
The light gathering power of a camera lens depends on
(a) its diameter only
(b) ratio of diameter and focal length
(c) product of focal length and diameter
(d) wavelength of the light used
Answer:
(a) its diameter only
Hint:
The light gathering power of a camera lens is proportional to its area or to the square of its diameter.

Question 44.
Amount of light entering into the camera depends upon
(a) focal length of objective lens
(b) product of focal length and diameter of the objective lens
(c) distance of object from camera
(d) aperture setting of the camera
Answer:
(d) aperture setting of the camera
Hint:
The amount of light entering into the camera depends upon the aperture setting of the camera.

Question 45.
Line spectrum can be obtained from
(a) sun
(b) candle
(c) mercury vapour lamp
(d) electric bulb
Answer:
(c) mercury vapour lamp

Question 46.
The Production of band spectra is caused by
(a) atomic nuclei
(b) hot metals
(c) molecules
(d) electrons
Answer:
(c) molecules

Question 47.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is
(a) six
(b) four
(c) five
(d) three
Answer:
(a) six
Hint:
Number of images formed, n = \(\frac { 360 }{ θ }\) -1 = \(\frac { 360 }{ 60 }\) -1 = 5

Question 48.
A point source kept at a distance of 1000 m has a illumination I. To change the illumination to 161, the new distance should become
(a) 250 m
(b) 500 m
(c) 750 m
(d) 800 m
Answer:
(a) 250 m
Hint:
\(\frac{I_{2}}{I_{1}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}\)
\(\frac{16 \mathrm{I}_{1}}{\mathrm{I}_{1}}=\frac{(1000)^{2}}{r_{2}^{2}}\) ; r2 \(\frac { 1000 }{ 4 }\) = 250 cm

Question 49.
A concave mirror of focal length 15 cm forms an image having twice the linear dimensions of the object. The position of the object, when the image is virtual will be
(a) 22.5 cm
(b) 7.5 cm
(c) 30 cm
(d) 45 cm
Answer:
(b) 7.5 cm
Hint:
For virtual images, m = \(\frac { -v }{ u }\) = + 2 or v = – 2u
As \(\frac { 1 }{ u }\) + \(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\)
∴ \(\frac { 1 }{ u }\) – \(\frac { 1 }{ 2u }\) = \(\frac { 1 }{ -15 }\) or \(\frac { 1 }{ 2u }\) = \(\frac { 1 }{ -15 }\) ⇒ u = -7.5 cm

Question 50.
when a ray of light enters a glass slab, then
(a) its frequency and velocity change
(b) only frequency changes
(c) its frequency and wavelength change
(d) its frequency does not change
Answer:
(d) its frequency does not change
Hint:
When a ray of light enters a glass slab, its velocity and wavelength change while frequency does not change.

Question 51.
A light wave of frequency u and wavelength A travels from air to glass. Then,
(a) y changes
(b) y does not change, A changes
(c) A does not change
(d) y and A change
Answer:
(b) y does not change, A changes
Hint:
Same reasoning as in the above question.

Question 52.
In refraction, light waves are bent on passing from one medium to the second medium, because in the second medium.
(a) the frequency is different
(b) the coefficient of elasticity is different
(c) the speed is different
(d) the amplitude is smaller
Answer:
(c) the speed is different
Hint:
Speed of light in second medium is different than that in first medium.

Question 53.
A ray of light having wavelength 720 nm enters in a glass of refractive index 1.5. The wavelength of the ray within the glass will be
(a) 360 nm
(b) 480 nm
(e) 720 nm
(d) 1080 nm
Answer:
(b) 480 nm
Hint:
\({ \lambda }_{ g }\) = \(\frac { { \lambda }_{ 0 } }{ \mu } \) = \(\frac { 790nm }{ 1.5 }\) = 480nm

Question 54.
Brilliance of a diamond is due to
(a) shape
(b) cutting
(c) reflection
(d) total internal reflection
Answer:
(d) total internal reflection
Hint:
Brilliance of a diamond is due to total internal reflection of light.

Question 55.
An endoscope is employed by a physician to view the internal parts of a body organ. If is based on the principle of
(a) refraction
(b) reflection
(c) total internal reflection
(d) dispersion
Answer:
(c) total internal reflection
Hint:
An endoscope is made of optical fibres which work on the principle of total internal reflection.

Question 56.
‘Mirage’ is a phenomenon due to
(a) reflection of light
(b) refraction of light
(c) total internal reflection of light
(d) diffraction of light Hint Mirage occurs due to total internal reflection of light.
Answer:
(c) total internal reflection of light

Question 57.
Two lenses of power +12D and -2D are combined together. What is their equivalent focal length?
(a) 10 cm
(b) 12.5 cm
(c) 16.6 cm
(d) 8.33 cm
Answer:
(a) 10 cm
Hint:
P = P1 + P2 = + 12 – 2 = 10 D
F = \(\frac { 1 }{ P }\) = \(\frac { 1 }{ 10 }\)m = 10 cm

Question 58.
If two lenses of power + 1.5 D and + 1.0 D are placed in contact, then the effective power of combination will be
(a) 2.5 D
(b) 1.5 D
(c) 0.5 D
(d) 3.25 D
Answer:
(a) 2.5 D
Hint:
P = P1 + P2 = + 1.5 + 1.0 = + 2.5D

Question 59.
The angle of a prism is 6° and its refractive index for green light is 1.5. If a green ray passes through it, the deviation will be
(a) 30°
(b) 15°
(c) 3°
(d) 0°
Answer:
(c) 3°
Hint:
5 = (p – 1) A = (1.5 – 1) x 6 = 3°

Question 60.
Sky appears to be blue in clear atmosphere due to light’s
(a) diffraction
(b) dispersion
(c) scattering
(d) polarisation
Answer:
(c) scattering
Hint:
Sky appears blue due to scattering of light by atmospheric modecules.

Question 61.
One can not see through fog, because
(a) fog absorbs the light
(b) light suffers total reflection at droplets
(c) refractive index of the fog is infinity
(d) light is scattered by the droplets
Answer:
(d) light is scattered by the droplets

Question 62.
Fraunhofer lines of the solar system is an example of
(a) emission lines spectrum
(b) emission band spectrum
(c) continuous emission spectrum
(d) line absorption spectrum
Answer:
(d) line absorption spectrum
Hint:
Fraunhofer lines is an example of line absorption spectrum.

Question 63.
A person using a lens as a sample microscope sees an
(a) inverted virtual image
(b) inverted real magnified image
(c) upright virtual image
(d) upright real magnified image
Answer:
(d) upright real magnified image
Hint:
A person sees an upright virtual image in a simple microscope.

Question 64.
Four lenses of focal length + 10 cm, + 50 cm, + 100 cm and + 200 cm are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eyepiece should be
(a) + 10 cm
(b) + 50 cm
(c) + 100 cm
(d) + 200 cm
Answer:
(a) + 10 cm
Hint:
To produce the largest magnification, the eyepiece should have minimum focal length.

Question 65.
The camera lens has an aperture of f and the exposure time is 1/60 s. What will be the new exposure time if the aperture become 1.4f?
(a) \(\frac { 1 }{ 42 }\) s
(b) \(\frac { 1 }{ 56 }\) s
(c) \(\frac { 1 }{ 72 }\) s
(d) \(\frac { 1 }{ 31 }\) s
Answer:
(d) \(\frac { 1 }{ 31 }\) s
Hint:
Time of exposure ∝ (f – number)2
\(\frac{t}{\left(\frac{1}{60}\right)}=\left(\frac{1.4}{1}\right)^{2}\) ⇒ t = \(\frac { 1.4×1.4 }{ 60 }\) ≈ \(\frac { 1 }{ 31 }\) s.

Question 66.
For a person near point of vision is 100 cm. Then the power of lens he must wear so as have normal vision, should be
(a) + ID
(b) – ID
(c) + 3D
(d) – 3D
Answer:
(c) + 3D
Hint:
f = \(\frac { yD }{ y-D }\) = \(\frac { 100×25 }{ 100-25 }\) = \(\frac { 100 }{ 3 }\) cm = \(\frac { 1 }{ 3 }\) cm ; P = \(\frac { 1 }{ f }\) = +3d

Question 67.
Ray optics is valid, when characteristic dimension ions are
(a) much smaller than the wavelength of light
(b) much larger than the wavelength of light
(c) of the same order as the wavelength of light
(d) of the order of one millimetre
Hint:
Ray optics is valid, when characteristic dimensions are much larger than the wavelength of
light.

Question 68.
A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirrorwillbe
(a) 12 feet
(b) 3 feet
(e) 6 feet
(d) any length
Hint:
inimum height of mirror required for seeing full image Height of the man = 3 feet.

Question 69.
The refractive index of water is 1.33. What will be the speed of light in water?
(a) 3 x 108 ms-1
(b) 2.26 x 108 ms-1
(c) 4 x 108 ms-1
(d) 1.33 x 108 ms-1
Answer:
(b) 2.26 x 108 ms-1
Hint:
As, μ = \(\frac { c }{ v }\) ⇒ v = \(\frac { c }{ μ }\) = \(\frac { 3\times { 10 }^{ 8 } }{ 1.33 } \) = 2.26 x 108 ms-1

Question 70.
A beam of monochromatic light is refracted from vacuum into a medium of refractive index
1.5. The wavelength of refracted light will be
(a) same
(b) dependent on intensity of refracted light
(c) larger
(d) smaller
Answer:
(d) smaller
Hint:
As light enters the medium, its wavelength decreases and becomes equal to \(\frac { λ }{ μ }\).

Question 71.
Optical fibers are based on
(a) total internal reflection
(b) less scattering
(c) refraGtion
(d) less absorption coefficient
Answer:
(a) total internal reflection

Question 72.
A convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the lens. Then, its focal length will
(a) become zero
(b) becomes infinite
(e) remain unchanged
(d) become small, but non-zero
Answer:
(b) becomes infinite
Hint:
\(\frac { 1 }{ { f }_{ 1 } } \) = \(\left(\frac{\mu_{g}}{\mu_{e}}-1\right)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) = (1 – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) = 0 ; f1 = ∞

Question 73.
A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power of the combination (in diopter) is
(a) zero
(b) 25
(c) 50
(d) infinite
Answer:
(a) zero
Hint:
\(\frac { 1 }{ F }\) = \(\frac { 1 }{ { f }_{ 1 } } \) + \(\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { 1 }{ +25 }\) + \(\frac { 1 }{ -25 }\) = 0 ; P = \(\frac { 1 }{ F }\) = 0

Question 74.
The focal length of a converging lens is measured for violet, green and red colours. If is fV, fG and fR respectively. We will get
(a) fV = fG
(b) fG = fR
(c) fV < fR
(d) fV > fR
Answer:
(c) fV < fR
Hint:
\(\frac { 1 }{ f }\) = (μ – 1) \(\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\) i.e., f ∝ \(\frac { 1 }{ μ – 1 }\)
As μV > μR, so, fV < fR

Question 75.
Rainbow is formed due to combination of
(a) refraction and scattering
(b) refraction and absorption
(c) dispersion and total internal reflection
(d dispersion and focusing
Answer:
(c) dispersion and total internal reflection
Hint:
Rainbow is formed due to dispersion and total internal reflection of sunlight by raindrops.

Question 76.
The blue colour of the sky is due to the phenomenon of
(a) scattering
(b) dispersion
(c) reflection
(d) refraction
Answer:
(a) scattering
Hint:
The blue colour of the sky is due to the scattering of sunlight by atmospheric molecules.

Question 77.
An astronomical telescope often fold angular magnification has a length of 44 cm. The focal length of the object is
(a) 4 cm
(b) 40 cm
(c) 44 cm
(d) 440 cm
Answer:
(b) 40 cm
Hint:
Here, f0 + fe = 44 cm
m =\(\frac { { f }_{ 0 } }{ { f }_{ e } } \) (or) f0 = 10 fe
∴ 10 fe + 10 fe = 44 cm (or) fe = 4 cm
Hence, f0 = 10 x 4 (or) f0 = 40 cm

Question 78.
Exposure time of a camera lens at the \(\frac { f }{ 2.8 }\) Setting is \(\frac { 1 }{ 200 }\) second. The correct time of exposure at \(\frac { f }{ 5.6 }\) is
(a) 0.20 second
(b) 0.40 second
(c) 0.02 second
(d) 0.04 second
Answer:
(c) 0.02 second
Hint:
Time of exposure, t ∝ (f- number)2
∴ \(\frac{t}{\left(\frac{1}{200}\right)}=\left(\frac{5.6}{2.8}\right)^{2}\) = 4 ⇒ t = 0.02s

Question 79.
Which of the following is not due to total internal reflection?
(a) Working of optical fibre
(b) Difference between apparent and real depth of a pond
(c) Mirage on hot summer day
(d) Brilliance of diamond
Answer:
(b) Difference between apparent and real depth of a pond
Hint:
Difference between apparent and real depth of a pond is due to refraction of light and the other three phenomena involve total internal reflection.

Question 80.
An object is placed at a distance of 0.5 m infront of a plane mirror. The distance between object and image will be
(a) 0.25 m
(b) 0.5 m
(c) 1.0 m
(d) 2.0 m
Answer:
(c) 1.0 m
Hint:
Distance between object and image = 0.5 + 0.5 = 1.0 m.

Question 81.
An observer moves towards a stationary plane mirror at a speed of 4 ms-1 with what speed – will his image move towards him?
(a) 2 ms-1
(b) 4 ms-1
(c) 8 ms-1
(d) the image will stay at rest
Answer:
(c) 8 ms-1
Hint:
Speed of the image towards the observer = 2 x 4 = 8 ms-1

Question 82.
If two mirrors are kept at 60° to each other and a body is placed at the middle, then total number of images formed is
(a) six
(b) four
(c) five
(d) three
Answer:
(c) five
Hint:
Number of images formed = \(\frac { 360 }{ θ }\) -1 = \(\frac { 360 }{ 60 }\) -1 = 5

Question 83.
If an object is placed at 10 cm infront of a concave mirror of focal length 15 cm. The magnification of image is
(a) -1.5
(b) 1.5
(c) -3
(d) 3
Answer:
d) 3
Hint:
m = \(\frac { f }{ f-u }\) = \(\frac { -15 }{ -15(-10) }\) = +3

Question 84.
An object of length 2.5 cm is placed at the principal axis of a concave mirror at a distance 1.5f. The image height is
(a) + 5 m
(b) -5 cm
(c) – 10 cm
(d) + 1 cm
Answer:
(b) -5 cm
Hint:
m = \(\frac { f }{ f-u }\) = \(\frac { f }{ f-1.5f}\) = -2cm
Height image = m x height of object -2 x 2.5 = – 5 cm

Question 85.
Which of the following mirror is used by a dentist to examine a small cavity?
(a) Concave mirror
(b) Convex mirror
(c) Combination of (a) and (b)
(d) None of these
Answer:
(a) Concave mirror
Hint:
A concave mirror, because it forms erect and enlarged image when held closer to the cavity.

Question 86.
When a ray of light enters from one medium to another, then which of the following does not change?
(a) Frequency
(b) Wavelength
(c) Speed
(d) Amplitude
Answer:
(a) Frequency
Hint:
Only frequency remains unchanged.

Question 87.
When light travels from one medium to the other medium of which the refractive index is different, then which of the following will change?
(a) Frequency, wavelength and velocity
(6) Frequency and wavelength
(c) Frequency and velocity
(d) Wavelength and velocity
Answer:
(d) Wavelength and velocity
Hint:
Wavelength and velocity will change while frequency remains unchanged.

Question 88.
The time taken by the light to cross a glass of thickness 4 mm and refractive index (μ = 3), will be
(a) 4 x 10-11 sec
(b) 16 x 10-11 sec
(c) 8 x 10-11 sec
(d) 24 x 10-10 sec
Answer:
(a) 4 x 10-11 sec
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 6 Optics-50
\(\frac { d }{ c/a }\) = \(\frac { μd }{ c }\) = \(\frac { 3×4×10^{-3} }{ 3×10^{8} }\) =  4 × 10-11 s

Question 89.
The critical angle of a medium with respect to air is 45°. The refractive index of medium is
(a) 1.41
(b) 1.2
(c) 1.5
(d) 2
Answer:
(a) 1.41
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { 1 }{ sin 45° }\) = \(\frac { 1 }{ 1/√2 }\) ≈ 1.41

Question 90.
If the critical angle for total internal reflection from a medium to vacuum is 30°, then velocity of light in the medium is
(a) 6 x 108 m/sec
(b) 2 x 108 m/sec
(c) 3 x 108 m/sec
(d) 1.5 x 108 m/sec
Answer:
(c) 3 x 108 m/sec
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { c }{ v }\)

Question 91.
When a ray of light enter from one medium to another, its velocity is doubled. The critical angle for the ray for two internal reflection will be
(a) 30°
(b) 60°
(c) 90°
(d) Information is incomplete
Answer:
(a) 30°
Hint:
μ = \(\frac{1}{\sin i_{c}}\) = \(\frac { { v }_{ 1 } }{ { v }_{ 2 } } \) =\(\frac { { 2v }_{ 2 } }{ { v }_{ 2 } } \) = 2 ; sin ic = \(\frac { 1 }{ 2 }\) ∴ic = 30°

Question 92.
A driver at a depth 12 m inside water (μ = 4/3) see the sky in a cone of semi-vertical angle is
(a) sin-1\(\left( \frac { 4 }{ 3 } \right) \)
(b) tan-1\(\left( \frac { 4 }{ 3 } \right) \)
(c) sin-1\(\left( \frac { 3 }{ 4 } \right) \)
(d) 90°
Answer:
(c) sin-1\(\left( \frac { 3 }{ 4 } \right) \)
Hint:
Required semi vertical angle = Critical angle ic = sin-1 \(\frac { 1 }{ μ }\) = sin-1\(\left( \frac { 3 }{ 4 } \right) \)

Question 93.
The principle behind optical fibres is
(a) total internal reflection
(b) total external reflection
(c) both (a) and (b)
(d) diffraction
Answer:
(a) total internal reflection
Hint:
Optical fibres work on the principle of total internal reflection.

Question 94.
Air bubble in water behaves as
(а) some times concave, sometimes convex lens
(b) concave lens
(c) convex lens
(d) always refracting surface
Answer:
(b) concave lens

Question 95.
A convex lens of 40 cm focal length is combined with a concave lens of focal length 25 cm. The power of combination is
(a) -1.5 D
(b) – 6.5 D
(c) + 6.6 D
(d) + 6.5 D
Answer:
(a) -1.5 D
Hint:
P = P1 + P2 = \(\frac { 1 }{ { f }_{ 1 } } \) + \(\frac { 1 }{ { f }_{ 2 } } \) = \(\frac { 100 }{ 40 }\) + \(\frac { 100 }{ -25 }\) = -1.5d.

Question 96.
Two thin lenses, one of focal length + 60 cm and the other of focal length – 20 cm are put in contact, the combined focal length is,
(a) 15 cm
(b) – 15 cm
(c) – 30 cm
(d) 30 cm
Answer:
(c) – 30 cm
Hint:
F = \(\frac { { f }_{ 1 }{ f }_{ 2 } }{ { f }_{ 1 }{ +f }_{ 2 } } \) = \(\frac { 60 ×(-20) }{ 60-20 }\) = 30 cm

Question 97.
How does refractive index (μ) of a material vary with respect to wavelength (λ). (A and B are constants).
(a) μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)
(b) μ = A + Bλ2
(c) μ = A + \(\frac { B }{ λ }\)
(d) μ = A + Bλ
Answer:
(a) μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)
Hint:
According to cauchy’s relation, μ = A + \(\frac { B }{ { \lambda }^{ 2 } } \)

Question 98.
a prism of a refracting angle 60° is made with a material of refractive index p. For a certain wavelength of light, the angle of minimum deviation is 30°. For this wavelength, the value of p of material is
(a) 1.820
(b) 1.414
(c) 1.503
(d) 1.231
Answer:
(b) 1.414
Hint:
μ = \(\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin 30^{\circ}}\) = \(\frac { sin 45° }{ sin 30° }\) = \(\frac { 1 }{ √2 }\) x \(\frac { 2 }{ 1 }\) √2 = 1.414

Question 99.
Refractive index of red and violet light are 1.52 and 1.54 respectively. If the angle of prism is 10°, the angular dispersion will be
(a) 0.02°
(b) 0.20°
(c) 3.06°
(d) 30.6°
Answer:
(b) 0.20°
Hint:
Angular dispersion = δV – δR = A(μV – μV) = 10° (1.54 – 1.52) = 0.20°

Question 100.
In a simple microscope, if the final image is located at 25 cm from the eye placed close to the lens, then magnifying power is
(a) \(\frac { 25 }{ f }\)
(b) 1 + \(\frac { 25 }{ f }\)
(c) \(\frac { f }{ 25 }\)
(d) \(\frac { f }{ 25 }\) + 1
Answer:
(b) 1 + \(\frac { 25 }{ f }\)
Hint:
When the final image is formed at the least distance of distinct vision in a simple microscope,
m = 1 + \(\frac { 25 }{ f }\)

Question 101.
Magnification at least distance of distinct vision of a simple microscope of focal length 5 cm is
(a) 2
(b) 5
(c) 4
(d) 6
Answer:
(d) 6
Hint:
m = 1 + \(\frac { 25 }{ f }\) = 1 + \(\frac { 25 }{ 5 }\) = 6

Question 102.
Magnification of a compound microscope is 30. Focal length of eyepiece is 5 cm and the image is formed at a distance of distinct vision of 25 cm. The magnification of the objective
lens is
(a) 6
(b) 5
(c) 7.5
(d) 10
Answer:
(b) 5
Hint:
me = \(\frac { v }{ u}\) = \(\frac { D }{ { u }_{ c } } \) = 1 + \(\frac { D }{ { f }_{ e } } \) = 1 + \(\frac { 25 }{ 5 }\) = 6
For the compound microscope, m = m0 x me ⇒ 30 = m0 x 6 (or) m0 = 5

Question 103.
The astronomical microscope consists of objective and eyepiece. The focal length of the objective is
(a) equal to that of the eyepiece
(b) shorter than that of the eyepiece
(c) greater than that of the eyepiece
(d) five times shorter than that of eyepiece
Answer:
(c) greater than that of the eyepiece
Hint:
For producing large magnification,f0 > fe

Question 104.
The number of lenses in terrestrial telescope is
(a) 2
(b) 4
(c) 3
(d) 6
Answer:
(c) 3
Hint:
A terrestrial telescope consists of three lenses: objective, erecting lens and eyepiece.

Question 105.
An achromatic combination of lenses is formed by joining
(a) 2 convex lens
(b) 1 convex, 1 concave lens
(c) 2 concave lenses
(d) 1 convex and 1 plane mirror
Answer:
(b) 1 convex, 1 concave lens
Hint:
An achromatic doublet should satisfy the condition \(\frac { { w }_{ 1 } }{ { f }_{ 1 } } \) + \(\frac { { w }_{ 2 } }{ { f }_{ 2 } } \) = 0.

Question 106.
The amount of light received by a camera depends upon
(a) diameter only
(b) ratio of focal length and diameter
(c) product of focal length and diameter
(d) only one of the focal length
Answer:
(b) ratio of focal length and diameter
Hint:
The amount of light received by a camera depends on the ratio of the focal length and diameter of the aperture.

Question 107.
Myopia is corrected by using a
(a) cylindrical lens
(b) bifocal lens
(c) convex lens
(d) concave lens
Answer:
(d) concave lens

Question 108.
The critical angle for total internal reflection in diamond is 24.5°. The angle refractive index of diamond is
(a) 2.41
(b) 1.41
(c) 2.59
(d) 1.59
Answer:
(a) 2.41
Hint:
μ = \(\frac { 1 }{ { sin i }_{ c } } \) = \(\frac { 1 }{ sin 24.5° }\) = 2.41

Question 109.
When a glass lens with μ = 1.47 is immersed in a trough of liquid, it looks to be disappeared. The liquid in the trough could be
(a) water
(b) kerosene
(c) glycerine
(d) alcohol
Answer:
(c) glycerine
Hint:
Glass lens will disappear, if µl = µg.

Question 110.
In optical fibres, the refractive index of the core is
(a) greater than that of the cladding
(b) equal to that of the cladding
(c) smaller than that of the cladding
(d) independent of that of the cladding
Answer:
(a) greater than that of the cladding
Hint:
In optical fibres, refractive index of core material > refractive index of the cladding.

Question 111.
For a wavelength of light ‘λ’ and scattering object of size ‘a’, all wavelength are scattered nearly equally, if
(a) a = λ
(b) a >> λ
(c) a << λ
(d) a ≥ λ
Answer: (b) a >> λ
Hint:
For a >> λ, the scattering power is not selective.

Question 112.
Two coherent monochromatic light beams of intensities I and 4I are supperposed. The maximum and minimum possible intensities in the resulting beams are
(a) 5I and I
(b) 9I and I
(c) 5I and 3I
(d) 9I and 3I
Answer:
(b) 9I and I
Hint:
Imax = \((\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}})^{2}\) = \((\sqrt{41}+\sqrt{1})^{2}\) = 9I
Imax = \((\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}})^{2}\) = \((\sqrt{41}-\sqrt{1})^{2}\) = I

Question 113.
screen is doubled. The fringe width is
(a) unchanged
(b) halved
(c) doubled
(d) quadrupled
Answer:
(d) quadrupled
Hint:
β = \(\frac { λD }{ d }\) ; β = \(\frac { λ ×2D }{ D/2 }\) = 4β

Question 114.
In a young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen, when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by
(a) 12
(b) 18
(c) 24
(d) 30
Answer:
(b) 18
Hint:
n1λ1= n2λ2 ⇒12 x 600 = n2 x 400 or n2 = 18

Question 115.
Consider ffaunhoffer diffraction pattern obtained with a single slit illuminated at normal incident. At the angular position of the first diffraction minimum the phase difference between the wavelets from the opposite edges of the slits is
(a) \(\frac { π }{ 4 }\)
(b) \(\frac { π }{ 2 }\)
(c) 2π
(d) π
Answer:
(c) 2π
Hint:
At the angular position of first minimum wavelets from opposite edges of the slits have a path difference of X and a phase difference of 2π radian.

Question 116.
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is
(a) 1.2 cm
(b) 1.2 mm
(c) 2.4 cm
(d) 2.4 mm
Answer:
(c) 2.4 cm
Hint:
Distance between the first dark fringes on either side = width of central maximum
= \(\frac { 2Dλ }{ d }\) = \(\frac{2 \times 2 \times 600 \times 10^{-9}}{1.00 \times 10^{-3}}\) m = 2.4 x 10-3 m = 2.4mm

Question 117.
A young’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is.
(a) hyperbola
(b) circle
(c) straight line
(d) parabola
Answer:
(a) hyperbola
Hint:
In young’s double slit experiment, the fringes obtained are hyperbolic in shape. But in a small interference pattern, the fringes appear straight.

Question 118.
The initial shape of the wavefront of the beam is
(a) planar
(b) convex
(c) concave
(d) convex near the axis and concave near the periphery
Answer:
(a) planar
Hint:
As the beam is initially parallel, the shape of wavefront is planar.

Question 119.
The angle of incident at which reflected light is totally polarised for reflection from air to glass (refractive index μ) is
(a) sin-1 (μ)
(b) sin-1 \(\left( \frac { 1 }{ μ } \right) \)
(c) tan-1 \(\left( \frac { 1 }{ μ } \right) \)
(d) tan-1 (μ)
Answer:
(d) tan-1 (μ)
According to Brewster’s law,
μ = tan ip ∴ ip = tan-1 (μ)

Question 120.
According to Huygen’s principle, light is a form of
(a) particle
(b) rays
(c) wave
(d) none of the above
Answer:
(c) wave
Hint:
According to Huygen’s principle, light travels in the form of a longitudinal wave.

Question 121.
Which one of the following phenomena is not explained by Huygen’s construction of wavefront?
(a) refraction
(b) reflection
(c) diffraction
(d) origin of spectra
Answer:
(d) origin of spectra
Hint:
Huygen’s construction of wavefront cannot explain origin of spectra which can be explained on the basis of quantum theory.

Samacheer Kalvi 12th Physics Optics Additional Problems

Question 1.
Light from a point source in air falls on a convex spherical glass surface (n = 1.5, radius of curvature = 20 cm). The distance of light source from the glass surface is 100 cm. At
What position is the image formed?
Solution:
n1 = 1; n2 = 1.5
u = 100cm ; R = + 20 cm
(R is + Ve for a convex refracting surfce)
As \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\)
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}\) = \(\frac{n_{2}-n_{1}}{R}\)
\(\frac{1.5}{v}+\frac{1}{100}\) = \(\frac{1.5-1}{20}=\frac{1}{40}\)
\(\frac { 3 }{ 2v }\) = \(\frac{1}{40}-\frac{1}{100}=\frac{5-2}{200}=\frac{3}{200}\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ 100 }\)
v = 100 cm
Thus the image is formed at a distance of 100 cm from the glass surface in the direction of incident light.

Question 2.
Find the value of critical angle for a material of refractive index √3 .
Solution:
Here, n = √3
sin ic = \(\frac { 1 }{ n }\) = \(\frac { 1 }{ √3 }\) = \(\frac { √3 }{ 3 }\)
∴ critical angle, ic = 35.3°

Question 3.
The radius of curvature of each face of biconcave lens, made of glass of refractive index 1.5 is 30 cm. Calculate the focal length of the lens in air.
Solution:
Here n = 1.5 ; R1 = – 30 cm ; R2 = 30 cm
Using len’s maker’s formula,
\(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= (1.5-1)\(\left[\frac{1}{-30}-\frac{1}{30}\right]\) = 0.5 × \(\left(\frac{-2}{30}\right)\)
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ 30 }\)
f = -30

Question 4.
The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length is 12 cm. What is the refractive index of glass?
Solution:
f = +12 cm ; R1 = 10 cm ; R2 = – 15 cm ; n = ?
As, \(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\frac { 1 }{ 12 }\) = (n – 1) \(\left(\frac{1}{10}+\frac{1}{15}\right)\) = (n – 1) x latex]\frac { 5 }{ 30 }[/latex]
(n – 1) = \(\frac { 6 }{ 12 }\) = 0.5
n = 0.5 +1
n = 1.5

Question 5.
A double convex lens made of glass of refractive index 1.5 has its both surfaces of equal radii of curvature of 20 cm each. An object of 5 cm height is placed at a distance of 10 cm from the lens. Find the position, nature and size of the image.
Solution:
Here n = 1.5 ; R1= +20 cm ; R2 = – 20 cm
Using lens maker’s formula,
\(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
(1.5-1) = \(\left[\frac{1}{20}-\frac{1}{-20}\right]\) 0.5 x \(\left(\frac{2}{20}\right)\)
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ 20 }\)
f = 20 cm
Now , u = -10 cm andf = + 20 cm
From thin lens formula, \(\frac { 1 }{ v }\) = \(\frac { 1 }{ f }\) + \(\frac { 1 }{ u }\) = \(\frac { 1 }{ 20 }\) – \(\frac { 1 }{ 10 }\) = \(\frac { 1 }{ 20 }\)
∴ v = -20 cm
Magnification, m = \(\frac{h_{2}}{h_{1}}\)= \(\frac { v }{ u }\)
\(\frac{h_{2}}{5cm}\) = \(\frac { -20 }{ -10 }\)
h2
Hence a virtual and erect image of height 10 cm is formed at a distance of 20 cm from the lens on the same side as the the object.

Question 6.
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances.
Solution.
Heref= 20 cm, m = + 4 for a virtual image.
To calculate u, we have
m = \(\frac { f }{ u+f }\)

Question 7.
The radius of curvature of each surface of a convex lens of refractive index 1.5 is 40 cm. Calculate its power.
Solution:
Here, n = 105 ; R1 = + 40cm = 0.40 m
R2 = – 40cm = – 0.40 m
Power (p) = \(\frac { 1 }{ f }\) = (n – 1) \(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
= (1.5 – 1)\(\left[\frac{1}{0.40}-\frac{1}{(-0.40)}\right]\) = 0.5 x \(\frac { 2 }{ 0.40 }\)
p = 2.5 D

Question 8.
A ray of light incident on an equilateral glass prism shows minimum deviation of 30°. Calculate the speed of light through the prism.
Solution:
Here, A = 60° ; D = 30°
Refractive index, n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\) = \(\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \left(\frac{60}{2}\right)}\)
n = \(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}\) = \(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\) = √2
n = 1.414
Velocity or light in glass, v = \(\frac { c }{ n }\) = \(\frac{3 \times 10^{8}}{1.414}\)
v = 2.12 x 108 ms-1

Question 9.
Two sources of intensity I and 41 are used in an interference experiment. Find the intensity at points where the waves from two sources superimpose with a phase
(i) Zero
(ii) \(\frac { π }{ 2 }\)
(iii) π
Solution:
The resultant intensity at a point where phase difference is Φ is
IR+I1+ I2 2\(+\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}\) cos Φ
As I1 + I and I2 = 4I, therefore,
IR = I + 4I + 2\(\sqrt{I.4I}\) cos Φ
IR = 5I + 4I cos Φ
(i) When Φ = 0 ; IR = 5I + 4I cos 0 = 9I
(ii) When Φ = \(\frac { π }{ 2 }\) ; IR = 5I + 4I cos \(\frac { π }{ 2 }\) = 5I
(iii) When Φ = π ; IR = 5I + 4I cos π = 51 – 41 = I
Φ = 0 ; IR = 9I
Φ = \(\frac { π }{ 2 }\) ; IR = 5I
Φ = π ; IR = I

Question 10.
Assume that light of wavelength 600 A is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Solution:.
The limit of resolution of a telescope, dθ = \(\frac { 1.22λ }{ D }\)
D = 100 inch = 254 cm
[∴ 1 inch = 2.54 cm]
λ = 6000 Å = 6000 x 10-10 m/s
dθ = \(\frac{1.22 \times 6000 \times 10^{-10}}{254 \times 10^{-2}}\) = 2.9 x 10-7
dθ = 2.9 x 10-7 rad.

Question 11.
Two polarising sheet have their polarising directions parallel so that the intensity of the transmitted light is maximum. Through what angle must the either sheet be turned if the intensity is to drop by one-half?
Solution:
Here, I = \(\frac{I_{o}}{2}\)
Using Malus law,,
I = Io cos2 θ
\(\frac{I_{o}}{2}\) = Io cos2 θ
cos θ ± \(\frac { 1 }{ √2 }\)
θ = ± 45°, ± 135°

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Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner

Students can Download Accountancy Chapter 6 Retirement and Death of a Partner Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner

Samacheer Kalvi 12th Accountancy Retirement and Death of a Partner Text Book Back Questions and Answers

I. Choose the Correct Answer

12th Accountancy 6th Chapter Solutions Question 1.
A partner retires from the partnership firm on 30th June. He is liable for all the acts of the firm up to the …………………..
(a) End of the current accounting period
(b) End of the previous accounting period
(c) Date of his retirement
(d) Date of his final settlement
Answer:
(c) Date of his retirement

12th Accountancy Chapter 6 Question 2.
On the retirement of a partner from a partnership firm, accumulated profits and losses are distributed to the partners on the basis of …………………
(a) New profit sharing ratio
(b) Old profit sharing ratio
(c) Gaining ratio
(d) Sacrificing ratio
Answer:
(b) Old profit sharing ratio

12th Accountancy Chapter 6 Solutions Question 3.
On the retirement of a partner, general reserve will be transferred to the …………………..
(a) Capital account of all the partners
(b) Revaluation account
(c) Capital account of the continuing partners
(d) Memorandum revaluation account
Answer:
(a) Capital account of all the partners

Samacheer Kalvi 12th Accountancy Guide Question 4.
On revaluation, the increase in liabilities leads to
(a) Gain
(b) Loss
(c) Profit
(d) None of these
Answer:
(b) Loss

Class 12 Accountancy Chapter 6 Solutions Question 5.
At the time of retirement of a partner, determination of gaining ratio is required …………………..
(a) To transfer revaluation profit or loss
(b) To distribute accumulated profits and losses
(c) To adjust goodwill
(d) None of these
Answer:
(c) To adjust goodwill

Class 12 Accountancy Chapter 6 Retirement Solutions Question 6.
The final amount due to a retiring partner is not paid immediately, it is transferred to …………………..
(a) Bank A/c
(b) Retiring partner’s capital A/c
(c) Retiring partner’s loan A/c
(d) Other partners’ capital A/c
Answer:
(c) Retiring partner’s loan A/c

12 Accountancy Book Samacheer Kalvi Question 7.
‘A’ was a partner in a partnership firm. He died on 31st March 2019. The final amount due to him is ₹ 25,000 which is not paid immediately. It will be transferred to …………………..
(a) A’s capital account
(b) A’s loan account
(c) A’s Executor’s account
(d) A’s Executor’s loan account
Answer:
(d) A’s Executor’s loan account

Samacheer Kalvi 12th Accountancy Solutions Question 8.
A, B and C are partners sharing profits in the ratio of 2:2:1. On retirement of B, goodwill of the firm was valued as ₹ 30,000. Find the contribution of A and C to compensate B:
(a) ₹ 20,000 and ₹ 10,000
(b) ₹ 8,000 and ₹ 4,000
(c) ₹ 10,000 and ₹ 20,000
(d) ₹ 15,000 and ₹ 15,000
Answer:
(b) ₹ 8,000 and ₹ 4,000

Samacheer Kalvi 12th Accountancy Book Question 9.
A, B and C are partners sharing profits in the ratio of 4:2:3. C retires. The new profit sharing ratio between A and B will be ………………….
(a) 4:3
(b) 3:4
(c) 2:1
(d) 1:2
Answer:
(c) 2:1

12th Accountancy Samacheer Kalvi Question 10.
X, Y and Z were partners sharing profits and losses equally. X died on 1st April 2019. Find out the share of X in the profit of 2019 based on the profit of 2018 which showed ₹ 36,000.
(a) ₹ 1,000
(b) ₹ 3,000
(c) ₹ 12,000
(d) ₹ 36,000
Answer:
(b) ₹ 3,000

II. Very Short Answer Questions

12th Accountancy Solutions Samacheer Kalvi Question 1.
What is meant by retirement of a partner?
Answer:
When a partner leaves from partnership firm it is known as retirement. The reasons for the retirement of a partner may be illness, old age and disagreement with other partners, etc.

Samacheer Kalvi 12 Accountancy Guide Question 2.
What is gaining ratio?
Answer:
Gaining ratio is the proportion of the profit which is gained by the continuing partner. Gaining ratio = Ratio of share gained by the continuing partners.
Share gained = New share – Old share

Samacheer Kalvi Guru 12th Accountancy Question 3.
What is the purpose of calculating gaining ratio?
Answer:
The purpose of finding the gaining ratio is to bear the goodwill to be paid to the retiring partner.

Samacheer Kalvi Accountancy 12th Question 4.
What is the journal entry to be passed to transfer the amount due to the deceased partner to the executor of the deceased partner?
Answer:
12th Accountancy 6th Chapter Solutions Retirement And Death Of A Partner Samacheer Kalvi

III. Short Answer Questions

Question 1.
List out the adjustments made at the time of retirement.
Answer:
(a) Distribution of accumulated profits, reserves and losses
(b) Revaluation of assets and liabilities
(c) Determination of new profit sharing ratio and gaining ratio
(d) Adjustment for goodwill
(e) Adjustment for current year’s profit or loss up to the date of retirement
(f) Settlement of the amount due to the retiring partner

Question 2.
Distinguish between sacrificing ratio and gaining ratio.
Answer:
12th Accountancy Chapter 6 Retirement And Death Of A Partner Samacheer Kalvi

Samacheer Kalvi 12th Accounts Guide Question 3.
What are the ways in which the final amount due to an outgoing partner can be settled?
Answer:
The amount due to the retiring partner may be settled in one of the following ways:

  • Paying the entire amount due immediately in cash
  • Transfer the entire amount due to the loan account of the partner
  • Paying part of the amount immediately in cash and transferring the balance to the loan account of the partner.

IV. Exercises

Samacheer Kalvi 12th Accountancy Question 1.
Dheena, Surya and Janaki are partners sharing profits and losses in the ratio of 5:3:2. On 31.3.2018, Dheena retired. On the date of retirement, the books of the firm showed a reserve fund of ₹ 50,000. Pass journal entry to transfer the reserve fund.
Answer:
Journal Entries
12th Accountancy Chapter 6 Solutions Retirement And Death Of A Partner Samacheer Kalvi

12th Samacheer Kalvi Accountancy Solution Book Question 2.
Rosi, Rathi and Rani are partners of a firm sharing profits and losses equally. Rathi retired from the partnership on 1.1.2018. On that date, their balance sheet showed accumulated loss of ? 45,000 on the asset side of the balance sheet. Give the journal entry to distribute the accumulated loss.
Answer:
Samacheer Kalvi 12th Accountancy Guide Chapter 6 Retirement And Death Of A Partner

Samacheer Kalvi 12 Accountancy Question 3.
Akash, Mugesh and Sanjay are partners in a firm sharing profits and losses in the ratio of 3:2:1. Their balance sheet as on 31st March, 2017 is as follows:
Class 12 Accountancy Chapter 6 Solutions Retirement And Death Of A Partner Samacheer KalviAnswer:
Pass journal entry to transfer accumulated Profit and prepare the capital account of the partners.
Class 12 Accountancy Chapter 6 Retirement Solutions Samacheer Kalvi
Capital Account
12 Accountancy Book Samacheer Kalvi Solutions Chapter 6 Retirement And Death Of A Partner

12th Accountancy Answer Pdf Question 4.
Roja, Neela and Kanaga are partners sharing profits and losses in the ratio of 4:3:3. On 1st April 2017, Roja retires and on retirement, the following adjustments are agreed upon:
(i) Increase the value of building by ₹ 30,000.
(ii) Depreciate stock by ₹ 5,000 and furniture by ₹ 12,000.
(iii) Provide an outstanding liability of ₹ 1,000 Pass journal entries and prepare revaluation account.
Answer:
Revaluation Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement And Death Of A PartnerJournal Entries
Samacheer Kalvi 12th Accountancy Book Solutions Chapter 6 Retirement And Death Of A Partner

Chapter 6 Retirement Of A Partner Solutions Question 5.
Vinoth, Karthi and Pranav are partners sharing profits and losses in the ratio of 2:2:1. Pranav retires from partnership on 1st April 2018. The following adjustments are to be made:
(i) Increase the value of land and building by ₹ 18,000
(ii) Reduce the value of machinery by ₹ 15,000
(iii) A provision would also be made for outstanding expenses for ₹ 8,000.
Give journal entries and prepare revaluation account.
Answer:
Revaluation Account
12th Accountancy Samacheer Kalvi Solutions Chapter 6 Retirement And Death Of A Partner

Question 6.
Chandru, Vishal and Ramanan are partners in a firm sharing profits and losses equally. Their balance sheet as on 31st March, 2018 is as follows:
Ramanan retired on 31 st March 2019 subject to the following conditions:
12th Accountancy Solutions Samacheer Kalvi Chapter 6 Retirement And Death Of A Partner
(i) Machinery is valued at ₹ 1,50,000
(ii) Value of furniture brought down by ₹ 10,000
(iii) Provision for doubtful debts should be increased to ₹ 5,000
(iv) Investment of ₹ 30,000 not recorded in the books is to be recorded now.
Pass necessary journal entries and prepare revaluation account and capital account of partners
Answer:
Revaluation Account
Samacheer Kalvi 12 Accountancy Guide Solutions Chapter 6 Retirement And Death Of A Partner
Capital Account
Samacheer Kalvi Guru 12th Accountancy Solutions Chapter 6 Retirement And Death Of A Partner

Question 7.
Kayal, Mala and Neela are partners sharing profits in the ratio of 2:2:1. Kayal retires and the new profit sharing ratio between Nila and Neela is 3:2. Calculate the gaining ratio.
Answer:
New Profit Sharing Ratio and Gaining Ratio
Gain Ratio = New Ratio – Old Ratio
Kayal –
Mala = \(\frac { 3 }{ 2 }\) – \(\frac { 2 }{ 5 }\) – \(\frac { 1 }{ 5 }\)
Neela = \(\frac { 2 }{ 5 }\) – \(\frac { 1 }{ 5}\) = \(\frac { 1 }{ 5 }\)
Gaining Ratio = 1:1

Question 8.
Sunil, Sumathi and Sundari are partners sharing profits in the ratio of 3:3:4. Sundari retires and her share is taken up entirely by Sunil. Calculate the new profit sharing ratio and gaining ratio.
Answer:
New Ratio – Old Ratio
Sunil = \(\frac { 3 }{ 10 }\) + \(\frac { 4 }{ 10 }\) = \(\frac { 7 }{ 10 }\)
(Sundari share is added with old ratio)
Sumathi = \(\frac { 3 }{ 10 }\)
New Ratio = 7 : 3
Sacrificing Ratio = 1:1

Question 9.
Ramu, Somu and Gopu are partners sharing profits in the ratio of 3:5:7. Gopu retires and the share is purchased by Ramu and Somu in the ratio of 3:1. Find the new profit sharing ratio and gaining ratio.
Answer:
Samacheer Kalvi Accountancy 12th Solutions Chapter 6 Retirement And Death Of A Partner

Question 10.
Navin, Ravi and Kumar are partners sharing profits in the ratio of 1/2,1/4 and 1/4 respectively, Kumar retires and his share is taken up by Navin and Ravi equally. Calculate the new profit sharing ratio and gaining ratio.
Answer:
Samacheer Kalvi 12th Accounts Guide Solutions Chapter 6 Retirement And Death Of A Partner

Question 11.
Mani, Gani and Soni are partners sharing the profits and losses in the ratio of 4:5:6. Mani retires from the firm. Calculate the new profit sharing ratio and gaining ratio.
Answer:
Since new profit sharing ratio, share gained and the proportion of share gained is not given, the new share is calculated by assuming that the share gained is in the proportion of old ratio.

Question 12.
Rajan, Suman and Jegan were partners in a firm sharing profits and losses in the ratio of 4:3:2. Suman retired from partnership. The goodwill of the firm on the date of retirement was valued at ₹ 45,000. Pass necessary journal entries for goodwill on the assumption that the fluctuating capital method is followed.
Answer:
Value of Goodwill = 45,000 x \(\frac { 3 }{ 9 }\) = 15,000
Journal Entries
Adjustment for goodwill
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement And Death Of A Partner

Question 13.
Balu, Chandru and Nirmal are partners in a firm sharing profits and losses in the ratio of 5:3:2. On 31st March 2018, Nirmal retires from the firm. On the date of Nirmal’s retirement, goodwill appeared in the books of the firm at ? 60,000. By assuming fluctuating capital account, pass the necessary journal entry if the partners decide to
(a) write off the entire amount of existing goodwill
(b) write off half of the existing goodwill.
Answer:
12th Samacheer Kalvi Accountancy Solution Book Chapter 6 Retirement And Death Of A Partner

Question 14.
Rani, Jaya and Rathi are partners sharing profits and losses in the ratio of 2:2:1. On 31.3.2018, Rathi retired from the partnership. Profit of the preceding years is as follows: 2014: 10,000; 2015: ₹ 20,000; 2016: ₹ 18,000 and 2017: ₹ 32,000
Find out the share of profit of Rathi for the year 2018 till the date of retirement if
(a) Profit is to be distributed on the basis of the previous year’s profit
(b) Profit is to be distributed on the basis of the average profit of the past 4 years Also pass necessary journal entries by assuming partners capitals are fluctuating.
Answer:
(a) If the profit is to be distributed on the basis of previous year profit (2017) Rathi’s share distributed 3 months = ₹ 32,000 x \(\frac { 1 }{ 5 }\) x \(\frac { 3 }{ 12 }\) = ₹ 1600
Samacheer Kalvi 12 Accountancy Solutions Chapter 6 Retirement And Death Of A Partner
(b) Average Profit
12th Accountancy Answer Pdf Solutions Chapter 6 Retirement And Death Of A Partner Samacheer Kalvi

Question 15.
Kavin, Madhan and Ranjith are partners sharing profits and losses in the ratio of 4:3:3, respectively. Kavin retires from the firm on 31st December, 2018. On the date of retirement, his capital account shows a credit balance of ₹ 1,50,000. Pass journal entries if:
(a) The amount due is paid off immediately.
(b) The amount due is not paid immediately.
(c) ₹ 1,00,000 is paid and the balance in future.
Answer:
Chapter 6 Retirement Of A Partner Solutions Samacheer Kalvi 12th Accountancy

Question 16.
Manju, Charu and Lavanya are partners in a firm sharing profits and losses in the ratio of 5:3:2. Their balance sheet as on 31st March, 2018 is as follows:
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 22
Manju retired from the partnership firm on 31.03.2018 subject to the following adjustments:
(i) Stock to be depreciated by ₹ 10,000
(ii) Provision for doubtful debts to be created for ₹ 3,000.
(iii) Buildings to be appreciated by ₹ 28,000
Prepare revaluation account and capital accounts of partners after retirement.
Answer:
Revaluation
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 23
Capital Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 24

Question 17.
Kannan, Rahim and John are partners in a firm sharing profit and losses in the ratio of 5:3:2. The balance sheet as on 31st December, 2017 was as follows:
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 25
John retires on 1st January 2018, subject to the following conditions:
(i) To appreciate building by 10%
(ii) Stock to be depreciated by 5%
(iii) To provide ₹ 1,000 for bad debts
(iv) An unrecorded liability of ₹ 8,000 have been noticed
(v) The retiring partner shall be paid immediately
Prepare revaluation account, partners’ capital account and the balance sheet of the firm after retirement.
Answer:
Revaluation A/c
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 26
Capital
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 27
Balance Sheet
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 28

Question 18.
Saran, Arun and Karan are partners in firm sharing profits and losses in the ratio of 4:3:3. Their balance sheet as of 31.12.2016 was as follows:
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 29
Karan retires on 1.1.2017 subject to the following conditions:
(i) Goodwill of the firm is valued at ₹ 21,000
(ii) Machinery to be appreciated by 10%
(iii) Building to be valued at ₹ 80,000
(iv) Provision for bad debts to be raised to ₹ 2,000
(v) Stock to be depreciated by ₹ 2,000
(vi) Final amount due to Karan is not paid immediately
Prepare the necessary ledger accounts and show the balance sheet of the firm after retirement.
Answer:
Revaluation Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 30

Capital Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 31
Balance Sheet
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 32

Question 19.
Rajesh, Sathish and Mathan are partners sharing profits and losses in the ratio of 3:2:1 respectively. Their balance sheet as on 31.3.2017 is given below.
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 33
Mathan retires on 31st March, 2017 subject to the following conditions:
(i) Rajesh and Sathish will share profits and losses in the ratio of 3:2
(ii) Assets are to be revalued as follows: Machinery ₹ 4,50,000, Stock ₹ 2,90,000, Debtors ₹ 1,52,000.
(iii) Goodwill of the firm is valued at ₹ 1,20,000
Prepare necessary ledger accounts and the balance sheet immediately after the retirement of Mathan.
Answer:
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 52
Revaluation Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 34

Capital Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 35
Balance Sheet as on 31.12.2017
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 36

Question 20.
Janani, Janaki and Jamuna are partners sharing profits and losses in the ratio of 3:3:1 respectively. Janaki died on 31st December, 2017. Final amount due to her showed a credit balance of ₹ 1,40,000. Pass journal entries if,
(a) The amount due is paid off immediately.
(b) The amount due is not paid immediately.
(c) ₹ 75,000 is paid and the balance in future.
Answer:
Journal Entries
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 45

Question 21.
Varsha, Shanthi and Madhuri are partners, sharing profits in the ratio of 5:4:3. Their balance sheet as on 31st December 2017 is as under:
Balance Sheet as on 31st December 2017
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 37
On 1.1.2018, Madhuri died and on her death the following arrangements are made:
(i) Stock to be depreciated by ₹ 5,000
(ii) Premises is to be appreciated by 20%
(iii) To provide ₹ 4,000 for bad debts
(iv) The final amount due to Madhuri was not paid
Prepare revaluation account, partners’ capital account and the balance sheet of the firm after death.
Answer:
Revaluation Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 38
Capital Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 39
Balance Sheet as on 1.1.2018
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 40

Question 22.
Vijayan, Sudhan and Suman are partners who share profits and losses in their capital ratio. Their balance sheet as on 31.12.2018 is as follows:
Balance Sheet as on 31.12.2018
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 41
Suman died on 31.3.2019. On the death of Suman, the following adjustments are made:
(i) Building is to be valued at ₹ 1,00,000
(ii) Stock to be depreciated by ₹ 5,000
(iii) Goodwill of the firm is valued at ₹ 36,000
(iv) Share of profit from the closing of the last financial year to the date of death on the basis of the average of the three completed years’ profit before death.
Profit for 2016, 2017 and 2018 were ₹ 40,000, ₹ 50,000 and ₹ 30,000, respectively.
Prepare the necessary ledger accounts and the balance sheet immediately after the death of Suman.
Answer:
Profit-Sharing Ratio:
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 53
Revaluation Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 42
Capital Account
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 43
Balance Sheet as on 31.3.19
Samacheer Kalvi 12th Accountancy Solutions Chapter 6 Retirement and Death of a Partner 44

Samacheer Kalvi 12th Accountancy Retirement and Death of a Partner Additional Questions and Answers

I. Choose the correct answer

Question 1.
On the retirement of a partner, profit on revaluation of assets and liabilities should be credited to the capital accounts of …………………
(a) Retiring partner in their old ratio
(b) All partners in their old ratio
(c) Remaining partners in new ratio
(d) Remaining partners in old ratio
Answer:
(b) All partners in their old ratio

Question 2.
On the retirement of a partner, reserves should be transferred to the capital accounts of …………………
(a) Retiring partner
(b) Remaining partner
(c) All partners
(d) None of these
Answer:
(c) All partners

Question 3.
Credit balance of Profit and Loss Account – appearing in the Balance sheet on the dealth of a partner is credited to …………………
(a) Deceased Partner’s Capital Account
(b) All partner’s capital account (including deceased partner’s capital account
(c) Remaining partner’s capital account
(d) None of the above
Answer:
(b) All partner’s capital account (including deceased partner’s capital account)

Question 4.
P, Q and R are partners sharing profits in the ratio of 4:3:1. P retires and his share is taken by Q and R equally. Calculate new profit sharing ratio of Q and R …………………
(a) 1:1
(b) 4:3
(c) 3:4
(d) 5:3
Answer:
(d) 5:3

Question 5.
In case of death of a partner, the whole amount standing to the credit of his capital account is transferred to ………………….
(a) Capital Accounts of all partners
(b) Capital Accounts of remaining partners
(c) His executor’s account
(d) Revenue Account of the Government
Answer:
(c) His executor’s account

Question 6.
A, B and C share profits in the ratio of \(\frac { 1 }{ 2 }\): \(\frac { 3 }{ 10 }\): \(\frac { 1 }{ 5 }\) C dies. The gaining ratio of A and B will be …………………..
(a) 1:1
(b) 1:3
(c) 5:3
(d) 3:1
Answer:
(c) 5:3

Question 7.
On retirement of a partner, the continuing partner’s capital accounts are debited with retiring partner’s share of goodwill in
(a) Old profit sharing ratio
(b) Gaining ratio
(c) New profit sharing ratio
(d) Equal ratio
Answer:
(b) Gaining ratio

Question 8.
N, S, and K have been sharing profit in the ratio of 3:5:7. K retires and his share is taken by N and S in the ratio of 3:2, the new ratio will be
(a) 12:13
(b) 3:5
(c) 2:1
(d) 3:2
Answer:
(a) 12:13

Question 9.
If at the time of retirement, there is some unrecorded liability, it will be …………………..
(a) Debited to Revaluation A/c
(b) Credited to Revaluation A/c
(c) Transferred to Old partners Capital A/Cs
(d) Transferred to All Partners Capital A/Cs
Answer:
(a) Debited to Revaluation A/c

Question 10.
The gain of remaining partners is equal to ………………….
(a) Their new share
(b) Their old share
(c) New Share – Old share
(d) Old share – New share
Answer:
(c) New Share – Old share

Question 11.
Which of the following is debited to partner’s capital at the time of retirement of a partner?
(a) General Reserve
(b) Profit on revaluation
(c) Accumulated losses
(d) Accumulated profits
Answer:
(c) Accumulated losses

Question 12.
At the time of retirement, of a partner, workmen compensation reserve after meeting the legal requirement is transferred to
(a) Revaluation Account
(b) All Partner’s Capital Account .
(c) Sacrificing Partner’s Capital A/cs
(d) Old Partner’s Capital Account
Answer:
(b) All Partner’s Capital Account

Question 13.
On the retirement of a partner, increase in the value of assets is recorded in ………………..
(a) Revaluation A/c
(b) Cash a/c
(c) Old Partner’s Capital A/c
(d) None of the above
Answer:
(a) Revaluation A/c

Question 14.
Undistributed profit and losses – transferred to all the partners at the time of retirement of a partner ………………….
(a) should be
(A) should not be
(c) maybe
Answer:
(a) should be

Samacheer Kalvi 12th Physics Solutions Chapter 5 Electromagnetic Waves

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Samacheer Kalvi 12th Physics Electromagnetic Waves Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electromagnetic Waves Multiple Choice Questions

12th Physics Chapter 5 Book Back Answers Question 1.
The dimension of \(\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \) is ……….
(a) [LT-1]
(b) [L2 T-2]
(c) [L-1 T]
(d) [L-2 T2]
Answer:
(b) [L2 T-2].

12th Physics Lesson 5 Book Back Answers Question 2.
If the amplitude of the magnetic field is 3 x 10-6 T, then amplitude of the electric field for a electromagnetic waves is ……….
(a) 100 V m-1
(b) 300 V m-1
(c) 600 V m-1
(d) 900 V m-1
Answer:
(d) 900 V m-1.

12th Physics 5th Lesson Book Back Answers Question 3.
Which of the following electromagnetic radiation is used for viewing objects through fog ……….
(a) microwave
(b) gamma rays
(c) X- rays
(d) infrared
Answer:
(d) infrared.

Samacheer Kalvi Guru 12th Physics Question 4.
Which of the following are false for electromagnetic waves ……….
(a) transverse
(b) mechanical waves
(c) longitudinal
(d) produced by accelerating charges
Answer:
(c) longitudinal.

12th Physics Samacheer Kalvi Question 5.
Consider an oscillator which has a charged particle and oscillates about its mean position with a frequency of 300 MHz. The wavelength of electromagnetic waves produced by this oscillator is ……….
(a) 1 m
(b) 10 m
(c) 100 m
(d) 1000 m
Answer:
(a) 1 m.

Physics Chapter 5 Class 12 Question 6.
The electric and the magnetic field, associated with an electromagnetic wave, propagating along X axis can be represented by ……….
(a) \(\vec { E } \) = E0 \(\hat{j}\) and \(\vec { B } \) = B0 \(\hat{k}\)
(b) \(\vec { E } \) = E0 \(\hat{k}\) and \(\vec { B } \) = B0 \(\hat{j}\)
(c) \(\vec { E } \) = E0 \(\hat{i}\) and \(\vec { B } \) = B0 \(\hat{j}\)
(d) \(\vec { E } \) = E0 \(\hat{j}\) and \(\vec { B } \) = B0 \(\hat{j}\)
Answer:
(b) \(\vec { E } \) = E0 \(\hat{k}\) and \(\vec { B } \) = B0 \(\hat{j}\).

Samacheer Kalvi Physics Question 7.
In an electromagnetic wave in free space the rms value of the electric field is 3 V m-1. The peak value of the magnetic field is ……….
(a) 1.414 x 10-8 T
(b) 1.0 x 10-8 T
(c) 2.828 x 10-8 T
(d) 2.0 x 10-8 T
Answer:
(a) 1.414 x 10-8 T.

Samacheer Kalvi 12th Physics Question 8.
A During the propagation of electromagnetic waves in a medium ……….
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density.

Physics Class 12 Samacheer Kalvi Question 9.
If the magnetic monopole exists, then which of the Maxwell’s equation to be modified ……….
12th Physics Chapter 5 Book Back Answers Electromagnetic Waves Samacheer Kalvi
Answer:
(b) \(\oint { \vec { E } } \).d\(\vec { A } \) = 0

Samacheerkalvi.Guru 12th Physics Question 10.
A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ……….
(a) \(\frac { E }{ c }\)
(b) 2\(\frac { E }{ c }\)
(c) Ec
(d) \(\frac { E }{{ c }^{2}}\)
Answer:
(b) 2\(\frac { E }{ c }\).

12 Physics Samacheer Kalvi Question 11.
Which of the following is an electromagnetic wave ……….
(a) α – rays
(b) β – rays
(c) γ – rays
(d) all of them
Answer:
(c) γ – rays.

Samacheer Kalvi Guru Physics Question 12.
Which one of them is used to produce a propagating electromagnetic wave ……….
(a) an accelerating charge
(b) a charge moving at constant velocity
(c) a stationary charge
(d) an uncharged particle
Answer:
(a) an accelerating charge.

Samacheer Kalvi.Guru 12th Physics Question 13.
Let E = E0 sin[106 x -ωt] be the electric field of plane electromagnetic wave, the value of to is ……….
(a) 0.3 x 10-14 rad s-1
(b) 3 x 10-14 rad s-1
(c) 0.3 x 1014 rad s-1
(d) 3 x 1014 rad s-1
Answer:
(d) 3 x 1014 rad s-1.

Physics Samacheer Kalvi Question 14.
Which of the following is NOT true for electromagnetic waves ……….
(a) it transport energy
(b) it transport momentum
(c) it transport angular momentum
(d) in vacuum, it travels with different speeds which depend on their frequency
Answer:
(d) in vacuum, it travels with different speeds which depend on their frequency.

Samacheer Kalvi 12th Physics Solutions Question 15.
The electric and magnetic fields of an electromagnetic wave are ……….
(a) in phase and perpendicular to each other
(b) out of phase and not perpendicular to each other
(c) in phase and not perpendicular to each other
(d) out of phase and perpendicular to each other
Answer:
(a) in phase and perpendicular to each other.

Samacheer Kalvi 12th Physics Electromagnetic Waves Short Answer Questions

Class 12 Physics Samacheer Kalvi Question 1.
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in which the electric field and the electric flux are changing with time.

Physics Class 12 Samacheer Kalvi Solutions Question 2.
What are electromagnetic waves?
Answer:
Electromagnetic waves are non-mechanical waves which move with speed equals to the speed of light (in vacuum).

12th Samacheer Kalvi Physics Question 3.
Write down the integral form of modified Ampere’s circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.
12th Physics Lesson 5 Book Back Answers Electromagnetic Waves Samacheer Kalvi

Physics Solution Class 12 Samacheer Kalvi Question 4.
Explain the concept of intensity of electromagnetic waves.
Answer:
The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, I = \(\left< u \right> \)c

Tn 12th Physics Solution Question 5.
What is meant by Fraunhofer lines?
Answer:
When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.

Samacheer Kalvi 12th Physics Electromagnetic Waves Long Answer Questions

12th Physics Solutions Samacheer Kalvi Question 1.
Write down Maxwell equations in integral form.
Answer:
Maxwell’s equations in integral form:
Electrodynamics can be summarized into four basic equations, known as Maxwell’s equations. These equations are analogous to Newton’s equations in mechanics. Maxwell’s equations completely explain the behaviour of charges, currents and properties of electric and magnetic fields. So we focus here only in integral form of Maxwell’s equations:
(i) First equation is nothing but the Gauss’s law. It relates the net electric flux to net electric charge enclosed in a surface. Mathematically, it is expressed as
12th Physics 5th Lesson Book Back Answers Electromagnetic Waves Samacheer Kalvi
Where \(\vec { E } \) is the electric field and Qenclosed is the charge enclosed. This equation is true for both discrete or continuous distribution of charges. It also indicates that the electric field lines start from positive charge and terminate at negative charge. This implies that the electric field lines do not form a continuous closed path. In other words, it means that isolated positive charge or negative charge can exist.

(ii) Second equation has no name. But this law is similar to Gauss’s law in electrostatics. So this law can also be called as Gauss’s law in magnetism. The surface integral of magnetic field over a closed surface is zero. Mathematically,
\(\oint { \vec { B } } \).d\(\vec { A } \) = 0
where \(\vec { B } \) is the magnetic field. This equation implies that the magnetic lines of force form a continuous closed path. In other words, it means that no isolated magnetic monopole exists.

(iii) Third equation is Faraday’s law of electromagnetic induction. This law relates electric field with the changing magnetic flux which is mathematically written as
\(\oint { \vec { E } } \).d\(\vec { l } \) = \(\frac { d }{ dt }\) ΦB
where \(\vec { E } \) is the electric field. This equation implies that the line integral of the electric field around any closed path is equal to the rate of change of magnetic flux through the closed path bounded by the surface.

(iv) Fourth equation is modified Ampere’s circuital law. This is also known as ampere- Maxwell’s law. This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.
Samacheer Kalvi Guru 12th Physics Solutions Chapter 5 Electromagnetic Waves
Where \(\vec { B } \) is the magnetic field. This equation shows that both conduction and also displacement current produces magnetic field. These four equations are known as Maxwell’s equations in electrodynamics.

Physics Practical Class 12 Samacheer Kalvi Question 2.
Write short notes on (a) microwave (b) X-ray (c) radio waves (d) visible spectrum.
Answer:
Microwaves:
It is produced by electromagnetic oscillators in electric circuits. The wavelength range is 1 x 10-3 m to 3 x 10-1 m and frequency range is 3 x 1011 Hz to 1 x 109 Hz. It obeys reflection and polarization. It is used in radar system for aircraft navigation, speed of the vehicle, microwave oven for cooking and very long distance wireless communication through satellites.

X-rays:
It is produced when there is a sudden deceleration of high speed electrons at high- atomic number target, and also by electronic transitions among the innermost orbits of atoms. The wavelength range 10-13 m to 10-8 m and frequency range are 3 x 1021 Hz to 1 x 1016 Hz. X-rays have more penetrating power than ultraviolet radiation.

X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.

Radio waves:
It is produced by oscillators in electric circuits. The wavelength range is 1 x 10-1 m to 1 x 104 m and frequency range is 3 x 109 Hz to 3 x 104 Hz. It obeys reflection and diffraction. It is used in radio and television communication systems and also in cellular phones to transmit voice communication in the ultra high frequency band.

Visible light:
It is produced by incandescent bodies and also it is radiated by excited atoms in gases. The wavelength range is 4 x 10-7 m to 7 x 10-7 m and frequency range are 7 x 1014 Hz to 4 x 1014Hz. It obeys the laws of reflection, refraction, interference, diffraction, polarization, photo-electric effect and photographic action. It can be used to study the structure of molecules, arrangement of electrons in external shells of atoms and sensation of our eyes.

Question 3.
Discuss briefly the experiment conducted by Hertz to produce and detect electromagnetic spectrum.
Answer:
Production of electromagnetic waves-Hertz experiment: Maxwell’s prediction was experimentally confirmed by Heinrich Rudolf Hertz in 1888. The experimental set up used is shown in figure.
12th Physics Samacheer Kalvi Solutions Chapter 5 Electromagnetic Waves
It consists of two metal electrodes which are made of small spherical metals. These are connected to larger spheres and the ends of them are connected to induction coil with very large number of turns. This is to produce very high electromotive force (emf). Since the coil is maintained at very high potential, air between the electrodes gets ionized and spark (spark means discharge of electricity) is produced. The gap between electrode (ring type – not completely closed and has a small gap in between) kept at a distance also gets spark.

This implies that the energy is transmitted from electrode to the receiver (ring electrode) as a wave, known as electromagnetic waves. If the receiver is rotated by 90° – then no spark is observed by the receiver. This confirms that electromagnetic waves are transverse waves as predicted by Maxwell. Hertz detected radio waves and also computed the speed of radio waves which is equal to the speed of light (3 x 108 m s-1).

Question 4.
Explain the Maxwell’s modification of Ampere’s circuital law.
Answer:
Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents are the usual sources of magnetic fields, a changing electric field must be associated with a current. Maxwell called this current as the displacement current.

If ‘A’ be the area of the capacitor plates and ‘q’ be the charge on the plates at any instant ‘t’ during the charging process, then the electric field in the gap will be
Physics Chapter 5 Class 12 Electromagnetic Waves Samacheer Kalvi
But \(\frac { dq }{ dt }\) is the rate of change of charge on the capacitor plates. It is called displacement current and is given by Id = \(\frac { dq }{ dt }\) = ε0 \(\frac {{ dΦ }_{E}}{ dt }\)
This is the missing term in Ampere’s Circuital Law. The total current must be the sum of the conduction current Ic and the displacement current ld. Thus, I = Ic + Id = Ic + ε0 \(\frac {{ dΦ }_{E}}{ dt }\).
Hence, the modified form of the Ampere’s Law is
Samacheer Kalvi Physics 12th Solutions Chapter 5 Electromagnetic Waves

Question 5.
Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:
(i) Electromagnetic waves are produced by any accelerated charge.

(ii) Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.

(iii) Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.

(iv) Electromagnetic waves travel with speed which is equal to the speed of light in vacuum or free space, c = \(\frac { 1 }{ \sqrt { { \varepsilon }_{ 0 }{ \mu }_{ 0 } } } \) 3 x 108 ms-1

(v) The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,
Samacheer Kalvi 12th Physics Solutions Chapter 5 Electromagnetic Waves
(vi) Electromagnetic waves are not deflected by electric field or magnetic field.
Physics Class 12 Samacheer Kalvi Solutions Chapter 5 Electromagnetic Waves

(vii) Electromagnetic waves can show interference, diffraction and can also be polarized.

(viii) The energy density (energy per unit volume) associated with an electromagnetic wave propagating in vacuum or free space is
u = \(\frac { 1 }{ 2 }\) ε0 E2 + \(\frac { 1 }{{ 2µ }_{0}}\) Bε2
Where, \(\frac { 1 }{ 2 }\) ε0 E2 = uE is the energy density in an electric field and \(\frac { 1 }{{ 2µ }_{0}}\) Bε2 = uB is the energy density in a managnetic field.
Since, E = Bc ⇒ uB = uE.
The energy density of the electromagnetic wave is = ε E2 = -B2

(ix) The average energy density for electromagnetic wave, \(\left< u \right> \) = \(\frac { 1 }{ 2 }\) ε0 E2 = \(\frac { 1 }{ 2 }\) \(\frac { 1 }{{ µ }_{0}}\) Bε2

(x) The energy crossing per unit area per unit time and perpendicular to the direction of propagation of electromagnetic wave is called the intensity.
Intensity, I = \(\left< u \right> \)c

(xi) Like other waves, electromagnetic waves also carry energy and momentum. For the electromagnetic wave of energy U propagating with speed c has linear momentum which is given by = \(\frac { Energy }{ speed }\) = \(\frac { U }{ c }\). The force exerted by an electromagnetic wave on unit area speed of a surface is called radiation pressure.

(xii) If the electromagnetic wave incident on a material surface is completely absorbed, then the energy delivered is U and momentum imparted on the surface is p = \(\frac { U }{ c }\).

(xiii) If the incident electromagnetic wave of energy U is totally reflected from the surface, then the momentum delivered to the surface is ∆p = \(\frac { U }{ c }\) – \(\left( -\frac { U }{ c } \right) \) = 2\(\frac { U }{ c }\).

(xiv) The rate of flow of energy crossing a unit area is known as pointing vector for electromagnetic waves, which is \(\vec { S } \) = \(\frac { 1 }{{ µ }_{0}}\) (\(\vec { E } \)×\(\vec { B } \)) = c2ε0 (\(\vec { E } \)×\(\vec { B } \)).The unit for pointing vector is W m-2. The pointing vector at any point gives the direction of energy transport from that point.

(xv) Electromagnetic waves carries not only energy and momentum but also angular momentum.

Question 6.
Discuss the source of electromagnetic waves.
Answer:
Sources of electromagnetic waves:
Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space dependent) around the conductor in which charge flows.

If the charged particle accelerates, in addition to electric field it also produces magnetic field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.

Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position, it produces electromagnetic waves.
Samacheerkalvi.Guru 12th Physics Solutions Chapter 5 Electromagnetic Waves
Suppose the electromagnetic field in free space propagates along z direction, and if the electric field vector points along y axis then the magnetic field vector will be mutually perpendicular to both electric field and the propagation vector direction,
which means
Ey = E0 sin (kz – ωt)
Bx = B0 sin (kz – ωt)
Where, E0 and B0 are amplitude of oscillating electric and magnetic field, k is a wave number, ω is the angular frequency of the wave and \(\hat{k}\) (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.

Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between E0 and B0 is equal to the speed of electromagnetic wave, which is equal to speed of light c.
c = \(\frac {{ E }_{0}}{ { B }_{0} }\)
In any medium, the ratio of E0 and B0 is equal to the speed of electromagnetic wave in that medium, mathematically, it can be written as
v = \(\frac {{ E }_{0}}{ { B }_{0} }\) < c
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.

Question 7.
What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into three types:

(i) Continuous emission spectra (or continuous spectra):
If the light from incandescent lamp (filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.
12 Physics Samacheer Kalvi Solutions Chapter 5 Electromagnetic Waves

(ii) Line emission spectrum (or line spectrum):
Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies.

Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc.
Samacheer Kalvi Guru Physics Solutions Chapter 5 Electromagnetic Waves

(iii) Band emission spectrum (or band spectrum):
Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end.

Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

Question 8.
What is absorption spectra? Give their types.
Answer:
Absorption spectra: When light is allowed to pass through a medium or an absorbing substance then the spectrum obtained is known as absorption spectrum. It is the characteristic of absorbing substance. Absorption spectrum is classified into three types:
(i) Continuous absorption spectrum:
When the light is passed through a medium, it is dispersed by the prism, we get continuous absorption spectrum. For instance, when we pass white light through a blue glass plate, it absorbs everything except blue. This is an example of continuous absorption spectrum.

(ii) Continuous absorption spectrum:
When light from the incandescent lamp is passed through cold gas (medium), the spectrum obtained through the dispersion due to prism is line absorption spectrum. Similarly, if the light from the carbon arc is made to pass through sodium vapour,-a continuous spectrum of carbon arc with two dark lines in the yellow region of sodium vapour is obtained.
Samacheer Kalvi.Guru 12th Physics Solutions Chapter 5 Electromagnetic Waves
(iii) Band absorption spectrum:
When the white light is passed through the iodine vapour, dark bands on continuous bright background is obtained. This type of band is also obtained when white light is passed through diluted solution of blood or chlorophyll or through certain solutions of organic and inorganic compounds.

Samacheer Kalvi 12th Physics Electromagnetic Waves Numerical problems

Question 1.
Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.
Solution:
The conduction current Ic = 5A
According to Gauss’s Law,
Electric flux, θE = \(\frac { q }{{ ε }{0}}\)
Physics Samacheer Kalvi 12th Solutions Chapter 5 Electromagnetic Waves

Question 2.
A transmitter consists of LC circuit with an inductance of 1 μH and a capacitance of 1 μF. What is the wavelength of the electromagnetic waves it emits?
Solution:
Inductance of LC circuit, L = 1 μH = 1 × 10-6 H
Capacitance of LC circuit, C = 1 μF = 1 × 10-6 F
Wave length of the electromagnetic wave X = λ = \(\frac { C }{ f }\)
Velocity of light C = 3 x 108 ms-1
Frequency of electromagnetic wave, f = \(\frac { 1 }{ 2\pi \sqrt { LC } } \)
Samacheer Kalvi 12th Physics Solutions Chapter 5 Electromagnetic Waves
= 0.1884 × 104
λ = 18.84 × 102 m

Question 3.
A pulse of light of duration 10-6 s is absorbed completely by a small object initially at rest. If the power of the pulse is 60 x 10-3 W, calculate the final momentum of the object.
Solution:
Duration of the absorption of light pulse, t = 10-6 s
Power of the pulse P = 60 x 10-3 W
Final momentum of the object, P = \(\frac { U }{ c }\)
Velocity of light, C = 3 x 108
Energy U = power x time
Momentum, P = \(\frac{60 \times 10^{-3} \times 10^{-6}}{3 \times 10^{8}}\)
P = 20 × 10-17 kg m s-1

Question 4.
Let an electromagnetic wave propagate along the x direction, the magnetic field oscillates at a frequency of 1010 Hz and has an amplitude of 10-5 T, acting along the y – direction. Then, compute the wavelength of the wave. Also write down the expression for electric field in this case.
Solution:
Frequency of electromagnetic wave, v = 1010 Hz
Amplitude of Oscillating magnetic field, B0 = 10-5 T
Wave length of the wave, λ = \(\frac { C }{ f }\) = \(\frac{3 \times 10^{8}}{10^{10}}\) = 3x 10-2 m
Amplitude of oscillating electric field, E0 = B0 C
C = \(\frac {{ E }_{0}}{ { B }_{0} }\)
E0 = 10-5 × 3 × 108
E0 = 3 × 103 = NC-1
Experession for electric field in Oscillataing wave
E = E0 sin (kx -wt)
K = \(\frac { 2π }{ λ }\) = \(\frac{2 \times 3.14}{3 \times 10^{-2}}\) = 209 x 102
W = 2πƒ = 2 × 3.14 x 1010 = 6.28 × 1010
\(\vec { E } \) = 3 x 103 sin (2.09 × 102 x – 6.28 × 1010 t) \(\hat{i}\) NC-1.

Question 5.
If the relative permeability and relative permittivity of the medium is 1.0 and 2.25, respectively. Find the speed of the electromagnetic wave in this medium.
Solution:
Relative permeability of the medium, μr = 1
Relative permitivity of the medium, εr = 2.25
Class 12 Physics Samacheer Kalvi Solutions Chapter 5 Electromagnetic Waves
Speed of electromagnetic wave, v = \(\frac { 1 }{ \sqrt { \mu \varepsilon } } \)
Physics Class 12 Samacheer Kalvi Solutions Chapter 5 Electromagnetic Waves

Samacheer Kalvi 12th Physics Electromagnetic Waves Additional Questions Solved

I Choose the Correct Answer from the Following

Question 1.
The speed of electromagnetic waves in vacuum is given by
(a) μ0ε0
(b) \(\sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \)
(c) \(\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \)
(d) \(\frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } \)
Answer:
(d) \(\frac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } \).

Question 2.
In an electromagnetic wave the electric field vector E and magnetic field vector B are
(a) Perpendicular to each other
(b) Parallel to each other
(c) at 45° to each other
(d) can have any angle between them
Answer:
(a) Perpendicular to each other.

Question 3.
If E and B be the electric and magnetic field vectors of an electromagnetic wave, then the propagation of the wave is along the direction of
(a) E
(b) B
(c) E x B
(d) Bx E
Answer:
(c) E x B.

Question 4.
In which of the following sequences are the electro magnetic radiations in decreasing order of wavelength
(a) Infrared, radio, X-rays, visible
(b) Radio, infrared, visible, X-rays
(c) Radio, visible, infrared, X-rays
(d) X-rays, visible, infrared, radio
Answer:
(b) Radio, infrared, visible, X-rays .

Question 5.
Ozone layer absorbs
(a) Infrared radiation
(b) Microwaves
(c) Radio waves
(d) UV radiation
Answer:
(d) UV radiation.

Question 6.
A RADAR beam consists of
(a) X-rays
(b) IR rays
(c) UV rays
(d) Microwaves
Answer:
(d) Microwaves.

Question 7.
Which of the following radiations has the longest wavelength?
(a) Radio waves
(b) IR radiation
(c) X-ray
(d) Visible light
Answer:
(a) Radio waves.

Question 8.
TV waves have a wavelength range of 1 – 10 metre. Their frequency range in MHz is
(a) 300 – 3000
(b) 3 – 3000
(c) 30 – 300
(d) 3 – 30
Answer:
(c) 30 – 300.

Question 9.
The electromagnetic radiation most prevalent in the atmosphere is
(a) Visible light
(b) Infrared
(c) UV
(d) Radio waves
Answer:
(b) Infrared.

Question 10.
Consider an electric charge oscillating with frequency of 10MHz. The radiation emitted will have a wavelength eqal to
(a) 20m
(b) 30m
(c) 40m
(d) 10m
Answer:
(b) 30m
Hint:
Wave length, λ = \(\frac { C }{ v }\) = \(\frac{3 \times 10^{8}}{10 \times 10^{6}}\)
= 0.3x 102 = 30m.

Question 11.
The frequencies of X-rays, v-rays and UV rays are respectively a, b and c. Then
(a) a < b, b < c
(b) a < b, b > c
(c) a > b, b > c
(d) a > b, b < c
Answer:
(b) a < b, b > c .

Question 12.
Greenhouse effect is caused by
(a) UV rays
(b) X-rays
(c) Gamma rays
(d) IR rays
Answer:
(d) IRrays.

Question 13.
If a capacitance C is connected across an inductance L, then the angular frequency is
(a) \( \sqrt{LC} \)
(b) LC
(c) \(\sqrt { \frac { L }{ C } } \)
(d) \(\sqrt { \frac { 1 }{ LC } } \)
Answer:
(d) \(\sqrt { \frac { 1 }{ LC } } \).

Question 14.
If ε0 and μ0 are the electric permitivity and permeability of free space, ε and μ are the corresponding quantities in a medium, then the index of refraction of the medium is
12th Samacheer Kalvi Physics Solutions Chapter 5 Electromagnetic Waves
Answer:
(d) \(\sqrt { \frac { \varepsilon \mu }{ { \varepsilon }_{ 0 }{ \mu }_{ 0 } } } \)
Hint:
Index of refraction
Physics Solution Class 12 Samacheer Kalvi Chapter 5 Electromagnetic Waves

Question 15.
Dimensions of \(\frac { 1 }{ { \mu }_{ 0 }{ \varepsilon }_{ 0 } } \) where symbols have their usual meaning, are
(a) [L-2 T2]
(b) [L2 T-2]
(c) [LT-1]
(d) [L-1 T]
Answer:
(b) [L2 T-2]
Hint:
Tn 12th Physics Solution Chapter 5 Electromagnetic Waves Samacheer Kalvi

Question 16.
If λv, λx, λm, represent the wavelength of visible light, X-rays and microwaves, respectively, then
(a) λmv > λx
(b) λm > λx > λv
(c) λv > λm > λx
(d) λv > λx > λm
Answer:
(a) λmv > λx.

Question 17.
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to
(a) the speed of light in vacuum
(b) reciprocal of the speed of light in vacuum
(c) the ratio of magnetic permeability to electric susceptibility in vacuum
(d) unity
Answer:
(b) reciprocal of the speed of light in vacuum.

Question 18.
Which of the following radiations forms part of the electromagnetic spectrum?
(a) alpha rays
(b) Beta rays
(c) Gamma rays
(d) Cathode rays
Answer:
(c) Gamma rays.

Question 19.
The ozone layer absorbs radiation of wavelegnths
(a) Less than 3 x 10-7 m
(b) More than 3 x 10-7 m
(c) Less than 3 x 10-5 m
(d) More than 3 x 10-5 m
Answer:
(a) Less than 3 x 10-7 m.

Question 20.
It is possible to take pictures of those objects which are not fully visible to the eye using camera films sensitive to
(a) UV rays
(b) IR rays
(c) Microwaves
(d) Radio waves
Answer:
(b) IRrays.

Question 21.
An electromagnetic wave has wavelength 10cm. It is in the
(a) Visible region
(b) Radio region
(c) UV region
(d) X-ray region
Answer:
(b) Radio region.

Question 22.
Electromagnetic radiation of frequency 3 x 105 MHz lies in the
(a) Radio wave region
(b) Visible region
(c) IR region
(d) Microwave region
Answer:
(d) Microwave region.

Question 23.
Radio waves and visible light in vacuum have
(a) Same wavelength but different velocities
(b) Same velocity but different wavelength
(c) Different velocities and different wavelengths
(d) Same velocity and same wavelength
Answer:
(b) Same velocity but different wavelength.

Question 24.
Which of the following has maximum frequency?
(a) X-rays
(b) IR rays
(c) UV rays
(d) Radio waves
Answer:
(a) X-rays.

Question 25.
Electromagnetic radiation of frequency 1 GHz lies in
(a) UV region
(b) IR region
(c) Visible region
(d) Microwave region
Answer:
(d) Microwave region.

Question 26.
An electromagnetic wave of frequency v = 3.0MHz passes from vacuum into a dielectric medium with permitivity s = 4.0 ε0. Then
(a) Wavelength is doubled and frequency becomes half
(b) Wavelength is doubled and the frequency remains unchanged.
(c) Wavelength and frequency both remain unchanged
(d) Wavelength is halved and frequency remains unchanged.
Answer:
(d) Wavelength is halved and frequency remain unchanged.
Hint:
When a wave goes from one medium to another, its frequency remains unchanged.
According to equation C = \(\frac { 1 }{ \sqrt { \mu \varepsilon } } \), the speed is halved. Therefore, the wavelength is also halved.

Question 27.
Frequency of a wave is 6 x 1015 Hz. The wave is
(a) Radio wave
(b) Microwave
(c) X-ray
(d) UV rays
Answer:
(d) UV rays.

Question 28.
In an electromagnetic wave the phase difference between electric field \(\vec { E } \) and magnetic field \(\vec { B } \) is
(a) \(\frac { π }{ 4 }\)
(b) \(\frac { π }{ 2 }\)
(c) π
(d) zero
Answer:
(d) Zero.

Question 29.
The energy in an electromagnetic wave is
(a) Wholly shared only by electric field vector
(b) Wholly shared only by magnetic field vector
(c) Equally divided between electric and magnetic field
(d) Zero
Answer:
(c) Equally divided between electric and magnetic field.

Question 30.
Electromagnetic waves are produced by
(a) Atoms and molecules in an electrical discharge
(b) Electric device
(c) Accelerated charges
(d) Molecules of hot bodies
Answer:
(c) Accelerated charges.

II Fill in The Blanks

Question 1.
Electromagnetic waves are ……………
Answer:
Transverse in nature

Question 2.
Accelerated charge is a source of ……………
Answer:
Electromagnetic radiation

Question 3.
In electromagnetic waves, angle between electric and magnetic field vectors is ……………
Answer:
90°

Question 4.
The electromagnetic waves travel in vacuum or free space with a velocity of ……………
Answer:
3 x 108 ms-1

Question 5.
The existance of electromagnetic waves was confirmed experimentally by ……………
Answer:
Hertz.

Question 6.
Radio frequency wave have wavelength ranging from ……………to ……………
Answer:
10 m to 104 m

Question 7.
The waves that are used in radio and television communication system is ……………
Answer:
Radio waves

Question 8.
The frequency range of X-rays is from ……………
Answer:
3 x 1018 Hz to 1016 Hz

Question 9.
Higher frequencies upto 54 mHz are used for ……………
Answer:
Short wave brands

Question 10.
The wave that is used in radar communication system is ……………
Answer:
Microwaves.

Question 11
…………… rays are used to detect forged documents.
Answer:
UV-rays.

Question 12
…………… is used for treatment of cancer.
Answer:
γ – rays (gamma rays).

Question 13.
Structure of atoms can be studied with ……………
Answer:
X-rays

Question 14.
When the light emitted directly from a source is examined with a spectrometer, the spectrum obtained is …………… spectrum.
Answer:
Emission

Question 15.
When the light emitted from a source is made to pass through an absorbing material, the spectrum obtained is called …………… spectrum.
Answer:
Absorption.

Question 16.
Electromagnetic disturbance can be propagated through space ……………
Answer:
without the help of material medium.

Question 17.
The velocity of electromagnetic wave is given by the relation ……………
Answer:
C = \(\frac { 1 }{ \sqrt { { { \mu }_{ 0 } }{ { \varepsilon }_{ 0 } } } } \)

III Match the following

Question 1.
(i) Light – (a) Medical surgery
(ii) Laser – (b) Law of induction
(iii) Radar – (c) Electromagnetic waves
(iv) Maxwell – (d) Defence (detection system)
Answer:
(i) → (c)
(ii) → (a)
(iii) → (d)
(iv) → (b)

Question 2.
(i) Gauss Law – (a) Electromagnetic induction
(ii) Faraday’s Law – (b) φ =\(\frac { q }{{ ε }_{0}}\)
(iii) Speed of light – (c) Electromagnetic oscillators
(iv) Microwaves – (d) 3 x 108 ms-1
Answer:
(i) → (b)
(ii) → (a)
(iii) → (d)
(iv) → (c)

Question 3.
(i) Gamma ray – (a) Structure of molecules
(ii) Visible light – (b) Treatment of cancer
(iii) UV radiation – (c) Detecting fractures
(iv) X-rays – (d) Destroy bacteria
Answer:
(i) → (b)
(ii) → (a)
(iii) → (d)
(iv) → (c)

Question 4.
(i) Continuous emission spectra – (a) Atomic hydrogen
(ii) Line emission spectra – (b) Ammonia gas
(iii) Band emission spectra – (c) Incandescent solids
(iv) Band absorption spectra – (d) Blood (or) chlorophyl
Answer:
(i) → (c)
(ii) → (a)
(iii) → (b)
(iv) → (d)

IV Assertion and Reason Questions

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not the correct explanation of the assertion.
(c) If assertion is true but-reason is false.
(d) If the assertion and reason both are false.
(e) If the assertion is false but reason is true.

Question 1.
Assertion: In Hertz experiment, the electric vector of radiation produced by the source gap is parallel to the gap.
Reason: Production of sparks between the detector gap is maximum when it is placed perpendicular to the source gap.
Answer:
(c) If assertion is true but reason is false.
Explanation:
Hertz experimentally observed that the production of spark between the detection gap is maximum when it is placed parallel to source gap. This means that the electric vector of radiation produced by the source gap is parallel to the two gaps.

Question 2.
Assertion: Ultraviolet radiation are of higher frequency waves are dangerous to human being.
Reason: Ultraviolet radiation are absorbed by the atmosphere.
Answer:
(b) If assertion and reason both are true but reason is not the correct explanation of the assertion.
Explanation:
The wavelength of these waves ranges between 6 x 10-10m to 4 x 10-7 m that is smaller wavelength and higher frequency. They are absorbed by atmosphere and convert oxygen into ozone. They cause skin diseases and they are harmful to eye and cause permanent blindness.

Question 3.
Assertion: Only microwaves are used in RADAR.
Reason: Because microwaves have very small wavelength.
Answer:
(a) If assertion and reason both are true and reason is the correct explanation of the assertion.
Explanation:
In a RADAR, a beam signal is needed in particular direction which is possible if wavelength of wave is very small. Since the wavelength of microwaves is a few millimeter, hence they are used in RADAR.

Question 4.
Assertion: Radio waves bend in a magnetic field.
Reason: Radio waves are electromagnetic in nature.
Answer:
(a) If assertion and reason both are true and reason is the correct explanation of the assertion.

Question 5.
Assertion:
In the visible spectrum of light, red light is more energetic than green light.
Reason:
The wavelength of red light is more than that of green light.
Answer:
(a) If assertion and reason both are true and reason is the correct explanation of the assertion.

Samacheer Kalvi 12th Physics Electromagnetic Waves Short Answer Questions

Question 1.
Name the scientist who first predicted the existence of electromagnetic waves.
Answer:
James Clerk Maxwell was first predicted the existence of EM waves.

Question 2.
Distinguish between conduction current and displacement current.
Answer:
Conduction Current:

  1. It is due to the flow of electrons in a circuit.
  2. It exists even if electrons flow at a uniform rate.

Displacement Current:

  1. It is due to time-varying electric field.
  2. It does not exist under steady conditions.

Question 3.
What oscillates in electromagnetic waves?
Answer;
In EM waves, electric and magnetic fields oscillate in mutually perpendicular directions. These waves are transverse in nature.

Question 4.
Name the basic source of electromagnetic waves.
Answer:
An electric dipole is a basic source of electro magnetic waves.

Question 5.
State Maxwell’s equations.
Answer:
The whole study of electricity and magnetism can be described mathematically with the help of four fundamental equations, called Maxwell’s equations. These are stated as follows:

  1. Gauss law of electrostatics : \(\oint { \vec { E } } \).\(\vec { d } \)s = 0 \(\frac { q }{{ ε }{0}}\)
  2. Gauss law of magnetism: \(\oint { \vec { E } } \).\(\vec { d } \)s = 0
  3. Faraday’s Law of electromagnetic induction: \(\oint { \vec { E } } \).\(\vec { d } \)l = – \(\frac{d \phi_{\mathrm{B}}}{d t}\) = –\(\frac { d }{ dt }\) \(\left[ \oint { \vec { B. } \vec { d } s } \right] \)
  4. Modified ampere’s circuital Law: \(\oint { \vec { E } } \).\(\vec { d } \)l = μ0 \(\left[\mathrm{I}_{\mathrm{C}}+\varepsilon_{0} \frac{d \phi_{\mathrm{E}}}{d t}\right]\)

Question 6.
What is electromagnetic spectrum.
Answer:
The orderly distribution of electromagnetic radiations of all types according to their wavelength or frequency into distinct groups having widely differing properties is called electromagnetic spectrum.

Question 7.
Define spectrum.
Answer:
The definite pattern of colours obtained on the screen after dispersion is called as spectrum.

Question 8.
Write down the concept of black body spectrum.
Answer:
When an object bums, it emits colours. That is, it emits electromagnetic radiation which depends on temperature. If the object becomes hot then it glows in red colour. If the temperature of the object is further increased then it glows in reddish-orange colour and becomes white when it is hottest. The spectrum usually called as black body spectrum.

Question 9.
How are radio waves produced?
Answer:
Radiowaves are produced due to accelerated motion of electrons in conducting wires or oscillating circuits.

Question 10.
How are X-rays produced?
Answer:
X-rays are produced due to sudden deceleration of fast moving electrons by a metal target.

Question 11.
How are microwaves produced?
Answer:
Microwaves are produced due to oscillating currents in special vacuum tubes like Klystroms.

Question 12.
What is ‘Greenhouse effect’?
Answer:
Greenhouse effect is the phenomenon which keeps the earth’s surface warm at night. The earth reflects back infra red part of solar radiation. Infra-red rays are reflected back by low lying clouds and lower atmosphere and keep the earth’s surface warm at night.

Question 13.
Why are microwaves used in RADAR?
Answer:
Microwaves have wavelength of the order of a few millimetres. Due to their short wavelengths, these are not diffracted (bent) much by objects of normal dimensions. So they can be transmitted as a beam in a particular direction.

Question 14.
Write down the uses of radiowaves?
Answer:
It is used in radio and television communication systems and also in cellular phones to transmit voice communication in the ultra high frequency band.

Question 15.
Write down the uses of radiowaves?
Answer:
It is used in radar system for aircraft navigation, speed of the vehicle, microwave oven for cooking and very long distance wireless communication through satellites.

Question 16.
Write uses of IR rays (or) infra-red radiations.
Answer:
It is used to produce dehydrated fruits; in green houses to keep the plants warm, heat therapy for muscular pain or sprain, TV remote as a signal carrier, to look through haze fog or mist and used in night vision or infrared photography.

Question 17.
Write down the uses of UV rays (or) Ultraviolet radiations.
Answer:
It is used to destroy bacteria, sterilizing the surgical instruments, burglar alarm, detect the invisible writing, finger prints and also in the study of molecular structure.

Question 18.
What are the uses of X-rays?
Answer:
X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting’fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.

Question 19.
What are emission spectrum? Write its types.
Answer:
When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum can be divided into
three types:

  1. Continuous emission spectra.
  2. Line emission spectrum (or line spectrum).
  3. Band emission spectrum (or band spectrum).

Question 20.
What are absorption spectrum? Write its types
Answer:
When light is allowed to pass through a medium or an absorbing substance then the spectrum .obtained is known as absorption spectrum. It is the characteristic of absorbing substance.
Absorption spectrum is classified into three types:

  1. Continuous absorption spectrum.
  2. Line absorption spectrum.
  3. Band absorption spectrum.

Samacheer Kalvi 12th Physics Electromagnetic Waves Long Answer Questions

Question 1.
Write short notes on (a) Infrared radiation (b) Ultraviolet radiation (c) Gamma radiation).
Answer:
(a) Infrared radiation:
It is produced from hot bodies (also known as heat waves) and also when the molecules undergo rotational and vibrational transitions. The wavelength range is 8 x 10-7 m to 5 x 103 m and frequency range are 4 x 1014 Hz to 6 x 1010 Hz. It provides electrical energy to satellites by means of solar cells. It is used to produce dehydrated fruits, in green houses to keep the plants warm, heat therapy for muscular pain or sprain, TV remote as a signal carrier, to look through haze fog or mist and used in night vision or infrared photography.

(b) Ultraviolet radiation:
It is produced by Sun, arc and ionized gases. The wavelength range is 6 x 10-10 m to 4 x 10-7 m and frequency range are 5 x 1017Hz to 7 x 1014 Hz. It has less penetrating power. It can be absorbed by atmospheric ozone and harmful to human body. It is used to destroy bacteria, sterilizing the surgical instruments, burglar alarm, detect the invisible writing, finger prints and also in the study of molecular structure.

(c) Gamma rays:
It is produced by transitions of atomic nuclei and decay of certain elementary particles. They produce chemical reactions on photographic plates, fluorescence, ionisation, diffraction. The wavelength range is 1 x 10-14 m to 1 x 10-10 m and frequency range are 3 x 1022 Hz to 3 x 1018 Hz. Gamma rays have high penetrating power than X-rays and ultraviolet radiations; it has no charge but harmful to human body. Gamma rays provide information about the structure of atomic nuclei. It is used in radio therapy for the treatment of cancer and tumour, in food industry to kill pathogenic microorganism.

Samacheer Kalvi 12th Physics Electromagnetic Waves Numerical Problems

Question 1.
A parallel plate capacitor has circular plates, each of radius 5.0 cm. It is being charged so that electric field in the gap between its plates rises steadily at the rate of 1012 V m-1 s-1.What is the displacement current?
Solution:
Radius, r = 5cm = 5 x 10-2 m
The rate okf electric frield, \(\frac { dE }{ dt }\) = 1012 V m-1 s-1
Displacement current, Id = ε0 \(\frac {{ dφ }_{E}}{ dt }\) =ε0 \(\frac { d }{ dt }\) (EA) = ε0 (πr2) \(\frac { dE }{ dt }\)
= 8.85 x 10-12 x 3.14 x (5 x 10-2)2 x 1012
Id= 0.069
Id = 0.07 (or) 70 mA

Question 2.
The voltage between the plates of a parallel – plate capacitor of capacitance 1 µF is changing at the rate of 5Vs-1. What is the displacement current in the capacitor?
Solution:
Capacitance of parallel plate capacitor, C = 1 µF
C= 1 x 10-6 F
The rate of voltage b/n the plate, \(\frac { dv }{ dt }\) = 5 Vs-1
Displacement current,
Id = ε0 \(\frac {{ dφ }_{E}}{ dt }\) =ε0 \(\frac { d }{ dt }\) (EA)
ε0 = A \(\frac { d }{ dt }\) \(\left( \frac { V }{ d } \right) \) E = \(\frac { V }{ d }\)
= \(\frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{d}\) \(\frac { dV }{ dt }\) = C. \(\frac { dV }{ dt }\)
Id = 1 x 10-6 × 5
Id = 5µA.

Question 3.
Electromagnetic waves travels in a medium at a speed of 2 x 108 ms-1. The relative permeability of the medium is 1. Find the relative permitivity.
Solution:
Speed of an em wave in a medium is given by
12th Physics Solutions Samacheer Kalvi Chapter 5 Electromagnetic Waves
Relative permitivity,
Physics Practical Class 12 Samacheer Kalvi Chapter 5 Electromagnetic Waves

Question 4.
A radiation of energy E falls normally on a perfectly reflecting surface. Find the momentum transferred to the surface.
Solution:
Momentum of radiation of energy E is P = \(\frac { E }{ C }\)
Since the radiation is completely reflected, its momentum changes by \(\frac { 2E }{ C }\)
Therefore, by the law of conservation of momentum the momentum transferred to the surface is \(\frac { 2E }{ C }\).

Question 5.
The energy of the EM wave is of the order of 15KeV. To which part of the spectrum does it belong?
Solution:
E =hv = \(\frac { hc }{ λ }\) ⇒ λ = \(\frac { hc }{ E }\)
h = 6.626 x 10-34 Js; c = 3 x 108 ms-1
E= 15 x 103 eV = 15 x 103 x 1.6 x 10-19 V
Samacheer Kalvi 12th Physics Solutions Chapter 5 Electromagnetic Waves-23
λ = 0.8282 x Å
Hence X-rays

Question 6.
Light With an energy flux of 25 x 104 Wm-2 falls on a perfectly reflecting surface at normal incidence. If surface area is 15cm2, to calculate the average force exerted on the surface.
Solution:
Average force = momentum transferred per second
Fav = \(\frac { P }{ T }\) = \(\frac { 2u }{ C }\)
Where U is the energy falling on th surface per second.
Fav = \(\frac{2 \times 25 \times 10^{4} \times 15 \times 10^{-4}}{3 \times 10^{8}}\) = 250 x 10-8
Fav = 2.5 x 10-6 N.

Common Errors And Its Rectifications
Common Errors :
1. Students do the mistakes most of times in unit of frequency. They write the units in improper ways. Eg. Hertz (or) H. This is the wrong way.
2. They may confuse the frequency range of radiations and wavelength range of radiation.

Rectifications:
1. The correct way of unit is hertz (or) Hz. The unit of frequency is hertz (or) Hz (or) s-1.
2. The easy way to understand frequency and wavelength range of radiations are, Frequency increases the order of gamma, X-ray, UV, visible, IR, microwave, radiowave. Wavelength increases the order of the reverse of frequency order.

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Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis

Students can Download Accountancy Chapter 8 Financial Statement Analysis Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Accountancy Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis

Samacheer Kalvi 12th Accountancy Financial Statement Analysis Text Book Back Questions and Answers

I. Choose the Correct Answer

12th Accountancy 8th Chapter Question 1.
Which of the following statements is not true?
(a) Notes and schedules also form part of financial statements
(b) The tools of financial statement analysis include common – size statement
(c) Trend analysis refers to the study of movement of figures for one year
(d) The common – size statements show the relationship of various items with some common base, expressed as percentage of the common base
Answer:
(c) Trend analysis refers to the study of movement of figures for one year

12th Accountancy Chapter 8 Question 2.
Balance sheet provides information about the financial position of a business concern ………………
(a) Over a period of time
(b) As on a particular date
(c) For a period of time
(d) For the accounting period
Answer:
(b) As on a particular date

12th Accountancy 8th Chapter Solutions Question 3.
Which of the following tools of financial statement analysis is suitable when data relating to several years are to be analysed?
(a) Cash flow statement
(b) Common size statement
(c) Comparative statement
(d) Trend analysis
Answer:
(d) Trend analysis

Class 12 Accountancy Chapter 8 Solutions Question 4.
The financial statements do not exhibit ………………
(a) Non – monetary data
(b) Past data
(c) Short term data
(d) Long term data
Answer:
(a) Non – monetary data

Accountancy Class 12 Chapter 8 Solutions Question 5.
Which of the following is not a tool of financial statement analysis?
(a) Trend analysis
(b) Common size statement
(c) Comparative statement
(d) Standard costing
Answer:
(d) Standard costing

Accounting Chapter 8 Answer Key Question 6.
The term ‘fund’ refers to ………………
(a) Current liabilities
(b) Working capital
(c) Fixed assets
(d) Non – current assets
Answer:
(b) Working capital

Class 12 Accountancy Chapter 8 Question 7.
Which of the following statements is not true?
(a) All the limitations of financial statements are applicable to financial statement analysis also.
(b) Financial statement analysis is only the means and not an end.
(c) Expert knowledge is not required in analysing the financial statements.
(d) Interpretation of the analysed data involves personal judgement.
Answer:
(c) Expert knowledge is not required in analysing the financial statements

Financial Statements Of A Company Class 12 Solutions Question 8.
A limited company’s sales has increased from ? 1,25,000 to? 1,50,000. How does this appear in comparative income statement?
(a) + 20%
(b) + 120%
(c) – 120 %
(d) – 20 %
Answer:
(a) + 20 %

Financial Statements Questions And Answers Pdf Question 9.
In a common-size balance sheet, if the percentage of non – current assets is 75, what would be the percentage of current assets?
(a) 175
(b) 125
(c) 25
(d) 100
Answer:
(c) 25

Financial Statement Analysis Solutions Pdf Question 10.
Expenses of a business for the first year were ₹ 80,000. In the second year, it was increased to ₹ 88,000. What is the trend percentage in the second year?
(a) 10%
(b) 110%
(c) 90%
(d) 11%
Answer:
(b) 110 %

II. Very Short Answer Questions

Financial Statements Solutions Question 1.
What are financial statements?
Answer:
Financial statements are the statements prepared by the business concerns at the end of the accounting period to ascertain the operating results and the financial position.

Financial Statement Analysis Chapter 8 Solutions Question 2.
List the tools of financial statement analysis.
Answer:

  1. Comparative Statement.
  2. Common Size Statement.
  3. Trend Analysis.
  4. Funds Flow Analysis.
  5. Cash Flow Analysis.

Financial Statement Analysis Solutions Question 3.
What is working capital?
Answer:
Working capital statement or schedule of changes in working is prepared to disclose net changes in working capitals on two specific dates (generally two balance sheet dates). It is prepared from current assets and current liabilities.
Working Capital = Current Assets – Current Liabilities

Financial Statement Analysis Problems And Solutions Pdf Question 4.
When is trend analysis preferred to other tools?
Answer:
Trend analysis discloses the changes in financial and operating data between specific periods when data for more than two years are to be analyzed. It may be difficult to use comparative statement.

III. Short Answer Questions

Question 1.
‘Financial statements are prepared based on the past data’. Explain how this is a limitation?
Answer:
The nature of financial statement is historical. Past cannot be the index of future and cannot be cent percent basis for future estimation, forecasting, budgeting and planning.

Question 2.
Write a short note on cash flow analysis.
Answer:
Cash flow analysis concerned with preparation of cash flow statement which shows the inflow and outflow of cash and cash equivalents in a given period of time. Cash includes cash in hand and demand deposit with banks. Cash equivalents denotes short term investments which can be realised easily within a short period of time without much loss in value. Cash flow analysis helps in assessing the liquidity and solvency of a business concern.

Question 3.
Briefly explain any three limitations of financial statements.
Answer:
1. Lack of qualitative information:
Qualitative information, that is non – monetary information is also important for business decisions. For examole Efficiency of the employees and efficiency of the management. But this is ignored in financial statements.

2. Record of historical data:
Financial statement are prepared based on historical data. They may not reflect the current position.

3. Ignores price level changes:
Adjustments for price level changes are not made in the financial statements. Hence financial statements may not reveal the current position.

Question 4.
Explain the steps involved in preparing comparative statements.
Answer:
Following are the steps to be followed in preparation of comparative statement.

  1. Column 1 : In this column, particulars of items of income statements or balance sheet are written.
  2. Column 2 : Enter absolute amount of year 1.
  3. Column 3 : Enter absolute amount of year 2.
  4. Column 4 : Show the difference in amounts between year 1 and year 2. If there is an increase in year 2, put plus sign and there is a decrease put minus sign.
  5. Column 5 : Show percentage increase or decrease of the difference amount shown in column 4 by dividing the amount shown in column 4 (absolute amount of increase or decrease) by column 2 (year 1 amount)

Percentage increase (or) decrease = 12th Accountancy 8th Chapter 8 Financial Statement Analysis Samacheer Kalvi

Question 5.
Explain the procedure for preparing common – size statement.
Answer:
Common – size statements can be prepared with three columns. Following are the steps to be followed in preparation of common – size statement.

  1. Column 1. In this column, particulars of items of income statement or balance sheet are written.
  2. Column 2. Enter absolute amount.
  3. Column 3. Choose a common base as 100.

For example Revenue from operations can be taken as the base for income statement and total of balance sheet can be taken as the base for balance sheet. Work out the percentage for all the items of column 2 in terms of the common base and enter them in column 3.

IV Exercises

Question 1.
From the following particulars, prepare comparative income statement of Arul Ltd.
12th Accountancy Chapter 8 Financial Statement Analysis Samacheer Kalvi
Answer:
Comparative Income Statement Analysis of Arul Ltd.
for the year ended 31.3.2016 to 31.3.2017
12th Accountancy 8th Chapter Solutions Financial Statement Analysis Samacheer Kalvi

Question 2.
From the following particulars, prepare comparative income statement of Barani Ltd.
Class 12 Accountancy Chapter 8 Solutions Financial Statement Analysis Samacheer Kalvi
Answer:
Comparative Income Statement of Barani Ltd. for the year ended 31st March 2017 to 31st March 2018
Accountancy Class 12 Chapter 8 Solutions Financial Statement Analysis Samacheer Kalvi

Question 3.
From the following particulars, prepare comparative income statement of Daniel Ltd.
Accounting Chapter 8 Answer Key Financial Statement Analysis Samacheer Kalvi 12th
Answer:
Comparative Income Statement of Daniel Ltd. for the year ended 31st March 2016 to 31st March 2017
Class 12 Accountancy Chapter 8 Financial Statement Analysis Samacheer Kalvi

Question 4.
From the following particulars, prepare comparative statement of financial position of Muthu Ltd.
Financial Statements Of A Company Class 12 Solutions Chapter 8 Samacheer Kalvi
Answer:
Comparative Balance Sheet of Muthu Ltd. as on 31st March 2017 to 31st March 2018
Financial Statements Questions And Answers Pdf Samacheer Kalvi 12th Accountancy Solutions Chapter 8

Question 5.
From the following particulars, prepare comparative statement of financial position of Kala Ltd.
Financial Statement Analysis Solutions Pdf Samacheer Kalvi 12th Accountancy Solutions Chapter 8
Comparative Balance Sheet of Kala Ltd. as on 31st March 2017 and 31st March 2018
Financial Statements Solutions Samacheer Kalvi 12th Accountancy Solutions Chapter 8

Question 6.
Prepare common – size income statement for the following particulars of Raja Ltd. for the year ended 31st March, 2017.
Financial Statement Analysis Chapter 8 Solutions Samacheer Kalvi 12th Accountancy
Answer:
Financial Statement Analysis Solutions Samacheer Kalvi 12th Accountancy Chapter 8

Question 7.
From the following particulars of Maria Ltd. and Kala Ltd. prepare a common – size income statement for the year ended 31st March, 2019.
Financial Statement Analysis Problems And Solutions Pdf Samacheer Kalvi 12th Accountancy Solutions Chapter 8
Answer:
Common Size Income statement for the year ended 31st March 2019
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 14

Question 8.
Prepare common – size income statement for the following particulars of Sam Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 16
Answer:
Common Size Income statement of Sam Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 17

Question 9.
Prepare Common – size balance sheet of Meena Ltd. as on 31st March, 2018.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 18
Answer:
Common Size Balance sheet of Meena Ltd. on 31st March 18
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 19

Question 10.
Prepare common – size statement of financial position for the following particulars of Rani Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 20
Answer:
Common Size Balance sheet of Rani Ltd. as on 31st March 2016 to 31st March 2017
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 21

Question 11.
Prepare common – size statement of financial position for the following particulars of Yasmin Ltd. and Sakthi Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 22
Common Size statement of Balance sheet
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 23

Question 12.
From the following particulars, calculate the trend percentages of Kala Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 42
Answer:
Trend Analysis of Kala Ltd Trend Percentage
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 25

Question 13.
From the following particulars, calculate the Trend percentages of Kavitha Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 26
Answer:
Trend Percentage of Kavitha Ltd (in thousands)
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 27

Question 14.
From the following particulars, calculate the trend percentages of Kumar Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 28
Answer:
Trend Percentage of Kumar Ltd (in thousands ₹)
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 29

Question 15.
From the following particulars, calculate the trend percentages of Anu Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 30
Answer:
Trend Percentage of Anu Ltd (In Thousands)
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 31

Question 16.
From the following particulars, calculate the trend percentages of Babu Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 32
Answer:
Trend Percentage of Babu Ltd (In Thousands)
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 33

Samacheer Kalvi 12th Accountancy Financial Statement Analysis Additional Questions and Answers

I. Choose the correct answer

Question 1.
Financial statements are meaningful and useful only when they are ……………..
(a) verified
(b) presented to owners
(c) Analysed and Interpreted
(d) Published
Answer:
(c) Analysed and Interpreted

Question 2.
Interpretation of Financial statements includes processes like ……………..
(a) Journalising
(b) Ledger writing
(c) Establishing relationship between the account data
(d) None of these
Answer:
(c) Establishing relationship between the account data

Question 3.
Trend analysis is significant for ……………..
(a) Profit planning
(b) Working capital management
(c) Capital rationing
(d) Forecasting and Budgeting
Answer:
(d) Forecasting and Budgeting

Question 4.
The three most useful general purpose financial statements for management are ……………..
(a) Income statement, Statement of Retained Earning and Balance Sheet
(b) Income statement, Balance sheet, and statement of changes in financial position
(c) Income statemen, Statement of Retained Earnings, and Funds flow statement
(d) Statement of Retained Earnings, Balance sheet and Funds flow statement
Answer:
(b) Income statement, Balance sheet, and statement of changes in financial position

Question 5.
In the case of limited company, the term financial statement includes ……………..
(a) Profit and Loss Account and Balance Sheet
(b) Profit and Loss Account, Profit and loss Appropriation Account and Balance Sheet
(c) Balance Sheet
Answer:
(b) Profit and Loss Account, Profit and loss Appropriation Account and Balance Sheet

Question 6.
The term Current Assets does not include ……………..
(a) Payments in Advance
(b) Bills Receivable
(c) Long – Term Deferred Changes
Answer:
(c) Long – Term Deferred Changes

Question 7.
The following is a recorded fact ……………..
(a) Market value of Investment
(b) Debtors
(c) Replacement cost of machinery
Answer:
(b) Debtors

Question 8.
The term Fixed Assets includes ……………..
(a) Stock – in – Trade
(b) Furniture
(c) Payments in Advance
Answer:
(b) Furniture

II. Fill in the blanks

Question 9.
The two statements which are generally included in the definition of financial statements are ……………..
Answer:
Income statement and Balance sheet.

Question 10.
Income statement …………….. the revenues and costs incurred in the process of earning revenues.
Answer:
Matches.

Question 11.
Balance Sheet is a statement of …………….. of a business at a specific moment of time.
Answer:
Financial position.

Question 12.
Assets and liabilities in a Balance sheet may be arranged either according …………….. order or …………….. order.
Answer:
Liquidity, permanency.

Question 13.
Financial statements disclose only …………….. facts.
Answer:
Monetary.

Question 14.
Profit and loss account also called as …………….. statement.
Answer:
Income.

Question 15.
Rearrangement of figures is necessary for …………….. and ………………
Answer:
Analysis, Interpretation.

Question 16.
Analysis of Financial Statements is meant for deriving additional information for various …………….. parties.
Answer:
interested.

III. True or False

Question 17.
Financial statement analysis is carried out as per Government regulation.
Answer:
False.

Question 18.
Financial statement analysis is performed with the help of various techniques and tools.
Answer:
True.

Question 19.
Common – size statements is a technique used in Financial statement analysis.
Answer:
True.

IV. Short answer questions

Question 1.
What are the features of financial statements?
Answer:

  1. Financial statements are generally prepared at the end of an accouning period based on transactions recorded in the books of accounts.
  2. These statements are prepared for the organisation as a whole.
  3. Information is presented in a meaningful way by grouping items of similar nature such as fixed assets and current assets.
  4. Financial statements are prepared based on historical cost.
  5. Financial statements are prepared based on accounting principles and Accounting standards, which make financial statements comparable and realistic.
  6. Financial statements involve personal judgement in certain cases. For example, selection of method of depreciation and percentage of reserves, etc.

Question 2.
Explain the significance of financial statements.
Answer:

  1. To Management: Financial statements provide information to the management to take decision and to have control over business activities, in various areas.
  2. To Shareholders: Financial statements help the shareholders to know whether the business has potential for growth and to decide to continue their share holding.
  3. To potential investors: Financial statements help to value the securities and compare it with those of other business concerns before making their investment decisions.
  4. To Creditors: Creditors can get information about the ability of the business to repay the debts from financial statements.
  5. To Bankers: Information given in the financial statements is significant to the bankers to assess whether there is a adequate security to cover the amount of the loan or overdraft.
  6. To Government: Financial statements are significant to government to assess the tax liability of business concerns and to frame and amend industrial policies.
  7. To Employees: Through the financial statements, the employees can assess the ability of the business to pay salaries and whether they have future growth in the concern.

Question 3.
Explain the objectives of financial statement analysis.
Answer:

  1. To analyse the profitability and earing capacity.
  2. To study the long term and short term solvency of the business.
  3. To determine the efficiency in operations and use of assets.
  4. To determine the efficiency of the management and employees.
  5. To determine the trend in sales and production, etc.
  6. To forecast for future and prepare budgets.
  7. To make inter – firm and intra – firm comparisons.

V. Exercise

Question 1.
From the following prepare “Comparative Statement of Proft and Loss” of Good Service Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 34
Answer:
Comparative Statement of Profit and Loss for the years ended 31st March 2012 and 2013
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 35

Question 2.
Prepare Common – size statement of Genius Paper work Ltd.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 36
Answer:
Common – size Income Statement for the year ended 31st March 2016 and 2017
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 37

Question 3.
From the following figures compute trend percentage using 2005 as the base year.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 38
Answer:
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 39

Question 4.
From the following balance sheet extracts, compute trend percentage and comment on the liquidity positon of X ltd. You may take 1990 as base year.
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 40
Answer:
Statement showing Trend Percentages
Samacheer Kalvi 12th Accountancy Solutions Chapter 8 Financial Statement Analysis 41

Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements – I

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements – I

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Samacheer Kalvi 12th Chemistry p-Block Elements – I TextBook Evalution

I. Choose the correct answer:

12th Chemistry Chapter 2 Book Back Answers Question 1.
An aqueous solution of borax is …………
(a) neutral
(b) acidic
(c) basic
(d) amphoteric
Answer:
(c) basic.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-1

12th Chemistry Lesson 2 Book Back Answers Question 2.
Boric acid is an acid because its molecule …………
(a) contains replaceable H+ ion
(b) gives up a proton
(c) combines with proton to form water molecule
(d) accepts OH from water, releasing proton.
Answer:
(d) accepts OH from water, releasing proton

Hint: B(OH)3 + H2O \(\rightleftharpoons\) [B(OH)4]+ H+

12th Chemistry Samacheer Kalvi Question 3.
Which among the following is not a borane?
(a) B2H6
(b) B3H6
(C) B4H10
(d) none of these
Answer:
(a) B2H6
Hint:

  • Nido borane – BnH4+n
  • aracno borane – BnH6+n , B3H6 is not a borane

Samacheer Kalvi 12th Chemistry Question 4.
Which of the following metals has the largest abundance in the earth’s crust?
(a) Aluminium
(b) calcium
(c) Magnesium
(d) Sodium
Answer:
(a) Aluminium

Samacheer Kalvi Guru 12th Chemistry Question 5.
In diborane, the number of electrons that accounts for banana bonds is …………
(a) six
(b) two
(c) four
(d) three
Answer:
(c) four

Hint: There are two 3c – 2e bonds i.e., the bonding in the bridges account for 4 electrons.

Samacheer Kalvi 12th Chemistry Solutions Question 6.
The element that does not show catenation among the following p-block elements is …………
(a) Carbon
(b) silicon
(c) Lead
(d) germanium
Answer:
(c) Lead

12th Chemistry Solutions Samacheer Kalvi Question 7.
Carbon atoms in fullerene with formula C60 have …………
(a) sp3 hybridised
(b) sp hybridised
(c) sp2 hybridised
(d) partially sp2 and partially sp3 hybridised
Answer:
(c) sp2 hybridised

12 Chemistry Samacheer Kalvi Question 8.
Oxidation state of carbon in its hydrides …………
(a) +4
(b) -4
(c) +3
(d) +2
Answer:
(a) +4
Hint: CH4+in which the oxidation state of carbon is 4.

Samacheer 12 Chemistry Solutions Question 9.
The basic structural unit of silicates is …………
(a) (SiO3)2-
(b) (SiO4)2-
(c) (SiO)
(d) (SiO4)4-
Answer:
(d) (SiO4)4-

Chemistry Class 12 Samacheer Kalvi Question 10.
The repeating unit in silicone is …………
12th Chemistry Chapter 2 Book Back Answers P-Block Elements - I Samacheer Kalvi
Answer:
12th Chemistry Lesson 2 Book Back Answers P-Block Elements - I Samacheer Kalvi

Class 12 Chemistry Samacheer Kalvi Question 11.
Which of these is not a monomer for a high molecular mass silicone polymer?
(a) Me3SiCl
(b) PhSiCl3
(c) MeSiCl3
(d) Me3SiCl3
Answer:
(a) Me3SiCl

Samacheerkalvi.Guru 12th Chemistry Question 12.
Which of the following is not sp2 hybridised?
(a) Graphite
(b) graphene
(c) Fullerene
(d) dry ice
Answer:
(a) dry ice
Hint: dry ice – solid CO2 in which carbon is in sp hybridized state

Question 13.
The geometry at which carbon atom in diamond are bonded to each other is …………
(a) Tetrahedral
(b) hexagonal
(c) Octahedral
(d) none of these
Answer:
(a) Tetrahedral

Question 14.
Which of the following statements is not correct?
(a) Beryl is a cyclic silicate
(b) Mg2SiO4 is an orthosilicate
(c) SiO44- is the basic structural unit of silicates
(d) Feldspar is not aluminosilicate
Answer:
(d) Feldspar is a three dimensional silicate

Question 15.
AlF3 is soluble in HF only in the presence of KF. It is due to the formation of ………… [NEET]
(a) K3[AlF3H3]
(b) K3[AlF6]
(C) AlH3
(d) K[AlF3H]
Answer:
(6)K3[AlF6]
Hint: AlF3 + 3KF → K3[AlF6]

Question 16.
Match items in column – I with the items of column – II ans assign the correct code
12th Chemistry Samacheer Kalvi Solutions Chapter 2 P-Block Elements - I
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 P-Block Elements - I
Answer:
(a) A – 2, B – 1, C – 4, D – 3

Question 17.
Duralumin is an alloy of …………
(a) Cu, Mn
(b) Cu, Al, Mg
(c) Al, Mn
(d) Al, Cu, Mn, Mg
Answer:
(d) Al, Cu, Mn, Mg
Hint: Al – 95% , Cu – 4% , Mn – 0.5% , Mg – 1.1 %

Question 18.
Thermodynamically the most stable form of carbon is …………
(a) Diamond
(b) graphite
(c) Fullerene
(d) none of these
Answer:
(b) graphite

Question 19.
The compound that is used in nuclear reactors as protective shields and control rods is …………
(a) Metal borides
(b) metal oxides
(c) Metal carbonates
(d) metal carbide
Answer:
(a) Metal borides

Question 20.
The stability of +1 oxidation state increases in the sequence …………
(a) Al < Ga < In < Tl
(b) Tl < In < Ga < Al
(c) In < Tl < Ga < Al
(d) Ga< In < Al < Tl
Answer:
(a) Al < Ga < In < Tl

II. Answer the following questions:

Question 1.
Write a short note on anamolous properties of the first element of p-block.
Answer:
In p-block elements the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour.

  1. Small size of the first member.
  2. High ionisation enthalpy and high electronegativity.
  3. Absence of d-orbitals in their valance shell.

The first member of the group-13, boron is a metalloid while others are reactive metals. Moreover, boron shows diagonal relationship with silicon of group -14. The oxides of boron and silicon are similar in their acidic nature.

Question 2.
Describe briefly allotropism in p- block elements with specific reference to carbon.
Answer:
Some elements exist in more than one crystalline or molecular forms in the same physical state. This phenomenon is called allotropism. Most common allotropes of carbon are,

  1. Graphite
  2. Diamond
  3. Fullerenes
  4. Carbon nanotubes
  5. Graphene.

1. Graphite:

  • It is the most stable allotropic form of carbon at normal temperature and pressure.
  • It is soft and conducts electricity.
  • It is composed of flat two dimensional sheets of carbon atoms.
  • Each sheet is a hexagonal net of sp2 hybridised carbon atoms with a C – C bond length of 1.41 A.
  • Structure of graphite,

Samacheer Kalvi Guru 12th Chemistry Solutions Chapter 2 P-Block Elements - I

2. Diamond:

  • It is very hard.
  • The carbon atoms in diamond are sp1 hybridised, with a C – C bond length of 1.54 A.
  • In the diamond, carbon atoms are arranged in tetrahedral manner.
  • Structure of Diamond,

Samacheer Kalvi 12th Chemistry Solutions Chapter 2 P-Block Elements - I

3. Fullerenes:

  • It is newly synthesised allotropes of carbon.
  • The C60 molecules have a soccer ball like structure and is called buckminster fullerene or buckyballs.
  • It has a fused ring structure consists of 20 six membered rings and 12 five membered rings.
  • Each carbon atom is sp2 hybridised.
  • The C – C bond distance is 1.44 A and C = C distance is 1.38 A.
  • Structure of fullerene,

12th Chemistry Solutions Samacheer Kalvi Chapter 2 P-Block Elements - I

4. Carbon nanotubes:

  1. It is recently discovered allotropes, have graphite like tubes with fullerene ends.
  2. These nanotubes are stronger than steel and conduct electricity.
  3. Structure of Carbon nanotubes.

12 Chemistry Samacheer Kalvi Solutions Chapter 2 P-Block Elements - I

5. Graphene:

  • It has a single planar sheet of sp2 hybridised carbon atoms that are densely packed in a honeycomb crystals lattice.
  • Structure of Graphene,

Samacheer 12 Chemistry Solutions Chapter 2 P-Block Elements - I

Question 3.
Boron does not react directly with hydrogen. Suggest one method to prepare diborane from BF3.
Answer:
Boron does not react directly with hydrogen. However it forms a variety of hydrides called boranes. Treatment of gaseous boron trifluoride with sodium hydride around 450 K gives diborane.
Chemistry Class 12 Samacheer Kalvi Chapter 2 P-Block Elements - I

Question 4.
Give the uses of Borax.
Answer:
Uses of borax:

  • Borax is used for the identification of coloured metal ions.
  • In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
  • It is also used as a flux in metallurgy and also acts as a good preservative.

Question 5.
What is catenation ? describe briefly the catenation property of carbon.
Answer:
Catenation:
It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to maximum extent because it is small in size and can form pn-pn multiple bonds to itself. The following conditions are necessary for catenation.

  1. The valency of element is greater than or equal to two.
  2. Element should have an ability to bond with itself.
  3. The self bond must be as strong as its bond with other elements.
  4. Kinetic inertness of catenated compound towards other molecules. Carbon possesses all the above properties and forms a wide range of compounds with itself.

Question 6.
Write a note on Fisher tropsch synthesis.
Answer:
The reaction of carbon monoxide with hydrogen at a pressure of less than 50 atm using metal catalysts at 500-700 K yields saturated and unsaturated hydrocarbons.
Class 12 Chemistry Samacheer Kalvi Chapter 2 P-Block Elements - I

Question 7.
Give the structure of CO and CO2.
Answer:
Structure of CO:
Samacheerkalvi.Guru 12th Chemistry Solutions Chapter 2 P-Block Elements - I
Structure of CO2:
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-14

Question 8.
Give the uses of silicones.
Answer:
Uses of silicones:

  1. Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc.
  2. They are used for making water proofing clothes.
  3. They are used as insulting material in electrical motor and other appliances
  4. They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.

Question 9.
AlCl3 behaves like a lewis acid. Substantiate this statement.
Answer:
In AlCl3, Al in electron deficient it needs two electrons to complete octet so it act as lewis acid. AlCl3 usually exist as a dimer to achieve octet by bridged Cl atom electron deficient compounds are lewis acids.

Question 10.
Describe the structure of diborane.
Answer:
In diborane two BH2 units are linked by two bridged hydrogens. Therefore, it has eight B-H bonds. However, diborane has only 12 valance electrons and are not sufficient to form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre – two electron bond or 2c-2e bond). The remaining four electrons have to used for the bridged bonds, i.e. two three centred B-H-B bonds utilise two electrons each.

Hence, these bonds are three centre – two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure. In dibome, the boron is sp3 hybridised. Three of the four sp3 hybridised orbitals contains single electron and the fourth orbital is empty.

Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre – two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled 1s orbital of hydrogen.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-15

Question 11.
Write a short note on hydroboration.
Answer:
Diborane adds on to alkenes and alkynes in ether solvent at room temperature. This reaction is called as hydroboration and is highly used in synthetic organic chemistry especially for anti-Markovnikov addition.
B2H6 + 3RCH = CHR → B( CH2 – CH2R )3+ 6H2

Question 12.
Give one example for each of the following:

  1. icosogens
  2. tetragen
  3. prictogen
  4. chalcogen

Answer:
1. Icosogens:

  • Boron
  • Aluminium
  • Gallium

2. Tetragen:

  • Carbon
  • Silicon
  • Germanium

3. Prictogen:

  • Oxygen
  • Sulfur
  • Selenium

4. Chalcogen:

  • Fluorine
  • Chlorine
  • Bromine

Question 13.
Write a note on metallic nature of p-block elements.
Answer:

  1. The tendency of an element to form a cation by loosing electrons is known as electro¬positive or metallic character.
  2. This character depends on the ionisation energy.
  3. Generally on descending a group the ionisation energy decreases and hence the metallic character increases.

In p-block, the elements present in lower left part are metals while the elements in the upper right part are non metals. Elements of group 13 have metallic character except the first element boron which is a metalloid, having properties intermediate between the metal and nonmetals. The atomic radius of boron is very small and it has relatively high nuclear charge and these properties are responsible for its nonmetallic character.

In the subsequent groups the non-metallic character increases. In group 14 elements, carbon is a nonmetal while silicon and germanium are metalloids. In group 15, nitrogen and phosphorus are non metals and arsenic & antimony are metalloids. In group 16, oxygen, sulphur and selenium are non metals and tellurium is a metalloid. All the elements of group 17 and 18 are non metals.

Question 14.
Complete the following reactions:
(a) B(OH)3 + NH3
(b) Na2B4O7 + H2SO4+ H2O →
(c) B2H6 + 2NaOH + 2H2O →
(d) B2H6 + CH3OH →
(e) BF3 + 9H2O →
(f) HCOOH + H2SO4
(g) SiCl4 + NH3
(h) SiCl4 + C2H5OH →
(i) B + NaOH →
(j) H2B4O7 \(\underrightarrow { Red\quad hot }\)

Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-16

Question 15.
How will you identify borate radical?
Answer:
When boric acid or borate salt is heated with ethyl alcohol in presence of concentrated H2SO4, an ester triethyl borate is formed. The Vapour of this ester burns with a green edged flame and this reaction is used to identify the presence of borate.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-17

Question 16.
Write a note on zeolites.
Answer:
Zeolites:

  1. Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxygen in their regular three dimensional framework.
  2. They are hydrated sodium alumino silicates with general formula, Na2O. (Al2O3). x(SiO2)y(H2O) (x = 2 to 10; y = 2 to 6)
  3. Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
  4. The Si and Al atoms are tetrahederally coordinated with each other through shared oxygen atoms.
  5. Zeolites structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
  6. Zeolite crystal to act as a molecular sieve. They helps to remove permanent hardness of water.

Question 17.
How will you convert boric acid to boron nitride?
Answer:
Fusion of urea with boric acid B(OH)3, in an atmosphere of ammonia at 800 – 1200 K gives
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-18

Question 18.
A hydride of 2nd period alkali metal
(A) on reaction with compound of Boron
(B) to give a reducing agent
(C) identify A, B and C.
Answer:

  1. A hydride of 2nd period alkali metal (A) is lithium hydride (LiH).
  2. Lithium hydride (A) reacts with diborane (B) to give lithiumborohydride (C) which is act as reducing agent.
    Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-19

(A) Lithium hydride – LiH
(B) Diborane – B2H6
(C) Lithium borohydride – LiBH4

Question 19.
A double salt which contains fourth period alkali metal
(A) on heating at 500K gives
(B). Aqueous solution of (B) gives white precipitate with BaCl2 and gives a red colour compound with alizarin. Identify A and B.
Answer:
1. A double salt which contains fourth period alkali metal (A) is potash alum
K2SO4. Al2(SO4)3. 24H2O

2. On heating potash alum (A) 500 k give anhydrous potash alum (or) burnt alum (B).
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-20

3. Aqueous solution of burnt alum, has sulphates ion, potassium ion and aluminium ion. Sulphate ion reacts with BaCl2 to form white precipitate of Barium Sulphate
(SO4)2 + BaCl2 → BaSO4 + 2Cl
Aluminium ion reacts with alizarin solution to give a red colour compound.

Question 20.
CO is a reducing agent. Justify with an example.
Answer:
Both thermodynamic and kinetic factors make carbon monoxide (CO) a better reducing agent. When CO is used to reduce a metal oxide, it gets oxidized to CO2 Thermodynamically, CO2 is much more stable than CO. For example,
CO + Fe2O3 → 2Fe + 3CO2

Samacheer Kalvi 12th Chemistry p-Block Elements – I Evaluate yourself

Question 1.
Why group 18 elements are called inert gases? Write the general electronic configuration of group 18 elements.
Answer:
The elements of group-18 have completely filled s and p orbitals, hence they are more stable and have least reactivity. Therefore group-18 elements are called inert gases. ns2np6 is the general electronic configuration of group elements.

Samacheer Kalvi 12th Chemistry p-Block Elements – I Additional Questions

Samacheer Kalvi 12th Chemistry p-Block Elements – I 1 Mark Questions and Answers

I. Choose the correct answer:

Question 1.
More common oxidation state for halogens is …………
(a) +1
(b) +2
(c) -1
(d) -2
Answer:
(c) -1

Question 2.
Electronic configuration of noble gases is …………
(a) ns2
(b) ns2np5
(c) ns1np6
(d) ns2np6
Answer:
(d) ns2np6

Question 3.
Noble gases are chemically inert. This is due to …………
(a) unstable electronic configuration
(b) stable electronic configuration
(c) only filled p-orbital
(d) only filled 5-orbital
Answer:
(b) stable electronic configuration

Question 4.
Consider the following statements.
(i) The first member of the group-13, boron is a metalloid while others are reactive metals.
(ii) The oxides of boron and silicon are similar in their acidic nature.
(iii) Both boron and silicon form metallic hydrides.

Which of the above statement(s) is/are not correct?
(a) (i) only
(b) (ii) only
(c) (ii) and (iii)
(d) (iii) only
Answer:
(d) (iii) only

Question 5.
Which element has a greater tendency to form a chain of bonds with itself?
(a) Boron
(b) Silicon
(c) Tin
(d) Carbon
Answer:
(d) Carbon

Question 6.
Which one of the following is the strongest oxidising agent?
(a) Fluorine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer:
(a) Fluorine

Question 7.
Some elements exist in more than one crystalline or molecular forms in the same physical state is called …………
(a) isomerism
(b) allotropism
(c) isomorphism
(d) isoelectronics
Answer:
(b) allotropism

Question 8.
How many allotropes possible for boron?
(a) 1
(b) 4
(c) 6
(d) 7
Answer:
(c) 6

Question 9.
Important ore of boron is …………
(a) bauxite
(b) borosilicate
(c) borax
(d) P-tetragonal boron
Answer:
(c) borax

Question 10.
Less reactive elements in boron family is …………
(a) Boron
(b) Aluminium
(c) Gallium
(d) Thallium
Answer:
(b) Aluminium

Question 11.
More toxic element in boron family is …………
(a) Boron
(b) Aluminium
(c) Gallium
(d) Thallium
Answer:
(d) Thallium

Question 12.
Boron does not …………
(a) Oxygen
(b) Hydrogen
(c) Acids
(d) Alkali
Answer:
(b) Hydrogen

Question 13.
Borontrifluoride reacts with sodium hydride at 450 K gives …………
(a) diborane
(b) tetraborane
(c) pentaborane
(d) decaborane
Answer:
(a) diborane

Question 14.
2B + N2 \(\underrightarrow { \triangle }\) A . Identify A
(a) BN3
(b) B3N
(c) (BN)3
(d) BN
Answer:
(d) BN

Question 15.
Boron reacts with fused sodium hydroxide to forms …………
(a) Borax
(b) Boric acid
(c) Sodium borate
(d) Sodium tetraborate
Answer:
(c) Sodium borate

Question 16.
Which isotope is used as moderator in nuclear reactors?
(a) 10B5
(b) nC6
(c) 4He2
(d) 40Ca2
Answer:
(a) 10B5

Question 17.
Borax is ………… in nature.
(a) basic
(b) acidic
(c) amphoteric
(d) chemically inert
Answer:
(a) basic

Question 18.
The compound used as a flux in metallurgy?
(a) Boron nitride
(b) Boron oxide
(c) Boron fluoride
(d) Borax
Answer:
(d) Borax

Question 19.
The trialkylborate on reaction with sodiumhydride in tetrahydrofuran to form …………
(a) NaBH4
(b) Na[BH(OR)3]
(c) Na[B(OR)3]
(d) Na[BH(OR)3]
Answer:
(b) Na[BH(OR)3]

Question 20.
Compounds used as an eye lotion …………
(a) H3BO3
(b) HBO2
(c) H2B4O7
(d) B2O3
Answer:
(a) H3BO3

Question 21.
Which one of the following is highly reactive compound?
(a) B2OH3
(b) H2BO3
(c) HBO2
(d) B2H6
Answer:
(d) B2H6

Question 22.
B2H6 + 6CH3OH → A Identify A
(a) B2O3
(b) CH3OB
(c) CH3OH
(d) B(OCH3)3
Answer:
(d) B(OCH3)3

Question 23.
Which one of the following is called as inorganic benzene?
(a) B2H6
(b) BN
(c) H2B4O7
(d) B3N3H6
Answer:
(d) B3N3H6

Question 24.
Compound contains two centred – two electron bond (2c-2e) is …………
(a) B6H11
(b) B5H9
(c) B2H6
(d) B10H11
Answer:
(c) B2H6

Question 25.
Diborane reacts with excess ammonia at high temperature to give …………
(a) Boron nitride
(b) Boron oxide
(c) Borazole
(d) Diborane diammonate
Answer:
(c) Borazole

Question 26.
Consider the following statements.
(i) Diborane contains two centre-two electron bond.
(ii) In diborane, the boron has sp3 hybridis ed.
(iii) Diborane has two terminal B – H bonds and four B – H – B bonds.

Which of the above statement(s) is/are correct.
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) only
(d) (i) and (ii)
Answer:
(d) (i) and (ii)

Question 27.
Compound used for propellant is …………
(a) BN
(b) H2B4O7
(c) B2H2
(d) Borax
Answer:
(c) B2H2

Question 28.
Which one of the following is double salt?
(a) Potash alum
(b) Potassium sulphate
(c) Aluminium Sulphate
(d) Ammonium sulphate
Answer:
(a) Potash alum

Question 29.
Consider the following statements.
(i) Alums are more soluble in hot water than in cold water.
(ii) Alums are more soluble in cold water than in hot water.
(iii) Potash alum is employed a styptic agent to arrest bleeding.

Which of the above statement(s) is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iii)
Answer:
(b) (ii) only

Question 30.
The structure of graphite is …………
(a) planner
(b) hexagonal
(c) octahedral
(d) bucky balls
Answer:
(b) hexagonal

Question 31.
Which one of the following is used as a lubricant?
(a) Graphite
(b) Diamond
(c) Fullerene
(d) Graphene
Answer:
(a) Graphite

Question 32.
Which one of the following carbon allotrope is very hard?
(a) Graphite
(b) Diamond
(c) Fullerene
(d) Graphene
Answer:
(b) Diamond

Question 33.
Recently discovered allotropes of carbon is …………
(a) Graphite
(b) Diamond
(c) Carbon nanotubes
(d) Fullerenes
Answer:
(c) Carbon nanotubes

Question 34.
CO and N2 mixture is …………
(a) natural gas
(b) producer gas
(c) water gas
(d) LPG
Answer:
(b) producer gas

Question 35.
Syn gas is …………
(a) CO + N2
(b) CO + H2
(c) CO2 + H2
(d) CO2 + N2
Answer:
(b) CO + H2

Question 36.
Ethene is mixed with carbon monoxide and hydrogen gas to produce propanal is known as …………
(a) Oxo process
(b) McAfee process
(c) Wacker process
(d) Haber process
Answer:
(a) Oxo process

Question 37.
………… is a good reducing agent.
(a) CO
(b) CO2
(c) [Cr(CO)6]
(d) [Ni(CO)4]
Answer:
(a) CO

Question 38.
Critical temperature of CO2 is ………..
(a) -31°C
(b) -13°C
(c) 31°C
(d) 13°C
Answer:
(c)31°C

Question 39.
Which one of the following compound is important for photosynthesis?
(a) CO
(b) CO2
(c) COCl2
(d) C
Anwer:
(b) CO

Question 40.
General empirical formula of silicone is ………..
(a) (R2SiO)
(b) (RSiO)
(c) (R2CO)
(d) (RSiH)
Answer:
(a) (R2SiO)

Question 41.
Consider the following statements.
(i) All Silicones are hydrophilic in nature.
(ii) Silicones are thermal and electrical insulators.
(iii) Chemically silicones are highly reactive.

Which of the above statement(s) is/are not correct?
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) (ii) only
Answer:
(b) (i) and (iii)

Question 42.
Silicate contains ……….. silicon and oxygen in units.
(a) [SiO2]4-
(b) [SiO2]
(c) [SiO2]2-
(d) [SiO2]
Answer:
(a) [SiO2]4-

Question 43.
Ortho silicates are also called as ………..
(a) Ino silicates
(b) Soro silicates
(c) Neso silicates
(d) Cyclic silicates
Answer:
(c) Neso silicates

Question 44.
Example of Ring silicate is ………..
(a) Olivine
(b) Beryl
(c) Spodumene
(d) Asbestos
Answer:
(b) Beryl

Question 45.
Pick out the three dimensional silicates?
(a) Talc
(b) Mica
(c) Quartz
(d) Asbestos
Answer:
(c) Quartz

Question 46.
Compound used to remove the permanent hardness of water is ………..
(a) Zeolite
(b) Feldspar
(c) Talc
(d) Mica
Answer:
(a) Zeolite

II. Fill in the blanks:

  1. …………… is the general electronic configuration of tetragen elements.
  2. Boron and silicon form …………… hydrides.
  3. …………… is most reactive element among the halogens.
  4. Most stable oxidation state of aluminium is ……………
  5. General formula of metal boride is ……………
  6. Boron has the capacity to absorb ……………
  7.  …………… is used as a rocket fuel igniter.
  8. …………… is essential for the cell walls of plants.
  9. …………… is a chemical formula of boron.
  10. …………… is used for the identification of coloured metal ions.
  11. …………… is a colourless transparent crystal.
  12. Boric acid has a …………… structure.
  13. Boric acid consists of …………… unit.
  14. On heating magnesium boride with HCl a mixture of volatile …………… are obtained.
  15. Diborane reacts with methylalcohol to give ……………
  16. Diborane has …………… B – H bonds.
  17. In diborane, the boron is …………… hybridised.
  18.  …………… is inorganic benzene.
  19. Boron trifluoride has a …………… geometry.
  20. Anhydrous aluminium chloride is a …………… substance.
  21. With excess of NaOH, aluminium chloride produces ……………
  22. …………… is used for the manufacture of petrol by cracking the mineral oils.
  23. …………… is used for water proofing and textiles.
  24. ……………is the most stable allotropic form of carbon at normal temperature and pressure.
  25. …………… is allotropic form of carbon has aromatic character.
  26. Graphene has …………… lattice.
  27. Carbonyl chloride is a …………… gas.
  28. Equimolar mixture of hydrogen and carbonmonoxide is called as ……………
  29. Producer gas is a mixture of …………… and ……………
  30. Aqueous solution of carbon dioxide forms ……………
  31. …………… are high temperature polymers.
  32. The viscosity of silicon oil remains ……………
  33.  …………… are used as insulating material.
  34. In beryl, each aluminium is surrounded by ……………
  35. …………… is carcinogenic silicates.
  36. Three dimensional by replacing units by ……………
  37. …………… crystal to act as a molecular sieve.
  38. Borax is a sodium salt of ……………

Answers:

  1. ns2np2
  2. covalent
  3. Fluorine
  4. +3
  5. MxBy
  6. neutrons
  7. Amorphous boron
  8. Boron
  9. Na2B4O7. 10H2O
  10. Borax
  11. Boric acid
  12. two dimensional layered
  13. [BO3]3-
  14. borones
  15. trimethylborate
  16. eight
  17. sp3
  18. Borazine
  19. planar
  20. hygroscopic
  21. Sodium meta aluminate
  22. AlCF
  23. Potash alum
  24. Graphite
  25. Fullerene
  26. honey comb nitrogen
  27. poisonous
  28. water gas
  29. carbonmonoxide, nitrogen
  30. carbonic acid
  31. silicones
  32. constant
  33. Silicones
  34. six oxygen
  35. Asbestos
  36. [SiO4]4-; [AlO4]5-
  37. Zeolite
  38. tetraboric acid

III. Match the following:

Question 1.
(i) Tetragens – (a) Oxygen
(ii) Icosagens – (b) Carbon
(iii) Chalcogens – (c) Nitrogen
(iv) Pnictogens – (d) Boron
Answer:
(i) – (b)
(ii) – (d)
(iii) – (a)
(iv) – (c)

Question 2.
(i) Borax – (a) Na2B4O7
(ii) Prismatic form – (b) Na2B4O7. 5H2O
(iii) Jeweller borax – (c) Na2B4O7. 10H2O
(iv) Borax glass – (d) [B4O5(OH)4]2
Answer:
(i) – (c)
(ii) – (d)
(iii) – (b)
(iv) – (a)

Question 3.
(i) Boron – (a) Optical
(ii) Borax – (b) Neutron absorber
(iii) Boric acid – (c) Welding torches
(iv) Diborane – (d) Eye lotion
Answer:
(i) – (b)
(ii) – (a)
(iii) – (d)
(iv) – (c)

Question 4.
(i) Graphene – (a) Honeycomb crystal
(ii) Diamond – (b) Aromatic character
(iii) Fullerene – (c) Lubricant
(iv) Graphite – (d) Very hard
Answer:
(i) – (c)
(ii) – (d)
(iii) – (b)
(iv) – (a)

Question 5.
(i) Ortho silicate
(ii) Pyro silicate
(iii) Cyclic silicate
(iv) Tecto silicate
Answer:
(i) – (d)
(ii) – (a)
(iii) – (b)
(iv) – (c)

IV. Assertion and reason:

Note:
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
(a) A and R are correct and R explains A.
(b) A and R are correct and R not explains A.
(c) A is correct but R is wrong.
(d) A is wrong but R is correct.

Question 1.
Assertion (A) – Noble gases are least reactivity.
Reason (R) – Noble gases have completely filled s and p-orbital and attain stable electronic configuration.
Answer:
(a) A and R are correct and R explains A.

Question 2.
Assertion (A) – Both boron and silicon form covalent hydrides.
Reason (R) – Boron does not shows diagonal relationship with silicon of group 14.
Answer:
(c) A is correct but R is wrong.

Question 3.
Assertion (A) – Fluorine is most reactive element among the halogens.
Reason (R) – Fluorine has minimum bond dissociation energy.
Answer:
(a) A and R are correct and R explains A.

Question 4.
Assertion (A) – Boron combines with halogen to form trihalides at high temperatures.
Reason (R) – Boron does not reacts directly with hydrogen.
(b) A and R are correct and R not explains A.

Question 5.
Assertion (A) – Boron-10 isotope is used as moderator in nuclear reactors.
Reason (R) – Boron has the capacity to absorb neutrons.
Answer:
(a) A and R are correct and R explains A.

Question 6.
Assertion (A) – Diborane is highly reactive.
Reason (R) – At high temperatures, diborane forms higher boranes.
Answer:
(b) A and R are correct and R not explains A.

Question 7.
Assertion (A) – BF3 reacts with ammonia to form complex.
Reason (R) – BF3 is a electron deficient compound and accepts electron pairs to form coordinate covalent bonds.
Answer:
(a) A and R are correct and R explains A.

Question 8.
Assertion (A) – Fullerene has aromatic character.
Reason (R) – Some of the fullerenes have e-bonds and delocalised n-bonds.
Answer:
(a) A and R are correct and R explains A.

Question 9.
Assertion (A) – Carbon dioxide is non flammable gas.
Reason (R) – CO2 critical temperature is 31°C and can be readily liquefied.
Answer:
(b) A and R are correct and R not explains A.

Question 10.
Assertion (A) – Zeolites act as a molecular sieve.
Reason (R) – Zeolite structure is pore/ channel sizes are nearly uniform.
Answer:
(a) A and R are correct and R explains A.

V. Find the odd one out and given the reason:

Question 1.
(a) Icosagens
(b) Tetragens
(c) Alkali metals
(d) Chalcogens
Answer:
(c) Alkali metals
Reason: Alkali metals are s-block elements but other are p-block elements.

Question 2.
(a) Al
(b)B
(c) O
(d) Na
Answer:
(d) Na
Reason: Na is s-block element but others are p-block elements.

Question 3.
(a) Diborane
(b) Borax
(c) Carbonmonoxide
(d) Boric acid
Answer:
(c) Carbonmonoxide
Reason: Carbonmonoxide is not a compound of boron.

Question 4.
(a) B3N3H6
(b) B4H10
(c) B5H9
(d) B5H11
Answer:
(a) B3N3H6
Reason: B3N3H6 not a higher borane

Question 5.
(a) Potash alum
(b) Sodium alum
(c) Burnt alum
(d) Ammonium alum
Answer:
(c) Burnt alum
Reason: Burnt alum does not having water molecule.

Question 6.
(a) Graphite
(b) Borax
(c) Fullerene
(d) Diamond
Answer:
(b) Borax
Reason: Borax is not a allotropic form of carbon.

Question 7.
(a)B
(b) Ga
(c) In
(d) N
Answer:
(d) N
Reason: N is not a icosagens.

VI. Find out the correct pair:

Question 1.
(a) Ortho sililcate – Soro silicate
(b) Pyro silicate – Neso silicate
(c) Chain silicate – Pyroxenes
(d) Double chain silicate – Ring silicate
Answer:
(c) Chain silicate – Pyroxenes

Question 2.
(a) Graphene – 1.54 A
(b) Diamond – 1.40 A
(c) Fullerene – 1.38 A
(d) Graphite – 1.40 A
Answer:
(d) Graphite – 1.40 A

Question 3.
(a) Diborane – welding torches
(b) AlCl3 – Eye lotion
(c) Borax – pigments
(d) Boric acid – optical
Answer:
(a) Diborane – welding torches

Question 4.
(a) Inert gas – ns2np2
(b) Chalcogens – ns2np1
(c) Pnictogens – ns2np4
(d) Tetragens – ns2np4
Answer:
(c) Pnictogens – ns2np3

Question 5.
(a) Icosagen – Oxygen
(b) Tetragen – Carbon
(c) Chalcogen – Fluorine
(d) Halogen – Boron
Answer:
(b) Tetragen – Carbon

VII. Find out the incorrect pair:

Question 1.
(a) Tetragens – ns2np2
(b) Icosagens – ns2np1
(c) Chalcogens – ns2np4
(d) Halogens – ns2np6
Answer:
(d) Halogens – ns2np6

Question 2.
(a) Boron is BF3 – +3
(b) Carbon in CO2 – +4
(c) Nitrogen in N2O5 – +5
(d) Fluorine in OF2 – +4
Answer:
(d) Fluorine in OF2 – +4

Question 3.
(a) Nitrogen – Tetragens
(b) Oxygen – Chalcogens
(c) Tin – Tetragens
(d) Gallium – Icosagens
Answer:
(a) Nitrogen – Tetragens

Question 4.
(a) Boron – Moderator
(b) Borax – Eye lotion
(c) Boric acid – Antiseptic
(d) Diborane – Propellant
Answer:
(b) Borax – Eye lotion

Question 5.
(a) McAfee process – AlCl3
(b) Burnt alum – K2SO4-Al2(SO4)3
(c) Oxo process – Propanal
(d) Fischer tropsch Synthesis – HCOOH
Answer:
(d) Fischer tropsch Synthesis – HCOOH

Question 6.
(a) Soro silicate – Thortveitite
(b) Ring silicate – Beryl
(c) Sheet silicate – Asbestos
(d) Neso silicate – Olivine
Answer:
(c) Sheet silicate – Asbestos

Question 7.
(a) Pyroxenes – Phenacite
(b) Amphiboles – Asbestos
(c) Phyllo silicate – Mica
(d) Tecto silicate – Quartz
Answer:
(a) Pyroxenes – Phenacite

Samacheer Kalvi 12th Chemistry p-Block Elements – I 2 Mark Questions and Answers

Question 1.
What are Wade’s Rule?
Answer:
Wade’s rules are used to rationalize the shape of borane clusters by calculating the total number of skeletal electron pairs (SEP) available for cluster bonding.

Question 2.
Why-1 oxidation state is more common in halogens. Explain.
Answer:
The halogens have a strong tendency to gain an electron to give a stable halide ion with completely filled electronic configuration (ns2sp6) and hence-1 oxidation state is more common in halogens.

Question 3.
Why boron has non-metallic character?
Answer:
The atomic radius of boron is very small and it has relatively high nuclear charge and these properties are responsible for its non-metallic character.

Question 4.
Define inert pair effect.
Answer:
In heavier post-transition metals, the outer s-electron(m) have a tendency to remain inert and show reluctance to take part in the bonding (only p-orbital involved in chemical bonding), which is known as inert pair effect.

Question 5.
Mention the allotropes of boron.
Answer:

  1. Amorphous boron
  2. a-rhombohedral boron
  3. p-rhombohedral boron
  4. γ – rhombohedral boron
  5. α – tetragonal boron
  6. β – tetragonal boron

Question 6.
List out the allotropes of tin.
Answer:

  1. Grey tin
  2. White tin
  3. Rhombic tin
  4. Sigma tin

Question 7.
Mention the allotropes of phosphorous?
Answer:

  1. White phosphorous
  2. Red phosphorous (v)
  3. Scarlet phosphorous
  4. Violet phosphorous
  5. Black phosphorous

Question 8.
Give the two allotropes of sulphur.
Answer:

  1. Rhombus
  2. Monoclinic

Question 9.
Why boron compounds are covalent in nature?
Answer:
Many of boron compounds are electron deficient and has unusual type of covalent bonding, which is due to its small size, high ionisation energy and similarity in electronegativity with carbon and hydrogen.

Question 10.
Define Borax is basic in nature.
Answer:
Borax is basic in nature and its solution in hot-water is alkaline as it dissociates into boric acid and sodium hydroxide,
Na2B4O7 + 7H2O → 4H3BO3 + 2NaOH

Question 11.
Explain action of heat on borax.
Answer:
On heating borax, it forms a transparent borax beads.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-21

Question 12.
What happen when borax treated with ammonium chloride?
Answer:
When borax treated with ammonium chloride, it forms boron nitride.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-22

Question 13.
What happen when boric acids reacts with sodium hydroxide?
Answer:
Boric acid reacts with sodium hydroxide to form sodium metaborate and sodium tetraborate.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-23

Question 14.
Identify A and B from the following reaction,
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-24
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-25

Question 15.
Explain the action of air on diborane.
Answer:
At room temperature pure diborane does not react with air or oxygen but in impure form it gives [B2O3]3- along with large amount of heat.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-26

Question 16
Explain the action of air on diborane.
Answer:
At room temperature pure diborane does not react with air or oxygen but in impure form it gives B2O3 along with large amount of heat.
B2H6+ 3O2 → B2O3 + 3H2O

Question 17.
How will you convert diborane into sodium borohydride?
Answer:
Diborane reacts with sodium hydride in the presence of diglyme to give sodim borohydride.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-27

Question 18.
Mention the uses o boron trifluoride.
Answer:

  1. Boron trifluoride is used for preparing HBF4, a catalyst in organic chemistry
  2. It is also used as a fluorinating reagent.

Question 19.
What is MCA fee process?
Answer:
Aluminium chloride is obtained by heating a mixture of alumina and coke in a current of chloride.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-28

Question 20.
What happen when potash alum is treated with ammonium hydroxide?
Answer:
Potash alum forms aluminium hydroxide, when, treated with ammonium hydroxide
K2SO4. A12(SO4)3. 24H2O + 6NH4OH → K2SO4 + 3(NH4)2SO4 + 24H2O + 3Al(OH)3

Question 21.
What is producer gas? How will you prepare producer gas?
Answer:
On industrial scale carbon monoxide is produced by the reaction of carbon with air. The carbon monoxide formed will contain nitrogen gas also and the mixture of nitrogen and carbon monoxide is called producer gas.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-29

Question 22.
What is synthetic gas?
Answer:
A mixture of carbonmonoxide and hydrogen is called synthetic gas.

Question 23.
What is phosgene?
Answer:
When carbon monoxide is treated with chlorine in presence of light or charcoal, it forms a poisonous gas carbonyl chloride, when is also known as phosgene.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-30

Question 24.
How will you prepare propanal by oxoprocess?
Answer:
In oxoprocess, ethene is mixed with carbon monoxide and hydrogen gas to produce propanal.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-31

Question 25.
What are water gas equilibrium?
Answer:
The equilibrium involved in the reaction between carbon dioxide and hydrogen, has many industrial applications and is called water gas equilibrium.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-32

Question 26.
Mention the uses of silicon tetrachloride.
Answer:

  1. Silicon tetrachloride is used in the production of semiconducting silicon.
  2. It is used as a starting material in the synthesis of silica gel, silicic esters, a binder for ceramic materials.

Question 27.
What are silicones?
Answer:
Silicones or poly siloxanes are organo silicon polymers with general empirical formula (R2SiO). Since their empirical formula is similar to that of ketone (R2CO), they were named silicones.

Question 28.
What are high-temperature polymers?
Answer:
Silicones are high-temperature polymers, because they have high thermal stability.

Question 29.
Why silicones are water repellent?
Answer:
All silicones are water repellent, this is due to the presence of organic side groups that surrounds the silicon which makes the molecule looks like an alkane. Therefore silicones are water repellent.

Samacheer Kalvi 12th Chemistry p-Block Elements – I 3 Marks Questions and Answers

Question 1.
Write a notes on ionisation enthalpy in p-block elements?
Answer:
1. As we move down a group, generally there is a steady decrease in ionisation enthalpy of elements due to increase in their atomic radius.

2. In p-block elements there are some minor deviations to this general trend. In group 13, from B to Al the ionisation enthalpy decreases as expected. But from Al to Tl there is only a marginal difference. This is due to the presence of inner d- and f-elements which has poor shielding effect compared to s and p electrons. As a result, the effective nuclear charge on the valance electrons increase.

3. A similar trend is also observed in group 14. The remaining groups (15-18) follows the general trend, in these groups the ionisation enthalpy decreases as we move down the group. Here poor shielding effect of d- and f-electrons are overcome by the increased shielding effect of the additional p-electrons.

4. The ionisation enthalpy of elements in successive groups is higher than the corresponding elements of the previous group as expected.

Question 2.
Give the uses of boron.
Answer:

  1. Boron has the capacity to absorb neutrons. Hence, its isotope 10B5 is used as moderator in nuclear reactors.
  2. Amorphous boron is used as a rocket fuel igniter.
  3. Boron is essential for the cell walls of plants.
  4. Compounds of boron have many applications. For example eye drops, antiseptics, washing powders etc..contains boric acid and borax. In the manufacture of Pyrex glass , boric oxide is used.

Question 3.
How will be prepare borax from colemanite?
Answer:
Borax is a sodium salt of tetraboric acid. It is obtained from colemanite ore by boiling its solution with sodium carbonate.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-33

Question 4.
Explain the extraction of boric acid from,

    1. Borax
    2. Colemanite

Answer:
1. Borax – Borax is treated with hydrochloric acid or sulphuric acid or nitric acid to give boric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-34

2. Colemanite – Colemanite is boiling with water, to give boric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-35

Question 5.
Explain the action of heat on boric acid.
Answer:
Boric acid when heated at 373 K gives metaboric acid and at 413 K, it gives tetraboric acid. When heated at red hot, it gives boric anhydride which is a glassy mass.

  • 4 H3BO3 \(\underrightarrow { 373k }\) 4HBOz + H2O
  • 4HBO2 \(\underrightarrow { 413k }\) H2B4O7 + H2O
  • H2B4O7 \(\underrightarrow { Red hot }\) 2B2O3 + H2O

Question 6.
How will you convert, boric acid into,

  1. Boron trifluoride
  2. Borax

Answer:
1. Boron trifluoride:
Boric acid reacts with calcium fluoride in presence of cone. Sulphuric acid and gives boron trifluoride.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-36

2. Borax:
Boric acid, when heated with soda ash it gives borax.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-37

Question 7.
Mention the use of boric acid.
Answer:

  1. Boric acid is used in the manufacture of pottery glazes, glass, enamels and pigments.
  2. It is used as an antiseptic and as an eye lotion.
  3. It is also used as a food preservative.

Question 8.
Explain the preparation of diborane.
Answer:
1. Diborane can also be obtained in small quantities by the reaction of iodine with sodium borohydride in diglyme.
2NaBH4 + I2 → B2H6 + 2NaI + H2

2. On heating magnesium boride with HCl a mixture of volatile boranes are obtained.

  • 2Mg3B2+ 12HCl → 6MgCl2 + B4H10 + H2
  • B4H10+ H2 → 2B2Hg (Diborane)

Question 9.
What happen when diborane heated at various temperature?
Answer:
Diboranc is a gas at room temperature with sweet smell and it is extremely toxic. It is also highly reactive. At high temperatures it forms higher boranes liberating hydrogen.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-38

Question 10.
Explain the reaction between diborane and ammonia?
Answer:
When treated with excess ammonia at low temperatures diborane gives diborane – diammonate on heating at higher temperature diborane gives borazole.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-39
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-40

Question 11.
Why borontrifluoride is act as lewis acid? Explain the react between BF3 and ammonia?
Answer:
BF3 is an electron deficient compound and accepts electron pairs to form coordinate covalent bonds. Therefore, borontrifluoride is act as lewis acid. When BF3 reacts with ammonia to form complex.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-41

Question 12.
How does AlCl3 reacts with following reagents?

  1. H2O
  2. NH4OH
  3. NaOH

Answer:
1. When aluminium chloride reacts with water to form aluminium hydroxide.
AlCl3 + 3H2O → Al(OH)3 + 3HCl

2. With ammonium hydroxide, aluminium chloride forms aluminium hydroxide.
AlCl3 + 3NH7OH → Al(OH)3 + 3NH4C1

3. With excess of sodium hydroxide AlCl3 produces sodium meta aluminate.
AlCl3 + 4NaOH → NaAlO2 + 2H2O + 3NaCl

Question 13.
Give the uses of aluminium chloride.
Answer:

  1. Anhydrous aluminium chloride is used as a catalyst in Friedels Crafts reactions.
  2. It is used for the manufacture of petrol by cracking the mineral oils.
  3. It is used as a catalyst in the manufacture on dyes, drugs and perfumes.

Question 14.
How will you prepare potash alum?
Answer:
The alunite the alum stone is the naturally occurring form and it is K2SO4. A12(SO4)3. 4Al(OH)3. When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation.

K2SO4. A12(SO4)3. 4Al(OH)3 + 6H2SO4 → K2SO4 + Al2(SO 4)3 + 12H2O
K2SO4 + A12(SO4)2 + 24H2O → K2SO4. A12(SO4)3. 24H2O

Question 15.
Mention the uses of the potash alum.
Answer:

  1. It is used for purification of water
  2. It is also used for water proofing and textiles
  3. It is used in dyeing, paper and leather tanning industries
  4. It is employed as a styptic agent to arrest bleeding.

Question 16.
What is burnt alum?
Answer:
When potash alum is heated at 365 K. It melts and form molten state. Again heated at 475 K they lose water of hydration and swells up. Thee swollen mass is known as burnt alum.

Question 17.
What are all the conditions are necessary for catenation?
Answer:
Essential condition for catenation:

  1. The valency of elements is greater than or equal to two.
  2. Element should have an ability to bond with itself.
  3. The self bond must be as strong as its bond with other elements.
  4. Kinetic inertness of catenated compound towards other molecules.

Question 18.
Give any three methods to prepare carbon monoxide.
Answer:
1. Carbon monoxide can be prepared by the reaction of carbon with limited amount of oxygen.
2C + O2 → 2CO

2. On industrial scale carbon monoxide is produced by the reaction of carbon with air.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-42

3. Pure carbon monoxide is prepared by warming methanoic acid with concentrated sulphuric acid which act as a dehydrating agent.
HCOOH + H2SO4 → CO + H2O + H2SO4

Question 19.
Mention the uses the carbon monoxide.
Answer:

  1. Equimolar mixture of hydrogen and carbon monoxide – water gas and the mixture of carbon monoxide and nitrogen – producer gas are important industrial fuels
  2. Carbon monoxide is a good reducing agent and can reduce many metal oxides to metals.
  3. Carbon monoxide is an important ligand and forms carbonyl compound with transition metals

Question 20.
Explain the methods to prepare carbon dioxide.
Answer:
1. On industrial scale it is produced by burning coke in excess of air.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-43

2. Calcination of lime produces carbon dioxide as by product.
CaCO3 \(\underrightarrow { \triangle }\) CaO + CO2

3. Carbon dioxide is prepared in laboratory by the action of dilute hydrochloric acid on metal carbonates.
CaCO2 + 2HCl → CaCl2 + H2O + CO2

Question 21.
Give the uses of carbon dioxide.
Answer:

  1. Carbon dioxide is used to produce an inert atomosphere for chemical processing.
  2. Biologically, it is important for photosynthesis.
  3. It is also used as fire extinguisher and as a propellent gas.
  4. It is used in the production of carbonated beverages and in the production of foam.

Question 22.
Explain the preparation of silicon tetrachloride.
Answer:
1. Silicon tetrachloride can be prepared by passing dry chlorine over an intimate mixture of silica and carbon by heating to 1675 K in a porcelain tube.
SiO2 + 2C + 2Cl2 → SiCl2 + 2CO

2. On commercial scale, reaction of silicon with hydrogen chloride gas occurs above 600 K.
Si + 4HCl → SiCl4 + 2H2

Question 23.
Complete the following reaction,

  1. SiCl4 + H2O → ?
  2. SiCl4 + C2HSOH → ?
  3. Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-44

Answer:

  1. SiCl4 + 4H2O → Si(OH)4 + 4HCl
  2. SiCl4 + C2H5OH → Si(OC2H5)4+ 2C12
  3. Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-45

Question 24.
Explain the types of silicones.
Answer:

  1. Linear silicones – They are obtained by the hydrolysis and subsequent condensation of dialkyl or diaryl silicon chlorides.
    • Silicone rubbers – These silicones are bridged together by methylene or similar groups.
    • Silicone resins – They are obtained by blending silicones with organic resins such as acrylic esters.
  2. Cyclic silicones – These are obtained by the hydrolysis of R2SiCl2.
  3. Cross linked silicones – They are obtained by hydrolysis of RSiClr

Question 25.
Mention the properties of silicones.
Answer:

  1. All silicones are water repellent.
  2. They are thermal and electrical insulators.
  3. Chemically they are inert.
  4. Lower silicones are oily liquids whereas higher silicones with long chain structure are waxy solids.
  5. The viscosity of silicon oil remains constant and doesn’t change with temperature and they don’t thicken during winter.

SamacheerKalvi.Guru

Question 26.
What are silicates and mention the types of silicates?
Answer:
Silicates:
The mineral which contains silicon and oxygen in tetrahedral [\({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) units linked together in different patterns are called silicates.
types of Silicates:

  1. Ortho silicates
  2. Pyro silicates
  3. Cyclic silicates
  4. Ino silicates
  5. Phyllo silicates
  6. Tecto silicates

Samacheer Kalvi 12th Chemistry p-Block Elements – I 5 Mark Questions and Answers

Question 1.
An element (A) extracted from kernite. A reacts with nitrogen at high temperature gives B. A reacts with alkali to form C. Find out A, B and C. Give the chemical equations.
Answer:
1. An element (A) extracted form kemite is boron.

2. Boron reacts with nitrogen at high temperature gives Boron nitride (B)
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-46

3. Boron reacts with alkali (NaOH) to gives sodium borate (C).
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-47

Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-48

Question 2.
Compound A is used in the manufacture of opticals. A on heating gives B. B further heating to form C. A reacts with hydrochloric acid to give D. Identify A, B and C. Explain the reaction.
Answer:
1. Compound (A) is borax, which is used in the manufacture of opticals.

2. Borax (A) on heating to give borax glass (B). Borax glass on further heating to give sodium metaborate (C).
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-48

3. Borax (A) reacts with hydrochloride acid to give boric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-49

Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-50

Question 3.
Complete the reaction.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-51
Answer:
Boric acid when heated at 373 K gives metaboric acid (A) and A heated at 413 K it gives tetraboric acid (B). B heated at red hot it gives boric anhydride (C).
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-52
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-53

Question 4.
Sodium borohydride reacts with iodine in the presence of diglyme to give A. A heated at 388 K give B. A heated at 373 K in sealed tube to form C. A further heated at red hot condition to give element D. Find out A, B, C and D. Give the reactions.
Answer:
1. Sodium borohydride reacts with iodine in the presence of diglyme to give diborane (A).
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-54

2. Diborane (A) heated at 388 K gives pentaborane (11) (B).
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-55

3. Diborane (A) heated at 373 K in the sealed tube to form decaborane
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-56

(14) (C).Diborane (A) heated at red hot conditions to give element boron (D) Diborane B2H6
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-57
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-58

Question 5.
How does carbonmonoxide reacts with following reagents,
(i) O2
(ii) Cl2
(iii) F2O3
(iv) H2
(v) ethene
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-59

Question 6.
Explain the preparation of silicones.
Answer:
Generally silicones are prepared by the hydrolysis of dialkyldichiorosilanes (R2SiCl2) or diaryldichlorosilanes Ar2SiCl2, which are prepared by passing vapours of RCl or ArCl over siliconat 570 K with copper as a catalyst.
2RCl + Si \(\underrightarrow { Cu/570k }\) R2SiCl2

The hydrolysis of dialkvlchloro silanes R2SiCl2 yields to a straight chain polymer which grown from both the sides
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-60

The hydrolysis of monoalkylchloro silanes RSiCl3, yields to a very complex cross linked polymer Linear silicones can be converted into cyclic or ring silicones when water molecules is removed
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-61

Question 7.
Discuss the ortho and pyro silicates.
Answer:
Ortho silicates:
The simplest silicates which contain discrete [SiO4]4- tetrahedral units are called ortho silicates or neso silicates.
Examples:
Phenacite – Be2SiO4(Be2+) ions are tetrahedrally surrounded by O2- ions). Olivine – (Fe/Mg)2SiO4( Fe2+ and Mg2+ cations are octahedrally surrounded by O2-ions).
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-62
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-63

Pyro silicates:
Silicates which contain [Si2O7]6- ions are called pyro silicates (or) Soro silicates They are formed by joining two [SiO4]4- tetrahedral units by sharing one oxygen atom at one comer, (one oxygen is removed while, joining).
Example:
Thortveitite – SC2Si2O7.

Question 8.
Explain:

  1. Cyclic silicates
  2. Ino silicates

Answer:
1. Cyclic silicates:
Silicates which contain \(({ Si{ O }_{ 3 }) }_{ n }^{ 2n- }\) ions which are formed by linking three or more tetrahedral \(({ Si{ O }_{ 3 }) }_{ n }^{ 4- }\) units cyclically are called cyclic silicates. Each silicate unit shares two of its oxygen atoms with other units.
Example:
Beryl [Be3Al2(SiO3)6] (an aluminosilicate with each aluminium is surrounded by 6 Structure of Cyclic silicates
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-64

2. Ino silicones:
Silicates which contain V number of silicate units liked by sharing two or more oxygen atoms are called inosilicates. They are further classified as chain silicates and double chain silicates. Chain silicates (or pyroxenes) – These silicates contain \(({ Si{ O }_{ 3 }) }_{ n }^{ 2n- }\) ions formed by linking ‘n’ number of tetrahedral \(({ Si{ O }_{ 3 }) }_{ n }^{ 4- }\) units linearly. Each silicate unit shares two of its oxygen atoms with other Structure of Chain silicates units.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-65
Example:
Spodumene – LiAl(SiO3)2 Double chain silicates (or amphiboles) – These silicates contains \([{ S{ i }_{ 4 }{ O }_{ 11 }] }_{ n }^{ 6n- }\) ions. In these silicates there are two different types of

3. Tetrahedra:

  1. Those sharing 3 vertices
  2. Those sharing only 2 vertices.

Examples.
Asbestos – These are fibrous and noncombustible silicates. Therefore they are used for thermal insulation material, brake linings, construction material and filters. Asbestos being carcinogenic silicates, their applications are restricted.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-66

Question 9.
Write a notes on

  1. Sheet silicates
  2. Three dimensional silicate

Answer:
1. Sheet silicates:
Silicates which contain \(({ S{ i }_{ 2 }{ O }_{ 5 }) }_{ n }^{ 2n- }\) are called sheet or phyllo silicates. In these, Each \({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) tetrahedron unit shares three oxygen atoms with others and thus by forming twodimensional sheets. These sheets, silicates form layered structures in which silicate sheets are stacked over each other. The attractive forces between these layers are very week, hence they can be cleaved easily just like graphite.
Samacheer Kalvi 12th Chemistry Solutions Chapter 2 p-Block Elements - I img-67
Example: Talc, Mica etc…

2. Three dimensional silicate:
Silicates in which all the oxygen atoms of \({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) tetrahedra are shared with other tetrahedra to form three-dimensional network are called three dimensional or tecto silicates. They have general formula (SiO2)n.
Examples:
Quartz. These tecto silicates can be converted into Three dimentional aluminosilicates by replacing [\({ { [S{ i }{ O }_{ 4 }] } }^{ 4- }\) – units by [AlO4]5- units. E.g. Feldspar, Zeolites etc.,

Question 10.
Explain – Boron neutron capture therapy.
Answer:
1. The affinity of boron-10 for neutrons is the basis of a technique known as born neutron capture therapy (BNCT) for treating patients suffering from brain tumours.

2. It is based on the nuclear reaction that occurs when boron-10 is irradiated with low- energy thermal neutoms to give high linear energy a-particles and a Li-particles.

3. Boron compounds are injected into a patient with a brain tumour and the compounds collect preferentially in the tumour. The tumour area is then irradiated with thermal neutom and results in the release of an alpha-particle that damages the tissue in the tumour each time a boron-10 nucleus captures a neutron.

3. In this way damages can be limited preferentially to the tumour, leaving the normal brain tissue less affected.

4. BNCT has also been studied as a treatment for several other tumours of the head and neck, The breast, the prostate, the bladder and the liver.

Commn Errors and its Rectifications
Common Errors:

  1. Ores formula may be difficult to remember.
  2. Allotropes of elements may confuse the students in writing equation.

Rectifications:

  1. Only chief ore from which the element is extracted can be easy to remember.
  2. Crystalline forms are more chemically reactive.

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Samacheer Kalvi 12th English Solutions Poem Chapter 4 Ulysses

Students can Download English Poem 4 Ulysses Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Solutions Poem Chapter 4 Ulysses

12th English Poem Ulysses Warm Up

Introduction:
The poem ‘Ulysses’ is a dramatic monologue that contains 70 lines of blank verse. Ulysses, the king of Ithaca, gathers his men together to prepare for the journey and exhorts them not to waste their time left on earth. Ulysses has grown old, having experienced many adventures at the battle of Troy and in the seas. After returning to Ithaca, he desires to embark upon his next voyage. His inquisitive spirit is always looking forward to more and more of such adventures.

12th English Poem Ulysses Chapter 4 Samacheer Kalvi

The poem can be divided into three parts:

  1. the thirst for adventure, which does not allow Ulysses to remain in his kingdom as a mere ruler
  2. Ulysses handing over the responsibility to his son Telemachus, with total confidence in his abilities
  3. Ulysses’ clarion call to his sailors, urging them to venture into unknown lands.

Samacheer Kalvi 12th Ulysses English Textual Questions

1. Complete the summary of the poem, choosing words from the list given below. Lines 1 to 32

It little profits that an idle king,
By this still hearth, among these barren crags,
Match’d with an aged wife, I mete and dole
Unequal laws unto a savage race,

That hoard, and sleep, and feed, and know not me.
I cannot rest from travel: I will drink
Life to the lees: All times I have enjoy’d
Greatly, have suffer’d greatly, both with those

That loved me, and alone, on shore, and when
Thro’ scudding drifts the rainy Hyades
Vext the dim sea: I am become a name;
For always roaming with a hungry heart

Much have I seen and known; cities of men
And manners, climates, councils, governments,
Myself not least, but honour’d of them all;
And drunk delight of battle with my peers,

Far on the ringing plains of windy Troy.
I am a part of all that I have met;
Yet all experience is an arch wherethro’
Gleams that untravell’d world whose margin fades

For ever and forever when I move.
How dull it is to pause, to make an end,
To rust unburnish’d, not to shine in use!
As tho’ to breathe were life! Life piled on life

Were all too little, and of one to me
Little remains: but every hour is saved
From that eternal silence, something more,
A bringer of new things; and vile it were

For some three suns to store and hoard myself,
And this gray spirit yearning in desire
To follow knowledge like a sinking star,
Beyond the utmost bound of human thought.

This is my son, mine own Telemachus,
To whom I leave the sceptre and the isle, –
Well-loved of me, discerning to fulfil
This labour, by slow prudence to make mild

A rugged people, and thro’ soft degrees
Subdue them to the useful and the good.
Most blameless is he, centred in the sphere
Of common duties, decent not to fail

offices of tenderness, and pay
Meet adoration to my household gods,
When I am gone. He works his work, I mine.
There lies the port; the vessel puffs her sail:

Ulysses Poem Questions And Answers Samacheer Kalvi 12th English Solutions Chapter 4

There gloom the dark, broad seas. My mariners,
Souls that have toil’d, and wrought, and thought with
me That ever with a frolic welcome took
The thunder and the sunshine, and opposed
Free hearts, free foreheads – you and I are old;

Old age hath yet his honour and his toil;
Death closes all: but something ere the end,
Some work of noble note, may yet be done,
Not unbecoming men that strove with Gods.

The lights begin to twinkle from the rocks:
The long day wanes: the slow moon climbs: the deep
Moans round with many voices. Come, my friends,
‘T is not too late to seek a newer world.

Push off, and sitting well in order smite
The sounding furrows; for my purpose holds
To sail beyond the sunset, and the baths
Of all the western stars, until I die.

It may be that the gulfs will wash us down:
It may be we shall touch the Happy Isles,
And see the great Achilles, whom we knew.
Tho’ much is taken, much abides; and tho’

We are not now that strength which in old days
Moved earth and heaven, that which we are, we are;
One equal temper of heroic hearts,
Made weak by time and fate, but strong in will
To strive, to seek, to find, and not to yield.

Ulysses Questions And Answers Samacheer Kalvi 12th English Solutions Poem Chapter 4

fullest unquenchable unattainable
experience knowledge king
matters rust adventure
unwilling travel breathing

Ulysses is (1) ______ to discharge his duties as a (2) ______ , as he longs for (3) ______ He is filled with an (4) ______ thirst for (5) ______ and wishes to live life to the (6) ______ He has travelled far and wide gaining (7) ______ of various places, cultures, men and (8) ______ He recalls with delight his experience at the battle of Troy. Enriched by his (9) ______ he longs for more and his quest seems endless. Like metal which would (10) ______ if unused, life without adventure is meaningless. According to him living is not merely (11) ______ to stay alive. Though old but zestful, Ulysses looks at every hour as a bringer of new things and yearns to follow knowledge even if it is (12) ______
Answers:

  1. unwilling
  2. king
  3. adventure
  4. unquenchable
  5. travel
  6. fullest
  7. experience
  8. matters
  9. knowledge
  10. rust
  11. breathing
  12. unattainable

Lines 33 to 42

prudence kingdom quest tender

Ulysses desires to hand over his (1) _____ to his son Telemachus, who would fulfil his duties towards his subjects with care and (2) _____ Telemachus possesses patience and has the will to civilise the citizens of Ithaca in a (3) _____ way. Ulysses is happy that his son would do his work blamelessly and he would pursue his (4) _____ for travel and knowledge.
Answer:

  1. kingdom
  2. prudence
  3. tender
  4. quest

Lines 44 to 70

world thunder meaningful
gather undaunted heaven

Ulysses beckons his sailors to (1) _____ at the port where the ship is ready to sail. His companions who have faced both (2) _____ and sunshine with a smile, are united by their undying spirit of adventure. Though death would end everything, Ulysses urges his companions to join him and sail beyond the sunset and seek a newer (3) _____ , regardless of consequences. These brave hearts who had once moved (4) _____ and earth, may have grown old and weak . physically but their spirit is young and (5) _____ His call is an inspiration for all those who seek true knowledge and strive to lead (6) _____ lives.
Answers:

  1. gather
  2. thunder
  3. world
  4. heaven
  5. undaunted
  6. meaningful

2. Answer the following questions in one or two sentences each.

Ulysses Poem Questions And Answers Question (а)
‘Ulysses is not happy to perform his duties as a king.’ Why?
Answer:
Ulysses is not happy to perform the ordinary duties of a king mainly because his heart is in voyages beyond horizon. He is bored with the task of enforcing law and order and giving reward and punishment to a savage race.

Ulysses Questions And Answers Question (b)
What does he think of the people of his kingdom?
Answer:
The citizen of Ithaca hoard, sleep and feed. They don’t understand the aspirations of the dreamer Ulysses.

12th English Poem Ulysses Paragraph Question (c)
What has Ulysses gained from his travel experiences?
Answer:
Ulysses has met people hailing from different cultural backgrounds. He has learnt much from their manners, climates, councils and governments. He learnt strategies of warfare in battles.

Ulysses Poem Erc Question (d)
Pick out the lines which convey that his quest for travel is unending.
Answer:
“How dull it is to pause, to make an end,
To rust unburnished, not to shine in use!
The lines quoted above convey his quest for travel is unending.

12th English Ulysses Paragraph Question (e)
‘As tho’ to breathe were life!’ – From the given line what do you understand of Ulysses’ attitude to life?
Answer:
Ulysses strongly believes that just breathing is not life. Life has to be adventurous and full of action.

Ulysses Poem Questions And Answers Pdf Question (J)
What does Ulysses yearn for?
Answer:
Ulysses yearns for following knowledge like a sinking star beyond the utmost bound of human thought.

Ulysses Question Answer Question (g)
Who does the speaker address in the second part?
Answer:
The speaker addresses the readers in the second part explaining the difference between his roles and that of Telemachus.

Ulysses Poem Short Questions And Answers Question (h)
Why did Ulysses want to hand over the kingdom to his son?
Answer:
Ulysses finds Telemachus discerning and prudent. Besides, Ulysses is wearing the crown uneasily as the call for adventure and desire to sail beyond sunset is obsessing his mind. So, he wants to hand over the kingdom to Telemachus.

Ulysses Short Questions And Answers Question (i)
How would Telemachus transform the subjects?
Answer:
Ulysses believes that his son Telemachus is wise and kind enough to transform rugged citizens into mild and civilized subjects by his tenderness and love.

Question (j)
‘He works his work, I mine’ – How is the work distinguished?
Answer:
Telemachus would do the work of ruling Ithaca with prudence and tenderness. Ulysses will pursue his dream of adventure and try to meet great Achilles in the other world.

Question (i)
In what ways were Ulysses and his mariners alike?
Answer:
Both Ulysses and his fellow sailors are now old. They no more have the strength they possessed in olden days moving earth and heaven. They are made weak by time and fate but strong in will “to strive, to seek, to find and not to yield.” They share the heroic temper and undying quest for knowledge and adventure.

Question (j)
What could be the possible outcomes of their travel?
Answer:
The sailors and Ulysses may be washed down by the gulfs or they could touch Greek paradise and meet their hero Achilles. They may die happily braving the elements of nature.

3. Identify the figures of speech employed in the following lines.

Question (a)
“Thro” scudding drifts the rainy Hyades
Vext the dim sea…”
Answer:
The figure of speech Personification is employed in the above lines.

Question (b)
“For always roaming with a hungry heart”
Answer:
Metaphor

Question (c)
“And drunk delight of battle with my peers;”
Answer:
Metaphor

Question (d)
“…..the deep
Moans round with many voices.”
Answer:
Personification

Question (e)
“To follow knowledge like a sinking star.”
Answer:
Simile

Question (f)
“ There lies the port the vessel puffs her sai”
Answer:
Personification

Additional Questions

Question (a)
“I will drink life to the lees'”
Answer:
Metaphor

Question (b)
“Vext the dim sea:”
Answer:
Personification

Question (c)
“Yet all experience is an arch wherethro”
Answer:
Metaphor

Question (d)
“Gleams that untravell’d world whose margin fades”
Answer:
Metaphor

Question (e)
“To rust unburnish’d, not to shine in use!”
Answer:
Metaphor

Question (f)
“There gloom the dark, broad seas. My mariners,”
Answer:
Metaphor

Question (g)
“Souls that have toil’d, and wrought, and thought with me”
Answer:
Synecdoche (part of the whole)

Question (h)
“The thunder and the sunshine, and opposed”
Answer:
Metaphor

Question (i)
“T is not too late to seek a newer world.”
Answer:
Synecdoche

Question (j)
“…in order smite The sounding furrows;”
Answer:
Metaphor

Question (k)
“To sail beyond the sunset, and the baths”
Answer:
Metaphor

Question (l)
“It may be we shall touch the Happy Isles,”
Answer:
Allusion (in Greek mythology the place is known as Greek paradise)

Question (m)
“And see the great Achilles, whom we knew.”
Answer:
Allusion (Greek mythology)

Appreciate The Poem

4. Read the sets of lines from the poem and answer the questions that follow.

(a) “…I mete and dole
Unequal laws unto a savage race,
That hoard, and sleep, and feed, and know not me.”

Question (i)
What does Ulysses do?
Answer:
Ulysses, like a grocery shop owner, measures and delivers rewards and punishments unto a large number of uncivilized citizens.

Question (ii)
Did he enjoy what he was doing? Give reasons.
Answer:
No, he did not enjoy is work. He does not like the idea of ministering variable justice to people who like “drones” or animals just eat, sleep and multiply their kind. He wants to leave such work to his son.

Question (b)
“Yet all experience is an arch wherethrough
Gleams that untravelled world, whose margin fades
For ever and for ever when I move. ”

Question (i)
What is experience compared to?
Answer:
Experience is compared to an arch.

Question (ii)
How do the lines convey that the experience is endless?
Answer:
Through the arch of experience one can see the untravelled world. But the experience in the untravelled has a margin whose border fades as one moves forward. Thus experience is endless.

Question (c)
“Little remains: but every hour is saved
From that eternal silence, something more,
A bringer of new things; and vile it were”

Question (i)
How is every hour important to Ulysses?
Answer:
One lives in this world for a limited time. Every hour can provide new knowledge. So, every hour is very important.

Question (ii)
What does the term ‘Little remains’ convey?
Answer:
Ulysses realizes that he has become old. He has not much time left. He doesn’t want to die resting in his kingdom. He states that his remaining lifetime is very limited.

Question (d)
“This is my son, mine own Telemachus,
To whom I leave the sceptre and the isle
Well-loved of me,”

Question (i)
Who does Ulysses entrust his kingdom to, in his absence?
Answer:
Ulysses entrusts his kingdom to his beloved son Telemachus in his absence.

Question (ii)
Bring out the significance of the ‘sceptre’.
Answer:
Sceptre is an ornamental staff carried by a King on ceremonial occasions as a symbol of sovereignity. It symbolizes the power of a king.

(e) ‘‘That ever with a frolic welcome took
The thunder and the sunshine, and opposed”

Question (i)
What do ‘thunder’ and ‘sunshine’ refer to?
Answer:
Thunder and sunshine refers to misfortunes and happy days. Ulysses and his comrades had undergone both kinds of experiences.

Question (ii)
What do we infer about the attitude of the sailors?
Answer:
The sailors shared the undying quest for exploration, adventure and for seeking newer knowledge in the untravelled world. They even welcomed dangers in fighting with Gods. They enjoyed the thrill of action and never worried about the outcome of battles or quests. They have equal temper of heroic hearts.

(f) ‘‘Death closes all: but something ere the end,
Some work of noble note, may yet be done,
Not unbecoming men that strove with Gods.”

Question (i)
The above lines convey the undying spirit of Ulysses. Explain.
Answer:
Ulysses is aware of ageing and substantial decrease in his physical strength. He knows that will close in on him sooner or later. But before that happens, he wants to sail beyond the sunset/horizon and if possible meet warriors like Achilles. He wants to achieve something worthy of those who challenged and fought with Gods. Thus these lines show the undying spirit of Ulysses.

Question (ii)
Pick out the words in alliteration in the above lines,
Answer:
ere, end, noble, note are the words that alliterate.

(g) “……… for my purpose holds
To sail beyond the sunset, and the baths Of all the western stars, until I die.”

Question (i)
What was Ulysses’ purpose in life?
Answer:
Ulysses proposes to sail beyond sunset and baths. His goal is not death but is in death. He seeks life in death. Ordinary mortals can’t reach ‘Happy isles’ or baths while they are alive.Ulysses wants to find direct evidence of spiritual reality after death. He wants to venture into the unknown.

Question (ii)
How long would his venture last?
Answer:
His venture would last until he confronts his death.

Question (h)
“One equal temper of heroic hearts,
Made weak by time and fate, but strong in will
To strive, to seek, to find, and not to yield. ”

Question (i)
Though made weak by time and fate, the hearts are heroic. Explain.
Answer:
Ulysses and his compatriots have visited many strange places in their previous voyages and enjoyed misfortunes and glorious triumphs with the same heroic temperament. They might have become old and may not have the same strength they had in youth. But they still share the thirst for travel and pursuit of knowledge in the unexplored world. Their bravery and spiritual strength are intact.

Question (ii)
Pick out the words in alliteration in the above lines.
Answer:
Strive, seek, heroic, hearts are the words that alliterate.

5. Explain with reference to the context the following lines.

Question (a)
“I cannot rest from travel: I will drink
Life to the lees:”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Ulysses, after spending many years in the seas returns to Ithaca and starts ruling his country. But his heart is not in the administration of his kingdom. He wants to sail again. At this context he says these words. He wishes to enjoy life to the fullest and so he can’t afford to idle away his remaining life as a king.

Question (b)
“I am become a name;
For always roaming with a hungry heart”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Ulysses says these words while discussing the reputation he has earned among the common multitude due to his daring adventures. He has roamed the world like a hungry lion. This line has a biblical alusion as well “Blessed are they which do hunger and thirst after righteousness” – Matthew V. 6.

Though Ulysses is aware of his fame, it doesn’t motivate him to stay or settle down in the kingdom of Ithaca. His inquisitive spirit is always looking for newer knowledge through ‘the arch’ to the untravelled world.

Question (c)
“How dull it is to pause, to make an end,
To rust unburnished, not to shine in use!”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words while discussing the mental agony of Ulysses who is unable to settle down with his ageing wife Penelope and son Telemachus. Ulysses finds doling out justice to a savage people as ‘boring’. He does not want to settle down and die in Ithaca. He compares himself to a sword which may rust if left unused. He wants to lead an active and adventurous life till his death.

Question (d)
“To follow knowledge like a sinking star,
Beyond the utmost bound of human thought”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words while describing the quest Ulysses has for adventure and fulfillment. Similar to a sinking star, Ulysses wants to pursue in his failing old age to pursue knowledge like the goal of Goethe’s Faust, his quest is defined by the pursuit of new and unique knowledge “beyond the utmost bound of human thought”.

Question (e)
“He works his work, I mine.’’’’
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words while justifying the decision of Ulysses to pass on his kingdom to Telemachus. Ulysses explains the polar difference between himself and his son Telemachus. His son will be a ‘fair’ and ‘decent’ ruler. Unlike Ulysses, Telemachus is rooted in regular political life. He enjoys leading “savage” population and the responsibility of showing the subject better moral codes of conduct and upholding justice. Whereas Ulysses finds this “slow” and intolerable. So, he wishes his son to rule Ithaca and for himself he wishes to set sail to the unknown.

Question (f)
“….you and I are old;
Old age hath yet his honour and his toil;”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Poet, in this part of the monologue describes the address made by Ulysses to his compatriots who were with him during “thunders and sunshine”. He admits the fact that they are growing old. But he does not want to retire like ordinary mortals. He accepts gracefully the honour befitting old age as a result of varied cultural experiences. Yet, he does not want old people to bow out of the field of action. He sincerely believes there is more work to be done, lands to be explored and newer knowledge to be acquired in old age before death.

Question (g)
“The long day wanes: the slow moon climbs: the deep
Moans round with many voices.”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet poet continues to discuss old age and the tantalising call of the oceans to conquer. Ulysses hints at the probable end of the cycle of life in the words “The long clay wanes”. The symbol of darkness or night is mostly associated with death. The lure of ocean to resume his voyages beyond the point of sunset is too tempting to resist. The dark unfathomable sea beckons him and his compatriots with mysterious voices.

Question (h)
“It may be we shall touch the Happy Isles,
And see the great Achilles, whom we knew.”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words through Ulysses as to the probable outcome of the daring voyage at the fag end of his life. Ulysses is uncertain about the probable outcome of his last voyage. But he infers that he might reach ‘Happy Isles’ and see the ‘great Achilles’ who was dipped in the river of life.

Question (i)
“We are not now that strength which in old days
Moved earth and heaven;”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: The poet says these words through Ulysses when he wants to justify the reasons for resuming the daring voyage. He admits the decline in the compatriots’ physical strength with which they were able to move heaven and earth in their youth. He asks his compatriots to ignore the infinity of age and draw on their inner spiritual strength to resume their voyage beyond sunset.

Question (j)
“To strive, to seek, to find, and not to yield.”
Answer:
Reference: These lines are from the poem ‘Ulysses” written by Alfred Tennyson.
Context and Explanation: Tennyson says these words through Ulysses who makes his motto loud and clear in these words. The final line of the poem is Ulysses’ enduring challenge to readers as well. The challenge if the aged ones could push ahead with vigour and strength of will no matter how fragile their bodies are. To yield to age or weakness is to be less than fully human. It might be honourable to live a peaceful settled life in the old age. But one would naturally miss out most exciting moments of life if one does not venture out, at least a little towards the unknown.

6. Answer the following questions in a paragraph of about 100 words each.

Question (a)
What makes Ulysses seek newer adventures?
Answer:
In the context of the poem, Ulysses has grown old. He has experienced all daring adventures. He has won the hearts of people during the battle at Troy. Back home, as per the prophecy of Tiresias, he rules Ithaca for a brief time. But he is fed up with a conventional duties of a king. He laments his own uselessness as a ruler of idle people who lead life like savages, just eating and sleeping. They don’t understand the over vaulting ambition of their adventurous king Ulysses who had moved earth and heavens in the past. He wishes to embark upon his next voyage. It might be his last. He is quite sensitive to the moans of the seas tantalising him and his compatriots to set sail quickly. He wants “to drink life to the lees”. Ulysses doesn’t want to bask on the glory he has earned in the past.

His inquisitive spirit is restless. He has seen much’ and acquired knowledge of various cultures of the world. But he considers all such experiences like an “arch” leading him to the unexplored or “untravelled world”. He wants to sail towards the area ‘beyond sunset’. He must shine in use like a sword but not “rust unbumished”. Yet at home, in the kingdom of Ithaca, he feels bored and yearns to truly engage with what is left of life. He is impatient for “new” experiences lamenting everyday and every hour to seek “something more”. His quest for adventure and fulfillment, like the goal of Goethe’s Faust is defined by the pursuit of new knowledge “beyond the utmost bound of human thought”.

Question (b)
List the roles and responsibilities Ulysses assigns to his son Telemachus, while he is away.
Answer:
The entire poem is a monologue. Yet the second part of the poem is an address to the readers justifying his decision to transfer the rule to his son Telemachus. The cloak of a king seems to be unfit for the temperament of Uly sses. He finds ruling Ithaca a boring thing. He finds Telemachus rooted to the political life of Ithaca. His role is merely to lead a ‘savage race’ to accept standard norms of behaviour in the society. He believes Telemachus fits well with the role of the ruler of “uninspired and imprudent citizens” and may discharge his duties with honour and grace. When he is away, he wants his son Telemachus to dispense variable justice to the subjects of Ithaca and guide them in the path of virtues and morals.

Question (c)
What is Ulysses’ clarion call to his sailors? How does he inspire them?
Answer:
In the third part of the poem, Ulysses makes a clarion call to his hearty compatriots (i.e.) mariners. They have been with him both during ‘thick and thin’ or thunders or sunshine. Similar to Ulysses they possess “free hearts and free foreheads” (i.e.) their hearts and brains are unburdened by domestic cares and resposibilities. They had frolicsome time fighting along with Ulysses against great warriors and Gods in the past. Ulysses does not want to live in memory of glory. He believes they need not waste away their precious time in nostalgic memories just recounting their escapades to younger generation. They can really do ‘ something of noble note’ before the end. He is conscious of the impending death in old age. But he tells it is not “too late to seek a newer world”.

The many “voices of the ocean” call out to the mariners to resume voyage. Ulysses is not content with having earned a name for himself. He has seen many countries and acquired knowledge of various cultures. Those experiences are not to be taken as accomplishments. They are just an ‘arch’leading them to “untravelled world” and constantly sailing to the ever expanding horizon. He does not want his compatriots to miss even an hour which could prrovide them novel experiences in their voyage. He persuades his compatriots to gather at the port as the sails are already puffing up welcoming them all. Their life would be one of fulfillment only when they venture out into the unkown on the seas. He uses an emotional bait to his mariners. He highlights the probable outcome of their voyage. They might reach the “Happy Isles” (i.e.) great paradise and meet Achilles, their war hero. No matter how much strength they have, they still have some “strength of will” left to strive, to seek, to find, and not to yield.

Listening Activity

Listen to the poem and fill in the blanks with appropriate words and phrases. If required listen to the poem again.

Wander-thirst

BEYOND the East the sunrise, beyond the West the sea, And East and West the wander-thirst that will not let me be; It works in me like madness, dear, to bid me say good-bye; For the seas call, and the stars call, and oh! the call of the sky!

I know not where the white road runs, nor what the blue hills are; But a man can have the sun for a friend, and for his guide a star; And there’s no end of voyaging when once the voice is heard, For the rivers call, and the roads call, and oh! the call of the bird!

Yonder the long horizon lies, and there by night and day The old ships draw to home again, the young ships sail away; And come I may, but go I must, and, if men ask you why, You may put the blame on the stars and the sun and the white road and the sky.

Choose the best option and complete the sentences:

Question 1
works like madness in the poet.
(a) Wander – Thirst
(b) Bidding Farewell
(c) Eastern Sunrise
(d) Western Seas
Answer:
(a) Wander – Thirst

Question 2.
A man could choose as his guide.
(a) the sun
(b) the hills
(c) a star
(d) a bird
Answer:
(c) a star

Question 3.
There is no end of once the voice is heard.
(a) walking
(b) roaming
(c) talking
(d) voyaging
Answer:
(d) voyaging

Question 4.
The old ships return, while the young ships
(a) drift
(b) move
(c) sail
(d) wander
Answer:
(c) sail

Question 5.
The blame is on the sun, stars, the road and the
(a) hills
(b) trees
(c) seas
(d) sky
Answer:
(d) sky

Ulysses About The Poet

Samacheer Kalvi 12th English Solutions Poem Chapter 4 Ulysses img-4

Alfred Tennyson was the poet laureate of England and Ireland during Queen Victoria ‘s reign. Tennyson excelled in writing short lyrics such as “Break, Break, Break“, “ Tears Idle Tears“, “The Charge of the Light Brigade” and.‘Crossing the Bar’. Tennyson’s use of the musical qualities of words to emphasise his rhythms and meanings is sensitive.

Ulysses Summary in English

Introduction
‘Ulysses’ is a dramatic monologue in which Ulysses, the king of Ithaca expresses his undying thirst of adventure overseas. Tennyson has written this poem in memory of Arthur Henry Hallam who died young. Death is not the end for both Arthur and Ulysses.

Thirst for adventure
Ulysses addresses mariners in the third part. Ulysses does not see any worth in staying in the comfort of family life with his wife. As a king, he listens to the complaints of people and gives rewards and punishments for the citizens of his country. He discloses his inner nature, “cannot rest from travel!” He has become a living icon of travel overseas. He wants “to drink life to the lees”. He confesses that he has both enjoyed and suffered a lot during his travels. He has gained profound wisdom during his travels and battles. Enriched by his new gained cultural knowledge, he longs to resume his voyages. He believes that to rest is rust. Every hour has the •potential to bring new knowledge. So, it is meaningless to stay.

Passing on the inheritance
In the second part addresses the reader explaining why Ulysses doesn’t want to continue to rule. Ulysses wishes to hand over his kingdom to his son Telemachus. He believes he would rule the kingdom and render appropriate justice to his subjects. Telemachus, unlike Ulysses, is rooted to the soil. He is kind to the subjects and addresses their needs. Besides, Ulysses hopes that Penelope, his aging wife would be happy to spend her last years with her son and grandchildren. So, Ulysses can resume his voyage along with his old friends.

Call to set sail
Ulysses called his old companions to the port where the ship awaits them all. His old companions have fought many battles alongside him and share the undying quest for adventure. Ulysses persuades his friends to join him on his voyage to the edge of the world and beyond to find a new world. He knows the limitations of his companions. They may have grown old and weak due to age. Their physical powers may not be the same as when they had once moved heaven and earth. But he is confident that their spirit is young and undaunted. His clarion call would inspire invariably all those who seek knowledge and strive to lead meaningful lives.

Conclusion
People who are endowed with an unquenchable thirst for ever-expanding knowledge and incurable love for long distances Ulysses continues to be a source of inspiration.

Ulysses Summary in Tamil

முடிவுரை:
தன் அறிவு வேட்கையை அதிகப்படுத்துவதில் | ஆர்வம் கொண்டவர்களுக்கும் மற்றும் மாறாத தொலைதூரப் பயணத்தில் காதல் கொண்டவர்களுக்கும் | யுலிசஸ் எப்போதும் தூண்டுதலாக இருக்கிறார்.

சாகசத்தின் மேல் ஆவல்
யுலிசஸ் மூன்றாம் பாகத்தில் கப்பலோட்டிகளிடத்தில் சொற்பொழிவாற்றுகிறார். தன் குடும்பத்தாருடன் சொகுசு வாழ்க்கை வாழ்வதில் எந்தப் பயனும் இல்லை என எண்ணுகிறார். ஒரு அரசனாக குடிமக்களின் கூற்றுகளைக் கேட்டு அதற்கு ஏற்ற பரிசையும் மற்றும் தண்டனையும் வழங்குகிறார். கடற் பயணம் செய்யாமல் இருக்க முடியாது என்ற தன்னுடைய உள்ளுணர்வை வெளிப்படுத்துகிறார். கடல் கடந்த பயணங்களை மேற்கொள்ளும் உயிரோவியமாகத் திகழ்ந்தார். வாழ்க்கை முழுவதும் பயணிக்க விரும்புகிறார். அவர் பயணத்தில் மேற்கொண்ட இன்ப துன்பங்களை ஒப்புக் கொள்கிறார். அவர் பயணத்தின் போது ஏற்பட்ட அனுபவங்களாலும் மற்றும் எதிர்கொண்ட போர்களினாலும் ஆழ்ந்த ஞானத்தை அடைந்திருந்தார். புதிதாக பெறப்பெற்ற பண்புகளின் புலமையால், அவர் பயணங்களை மேற்கொள்வதில் ஆர்வத்தைக் காட்டினார். சோம்பலாக கிடந்தால் உடல் துருப்பிடித்து விடும் என எண்ணினார். ஒவ்வொரு கணமும் புதுப் பொலிவைக் கொண்டு வரும் ஆற்றல் மிக்கது. ஆதலால், ஓரிடத்தில் ஒண்டிக் கிடப்பதில் பயனில்லை என எண்ணினார்.

பரம்பரை சொத்துக்களை கைமாற்றம் செய்தல்:
இரண்டாம் பாக சொற்பொழிவில் அவருக்கு அரசாள்வதில் ஏன் விருப்பம் இல்லை என்பதைக் குறித்துக் கூறுகிறார். யுலிசஸ் தன் இராஜ்ஜியத்தை தன் புதல்வன் டெலிமேக்கஸ் (Tele Macus) இடம் கைமாற்றுகிறார். அவர் தன் புதல்வன் நல்ல முறையில் அரசாண்டு தன் குடிமக்களுக்கு தக்க நீதி வழங்குவான் என எண்ணினார். டெலிமாக்கஸ், யுலிசஸ் போல் அல்லாமல் ஒரே இடத்தில் ஊன்றி நிற்பவராய் இருந்தார். அவர்தம் குடிமக்களுக்கு அன்பார்ந்தவராகவும் அவர்கள் குறைகளைக் கேட்டறிபவராகவும் இருந்தார். அது தவிர யுலிசஸ் தன் வயது முதிர்ந்த மனைவி பெனைலோப்பிம், மகனுடனும் பேரப் பிள்ளைகளுடனும் மகிழ்ச்சியோடு இருப்பார் என நம்பினார். ஏனெனில், அது அவர் தன் தோழர்களுடன் கடல் பயணம் மேற்கொள்ளத் தக்கவாறு அமையும்.

கடல் பயணம் மேற்கொள்ளல்:
யுலிசஸ் தன் நண்பர்களைத் தமக்காக காத்துக் கொண்டிருக்கும் துறைமுகத்திற்கு வருமாறு அழைப்பு விடுகிறார். அவர் தோழர்களும் கடல் பயணத்தின் போது பல கடற்போர்களில் ஈடுபட்டு தனக்கு உள்ள கடல் பயணத்தின் மேல் உள்ள ஆவலைப் பகிர்ந்து கொண்டனர். உலகின் விளிம்பை அடைந்து அதற்கு அப்பாலும் செல்வதற்கு தன்னுடன் கடல் பயணம் மேற்கொள்ள நண்பர்களைத் தூண்டுகிறார். அவர் தம் நண்பர்களின் குறைகளை அறிந்திருந்தார். அவர்கள் வயது முதிர்ச்சியால் பலவீனமாகக் காணப்பட்டனர். அவர்கள், முன்னர் வானத்தையும், பூமியையும் அளந்தது போல் அவர்கள் உடல் வலிமை தற்போது இல்லை. ஆனால், அவர்கள் ஊக்கம் குன்றாமல், சளைக்கா வண்ணம் தொடரவேண்டும் என்பதில் உறுதியாக இருந்தார். அவரின் வீர முழக்கம் புதிய அனுபவத்தை நாடி மற்றும் அர்த்தமுள்ள வாழ்க்கையை வாழ முயற்சிக்கும் அனைவருக்கும் தூண்டுதலாக இருக்கும்.

முன்னுரை:
யூலிஸ் என்பது வியத்தகு மோனோலாலஜி ஆகும். இதில் இதாகா மன்னனான யுலிசஸ் வெளிநாடுகளில் சாகசத் தன்மையற்ற தாகத்தை வெளிப்படுத்துகிறார். டென்னிசன் இளம் கவிஞரான ஆர்தர் ஹென்றி ஹாலமின் நினைவாக இந்தக் கவிதை எழுதினார். ஆர்தர் மற்றும் யுலிசஸ் இருவருக்கும் முடிவே இல்லை.

Ulysses Glossary

Textual:

12th English Poem Ulysses Paragraph Chapter 4 Samacheer Kalvi

Additional:

Ulysses Poem Erc Samacheer Kalvi 12th English Solutions Poem Chapter 4

Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current

Students those who are preparing for the 12th Physics exam can download this Samacheer Kalvi 12th Physics Book Solutions Questions and Answers for Chapter 4 Electromagnetic Induction and Alternating Current from here free of cost. These Tamilnadu State Board Solutions for Class 12th Physics PDF cover all Chapter 4 Electromagnetic Induction and Alternating Current Questions and Answers. All these concepts of Samacheer Kalvi 12th Physics Book Solutions Questions and Answers are explained very conceptually as per the board prescribed Syllabus & guidelines.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current

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Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Multiple Choice Questions

12th Physics Chapter 4 Book Back Answers Question 1.
An electron moves on a straight, line path XY as shown in the figure. The coil abed is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil? (NEET – 2015)
12th Physics Chapter 4 Book Back Answers Electromagnetic Induction And Alternating Current Samacheer Kalvi
(а) The current will reverse its direction as the electron goes past the coil
(b) No current will be induced Electron
(c) abcd
(d) adcb
Answer:
(a) The current will reverse its direction as the electron goes past the coil

12th Physics Lesson 4 Book Back Answers Question 2.
A thin semi-circular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed v, is- (NEET 2014)
12th Physics Lesson 4 Book Back Answers Electromagnetic Induction And Alternating Current Samacheer Kalvi
(a) Zero
(b) \(\frac {{ Bvπr }^{2}}{ 2 }\)
(c) πrBv and R is at higher potential
(d) 2rBv and R is at higher potential
Answer:
(d) 2rBv and R is at higher potential

Samacheerkalvi.Guru 12th Physics Question 3.
The flux linked with a coil at any instant t is given by ΦB = 10t2 – 50t + 250. The induced emf at t = 3s is-
(a) -190 V
(b) -10 V
(c) 10 V
(d) 190 V
Answer:
(b) -10 V

Samacheer Kalvi 12th Physics Solutions Question 4.
When the current changes from +2A to -2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is-
(a) 0.2 H
(b) 0.4 H
(c) 0.8 H
(d) 0.1 H
Answer:
(d) 0.1 H

Samacheer Kalvi 12th Physics Solution Book Question 5.
The current i flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be- (NEET-2011)
Samacheerkalvi.Guru 12th Physics Solutions Chapter 4 Electromagnetic Induction And Alternating Current
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction And Alternating Current
Answer:
Samacheer Kalvi 12th Physics Solution Book Chapter 4 Electromagnetic Induction And Alternating Current

Tn 12th Physics Solution Question 6.
A circular coil with a cross-sectional area of 4 cm2 has 10 turns. It is placed at the centre of a long solenoid that has 15 turns/cm and a cross-sectional area of 10 cm2. The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?
(a) 7.54 μH
(b) 8.54 μH
(c) 9.54 μH
(d) 10.54 μH
Answer:
(a) 7.54 μH

Samacheer Kalvi Guru 12th Physics Question 7.
In a transformer, the number of turns in the primary and the secondary are 410 and 1230, respectively. If the current in primary is 6A, then that in the secondary coil is-
(a) 2 A
(b) 18 A
(c) 12 A
(d) 1 A
Answer:
(a) 2 A

Physics Class 12 Chapter 4 Notes Question 8.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is-
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Alternating Current Class 12 Notes Pdf Download Question 9.
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac { 1 }{ 2 }\). Instead, if C is removed from the circuit, the phase difference is again \(\frac { π }{ 3 }\). The power factor of the circuit is- (NEET 2012)
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ √ 2 }\)
(c) 1
(d) \(\frac { √ 3 }{ 2 }\)
Answer:
(c) 1

Physics Solution Class 12 Samacheer Kalvi Question 10.
In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is-
(a) \(\frac { π }{ 4 }\)
(b) \(\frac { π }{ 2 }\)
(c) \(\frac { π }{ 6 }\)
(d) zero
Answer:
(a) \(\frac { π }{ 4 }\)

Samacheer 12th Physics Solutions Question 11.
In a series resonant RLC circuit, the voltage across 100 Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 μF, then the voltage across L is-
(a) 600 V
(b) 4000 V
(c) 400 V
(d) 1 V
Answer:
(c) 400 V

Question 12.
An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf v = 10 sin 340 t. The power loss in AC circuit is-
(a) 0.76 W
(b) 0.89 W
(c) 0.46 W
(d) 0.67 W
Answer:
(c) 0.46 W

Questions 13.
The instantaneous values of alternating current and voltage in a circuit are i = \(\frac { 1 }{ √ 2 }\) = sin(100πt) A and v = \(\frac { 1 }{ √ 2 }\) sin \(\left(100 \pi t+\frac{\pi}{3}\right)\) V. The average power in watts consumed in the circuit is-
(IIT Main 2012)
(a) \(\frac { 1 }{ 4 }\)
(b) \(\frac { √3 }{ 4 }\)
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ 8 }\)
Answer:
(d) \(\frac { 1 }{ 8 }\)

Question 14.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is-
(a) \(\frac { Q }{ 2 }\)
(b) \(\frac { Q }{ √3 }\)
(c) \(\frac { Q }{ √2 }\)
(d) \(\frac { Q }{ 2 }\)
Answer:
(c) \(\frac { Q }{ √2 }\)

Question 15.
\(\frac { 20 }{{ π }^{2}}\) H inductor is connected to a capacitor of capacitance C. The value of C in order to impart maximum power at 50 Hz is-
(a) 50 μF
(b) 0.5 μF
(c) 500 μF
(d) 5 μF
Answer:
(d) 5 μF

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Short Answer Questions

Question 1.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.

Question 2.
State Faraday’s laws of electromagnetic induction.
Answer:
First law:
Whenever magnetic flux linked with a closed circuit changes, an emf is induced in the circuit.

Second law:
The magnitude of induced emf in a closed circuit is equal to the time rate of change of magnetic flux linked with the circuit.

Question 3.
State Lenz’s law.
Answer:
State Lenz’s law:
Lenz’s law states that the direction of the induced current is such that it always opposes the cause responsible for its production.

Question 4.
State Fleming’s right hand rule.
Answer:
The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the Electromagnetic Induction and Alternating Current thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.

Question 5.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 6.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:

  1. By changing the magnetic field B
  2. By changing the area A of the coil and
  3. By changing the relative orientation 0 of the coil with magnetic field

Question 7.
What for an inductor is used? Give some examples.
Answer:
Inductor is a device used to store energy in a magnetic field when an electric current flows through it. The typical examples are coils, solenoids and toroids.

Question 8.
What do you mean by self-induction?
Answer:
If the magnetic flux is changed by changing the current, an emf is induced in that same coil. This phenomenon is known as self-induction.

Question 9.
What is meant by mutual induction?
Answer:
When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction.

Question 10.
Give the principle of AC generator.
Answer:
Alternators work on the principle of electromagnetic induction. The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf. The magnitude of the induced emf is given by Faraday’s law of electromagnetic induction and its direction by Fleming’s right hand rule.

Question 11.
List out the advantages of stationary armature-rotating field system of AC generator.
Answer:

  1. The current is drawn directly from fixed terminals on the stator without the use of brush contacts.
  2. The insulation of stationary armature winding is easier.
  3. The number of sliding contacts (slip rings) is reduced. Moreover, the sliding contacts are used for low-voltage DC Source.
  4. Armature windings can be constructed more rigidly to prevent deformation due to any mechanical stress.

Question 12.
What are step-up and step-down transformers?
Answer:
If the transformer converts an alternating current with low voltage into an alternating current with high voltage, it is called step-up transformer. On the contrary, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer.

Question 13.
Define average value of an alternating current.
Answer:
The average value of alternating current is defined as the average of all values of current over a positive half-cycle or negative half-cycle.

Question 14.
How will you define RMS value of an alternating current?
Answer:
RMS value of alternating current is defined as that value of the steady current which when flowing through a given circuit for a given time produces the same amount of heat as produced by the alternating current when flowing through the same circuit for the same time.

Question 15.
What are phasors?
Answer:
A sinusoidal alternating voltage (or current) can be represented by a vector which rotates about the origin in anti-clockwise direction at a constant angular velocity ω. Such a rotating vector is called a phasor.

Question 16.
Define electric resonance.
Answer:
When the frequency of the applied alternating source is equal to the natural frequency of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance.

Question 17.
What do you mean by resonant frequency?
Answer:
When the frequency of the applied alternating source (ωr) is equal to the natural frequency \(\left[\frac{1}{\sqrt{L C}}\right]\) of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance. The frequency at which resonance takes place is called resonant frequency. Resonant angular frequency, ωr = \(\frac { 1 }{ \sqrt { LC } } \)

Question 18.
How will you define Q-factor?
Answer:
It is defined as the ratio of voltage across L or C to the applied voltage.
Tn 12th Physics Solution Chapter 4 Electromagnetic Induction And Alternating Current Samacheer Kalvi

Question 19.
What is meant by wattles current?
Answer:
The component of current (IRMS sin φ), which has a phase angle of \(\frac { π }{ 2 }\) with the voltage is called reactive component. The power consumed is zero. So that it is also known as ‘Wattless’ current.

Question 20.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit. It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.
Samacheer Kalvi Guru 12th Physics Solutions Chapter 4 Electromagnetic Induction And Alternating Current

Question 21.
What are LC oscillations?
Answer:
Whenever energy is given to a LC circuit, the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations. During LC oscillations, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Long Answer Questions

Question 1.
Establish the fact that the relative motion between the coil and the magnet induces an emf in the coil of a closed circuit.
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.
The relative motion between the coil and the magnet induces:
In the first experiment, when a bar magnet is placed close to a coil, some of the magnetic field lines of the bar magnet pass through the coil i.e., the magnetic flux is linked with the coil. When the bar magnet and the coil approach each other, the magnetic flux linked with the coil increases. So this increase in magnetic flux induces an emf and hence a transient electric current flows in the circuit in one direction (Figure(a)).
Physics Class 12 Chapter 4 Notes Electromagnetic Induction And Alternating Current Samacheer Kalvi
At the same time, when they recede away from one another, the magnetic flux linked with the coil decreases. The decrease in magnetic flux again induces an emf in opposite direction and hence an electric current flows in opposite direction (Figure (b)). So there is deflection in the galvanometer when there is a relative motion between the coil and the magnet.
Alternating Current Class 12 Notes Pdf Download Chapter 4 Samacheer Kalvi
In the second experiment, when the primary coil P carries an electric current, a magnetic field is established around it. The magnetic lines of this field pass through itself and the neighbouring secondary coil S.
Physics Solution Class 12 Samacheer Kalvi Chapter 4 Electromagnetic Induction And Alternating Current
When the primary circuit is open, no electric current flows in it and hence the magnetic flux linked with the secondary coil is zero (Figure(a)).
However, when the primary circuit is closed, the increasing current builds up a magnetic field around the primary coil. Therefore, the magnetic flux linked with the secondary coil increases. This increasing flux linked induces a transient electric current in the secondary coil (Figure(b)).

When the electric current in the primary coil reaches a steady value, the magnetic flux linked with the secondary coil does not change and the electric current in the secondary coil will disappear. Similarly, when the primary circuit is broken, the decreasing primary current induces an electric current in the secondary coil, but in the opposite direction (Figure (c)). So there is deflection in the galvanometer whenever there is a change in the primary current

Question 2.
Give an illustration of determining direction of induced current by using Lenz’s law.
Answer:
Illustration 1:
Consider a uniform magnetic field, with its field lines perpendicular to the plane of the paper and pointing inwards. These field lines are represented by crosses (x) as shown in figure (a). A rectangular metallic frame ABCD is placed in this magnetic field, with its plane perpendicular to the field. The arm AB is movable so that it can slide towards right or left.
Samacheer 12th Physics Solutions Chapter 4 Electromagnetic Induction And Alternating Current
If the arm AB slides to our right side, the number of field lines (magnetic flux) passing through the frame ABCD increases and a current is induced. As suggested by Lenz’s law, the induced current opposes this flux increase and it tries to reduce it by producing another magnetic field pointing outwards i.e., opposite to the existing magnetic field.

The magnetic lines of this induced field are represented by circles in the figure (b). From the direction of the magnetic field thus produced, the direction of the induced current is found to be anti-clockwise by using right-hand thumb rule.
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The leftward motion of arm AB decreases magnetic flux. The induced current, this time, produces a magnetic field in the inward direction i.e., in the direction of the existing magnetic field (figure (c)). Therefore, the flux decrease is opposed by the flow of induced current. From this, it is found that induced current flows in clockwise direction.

Illustration 2:
Let us move a bar magnet towards the solenoid, with its north pole pointing the solenoid as shown in figure (b). This motion increases the magnetic flux of the coil which in turn, induces an electric current. Due to the flow of induced current, the coil becomes a magnetic dipole whose two magnetic poles are on either end of the coil.
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In this case, the cause producing the induced current is the movement of the magnet. According to Lenz’s law, the induced current should flow in such a way that it opposes the movement of the north pole towards coil. It is possible if the end nearer to the magnet becomes north pole (figure (b)).

Then it repels the north pole of the bar magnet and opposes the movement of the magnet. Once pole ends are known, the direction of the induced current could be found by using right hand thumb rule. When the bar magnet is withdrawn, the nearer end becomes south pole which attracts north pole of the bar magnet, opposing the receding motion of the magnet (figure (c)). Thus the direction of the induced current can be found from Lenz’s law.

Question 3.
Show that Lenz’s law is in accordance with the law of conservation of energy.
Answer:
Conservation of energy:
The truth of Lenz’s law can be established on the basis of the law of conservation of energy. According to Lenz’s law, when a magnet is moved either towards or away from a coil, the induced current produced opposes its motion. As a result, there will always be a resisting force on the moving magnet.

Work has to be done by some external agency to move the magnet against this resisting force. Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into Joule heat in the coil i.e., energy is converted from one form to another.

Question 4.
Obtain an expression for motional emf from Lorentz force.
Answer:
Motional emf from Lorentz force:
Consider a straight conducting rod AB of length l in a uniform magnetic field \(\vec { B } \) which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity \(\vec { v } \) towards right side.
When the rod moves, the free electrons present in it also move with same velocity \(\vec { v } \) in \(\vec { B } \). As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation
\(\vec { F } \)B = -e(\(\vec { v } \) x \(\vec { B } \) ) ……. (1)
The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field E, the coulomb force starts acting on the free electrons along AB and is given by
\(\vec { F } \)E = -e\(\vec { E } \) ……. (2)
The magnitude of the electric field \(\vec { E } \) keeps on increasing as long as accumulation of electrons at the end A continues. The force \(\vec { F } \)E also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force \(\vec { F } \)B and the coulomb force \(\vec { F } \)E balance each other and no further accumulation of free electrons at the end A takes place, i.e.,
\(\left| \vec { { F }_{ B } } \right| \) = \(\left| \vec { { F }_{ E } } \right| \)
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vB sin 90° = E
vB = E ……. (3)
The potential difference between two ends of the rod is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-15
Figure: Motional emf from Lorentz force
V = El
V = vBl
Thus the Lorentz force on the free electrons is responsible to maintain this . potential difference and hence produces an emf
ε = Blv ….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.

Question 5.
Using Faraday’s law of electromagnetic induction, derive an equation for motional emf.
Answer:
Motional emf from Faraday’s law:
Let us consider a rectangular conducting loop of width l in a uniform magnetic field \(\vec { B } \) which is perpendicular to the plane of the loop and is directed inwards. A part of the loop is in the magnetic field while the remaining part is outside the field.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-16
Figure: Motional emf from Faraday’s law

When the loop is pulled with a constant velocity \(\vec { v } \) to the right, the area of the portion of the loop within the magnetic field will decrease. Thus, the flux linked with the loop will also decrease. According to Faraday’s law, an electric current is induced in the loop which flow’s in a direction so as to oppose the pull of the loop.
Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-17
As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by
ε = \(\frac {{ dΦ }_{B}}{ dt }\) = \(\frac { d }{ dt }\) (Blx)
Here, both B and l are constants. Therefore,
ε = Bl \(\frac { dx}{ dl }\) = Blv …… (2)
where v = \(\frac { dx}{ dt }\) is the velocity of the loop. This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.

Question 6.
Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are

  1. Induction stove
  2. Eddy current brake
  3. Eddy current testing
  4. Electromagnetic damping

1. Induction stove:
Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone.

When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

2. Eddy current brake:
This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.

In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.

3. Eddy current testing:
It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field.

When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

4. Electro magnetic damping:
The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 7.
Define self-inductance of a coil interns of (i) magnetic flux and (ii) induced emf.
Answer:
Self-inductance or simply inductance of a coil is defined as the flux linkage of the coil when 1A current flows through it.
When the current i changes with time, an emf is induced in it. From Faraday’s law of electromagnetic induction, this self-induced emf is given by
ε = –\(\frac{d\left(\mathrm{N} \Phi_{\mathrm{B}}\right)}{d t}\) = –\(\frac { d(Li)}{ dt }\)
∴ ε = -L\(\frac { di}{ dt }\) or L = \(\frac { -ε}{ di/dt }\)
The negative sign in the above equation means that the self-induced emf always opposes the change in current with respect to time. If \(\frac { di}{ dt }\) = 1 As-1, then L= -ε. Inductance of a coil is also defined as the opposing emf induced in the coil when the rate of change of current through the coil is 1 A s-1.

Question 8.
How will you define the unit of inductance?
Answer:
Unit of inductance: Inductance is a scalar and its unit is Wb A-1 or V s A-1. It is also measured in henry (H).
1 H = 1 Wb A-1 = 1 V s A-1
The dimensional formula of inductance is M L2 T-2A-2.
If i = 1 A and NΦB = 1 Wb turns, then L = 1 H.
Therefore, the inductance of the coil is said to be one henry if a current of 1 A produces unit flux linkage in the coil.
If \(\frac { di}{ dt }\) = 1 As-1 and ε = -1 V, then L = 1 H.
Therefore, the inductance of the coil is one henry if a current changing at the rate of 1 A s-1 induces an opposing emf of 1 V in it.

Question 9.
What do you understand by self-inductance of a coil? Give its physical significance.
Answer:
Self-inductance or simply inductance of a coil is defined as the flux linkage of the coil when 1A current flows through it.
When the current i changes with time, an emf is induced in it. From Faraday’s law of electromagnetic induction, this self-induced emf is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-18
Physical significance of inductance:
When a circuit is switched on, the increasing current induces an emf which opposes the growth of current in a circuit. Likewise, when circuit is broken, the decreasing current induces an emf in the reverse direction. This emf now opposes the decay of current.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-19
Figure: Induced emf ε opposes the changing current i

Thus, inductance of the coil opposes any change in current and tries to maintain the original state.

Question 10.
Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance.
Answer:
Self-inductance of a long solenoid:
Consider a long solenoid of length l and cross-sectional area A. Let n be the number of turns per unit length (or turn density) of the solenoid. When an electric current i is passed through the solenoid, a magnetic field is produced by it which is almost uniform and is directed along the axis of the solenoid. The magnetic field at any point inside the solenoid is given by
B = μ0ni
As this magnetic field passes through the solenoid, the windings of the solenoid are linked by the field lines. The magnetic flux passing through each turn is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-20
The total magnetic flux linked or flux linkage of the solenoid with N turns (the total number of turns N is given by N = nl) is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-21
B = n (nl) (μ0ni)A
B = (μ0n2Al)i ….. (1)
From the self induction
B = LI ….. (2)
Comparing equations (1) and (2), we have L = μ0n2Al
From the above equation, it is clear that inductance depends on the geometry of the solenoid (turn density n, cross-sectional area A, length l) and the medium present inside the solenoid. If the solenoid is filled with a dielectric medium of relative permeability μr, then
L = μ0
L = μn0μrn2Al

Question 11.
An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?
Answer:
Energy stored in an inductor:
Whenever a current is established in the circuit, the inductance opposes the growth of the current. In order to establish a current in the circuit, work is done against this opposition by some external agency. This work done is stored as magnetic potential energy.

Let us assume that electrical resistance of the inductor is negligible and inductor effect alone is considered. The induced emf e at any instant t is
ε = -L\(\frac { di}{ dt }\) …… (1)
Let dW be work done in moving a charge dq in a time dt against the opposition, then
dW = -εdq = -εidi = εidi           [∵dq = idt]
Substituting for s value from equation (1)
= – \(\left(-\mathrm{L} \frac{d i}{d t}\right)\) idt
dW = Lid …… (2)
Total work done in establishing the current i is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-22
This work done is stored as magnetic potential energy.
UB = \(\frac { 1 }{ 2 }\) Li2 …….. (4)

Question 12.
Show that the mutual inductance between a pair of coils is same (M12 = M21).
Answer:
Mutual induction:
When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction and the emf is called mutually induced emf.
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Consider two coils which are placed close to each other. If an electric current i1 is sent through coil 1, the magnetic field produced by it is also linked with coil 2. Let Φ21 be the magnetic flux linked with each turn of the coil 2 of N2 turns due to coil 1, then the total flux linked with coil 2 (N2Φ21) is proportional to the current i1 in the coil 1.
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The constant of proportionality M21 is the mutual inductance of the coil 2 with respect to coil 1. It is also called as coefficient of mutual induction. If i1 = 1A, then M21 = N2Φ21.
Therefore, the mutual inductance M21 is defined as the flux linkage of the coil 2 when 1A current flows through coil 1. When the current changes with time, an emf ε2 is induced in coil 2. From Faraday’s law of electromagnetic induction, this mutually induced emf ε2 is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-25
The negative sign in the above equation shows that the mutually induced emf always opposes the change in current i, with respect to time. If \(\frac { di }{ dt }\) = 1 As-1, then M21 = -ε2. Mutual inductance M21, is also defined as the opposing emf induced in the coil 2 when the rate of change of current through the coil 1 is 1 As-1. Similarly, if an electric current i2 through coil 2 changes with time, then emf ε1 is induced in coil 1. Therefore,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-26
where M12 is the mutual inductance of the coil 1 with respect to coil 2. It can be shown that for a given pair of coils, the mutual inductance is same, i.e., M21 = M12 = M.
In general, the mutual induction between two coils depends on size, shape, the number of turns of the coils, their relative orientation and permeability of the medium.

Question 13.
How will you induce an emf by changing the area enclosed by the coil?
Answer:
Induction of emf by changing the area of the coil:
Consider a conducting rod of length 1 moving with a velocity v towards left on a rectangular metallic framework. The whole arrangement is placed in a uniform magnetic field \(\vec { B } \) whose magnetic lines are perpendicularly directed into the plane of the paper. As the rod moves from AB to DC in a time dt, the area enclosed by the loop and hence the magnetic flux through the loop decreases.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-27
The change in magnetic flux in time dt is
B = B x change in area
B x Area ABCD
= Blvdt since Area ABCD = l(vdt)
or \(\frac {{ dΦ }_{B}}{ dt }\) = Blv
As a result of change in flux, an emf is generated in the loop. The magnitude of the induced emf is
ε = \(\frac {{ dΦ }_{B}}{ dt }\) = Blv
This emf is called motional emf. The direction of induced current is found to be clockwise from Fleming’s right hand rule.

Question 14.
Show mathematically that the rotation of a coil in a magnetic field over one rotation induces an alternating emf of one cycle.
Answer:
Induction of emf by changing relative orientation of the coil with the magnetic field:
Consider a rectangular coil of N turns kept in a uniform magnetic field \(\vec { B } \) figure (a). The coil rotates in anti-clockwise direction with an angular velocity ω about an axis, perpendicular to the field. At time = 0, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value Φm = BA (where A is the area of the coil).
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In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux linked is Φm cos ωt, a component of Φm normal to the plane of the coil (figure (b)). The component parallel to the plane (Φm sin ωt) has no role in electromagnetic induction. Therefore, the flux linkage at this deflected position is NΦB = NΦm cos ωt. According to Faraday’s law, the emf induced at that instant is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-29
ε= \(\frac { d }{ dt }\) (NΦB ) = \(\frac { d }{ dt }\) (NΦm cos ωt)
= -NΦm (-sin ωt)ω = NΦm ω sin ωt
When the coil is rotated through 90° from initial position, sin ωt = 1, Then the maximum value of induced emf is
εm = NΦmω = NBAω since Φm = BA
Therefore, the value of induced emf at that instant is then given by
ε = εm sin ωt
It is seen that the induced emf varies as sine function of the time angle ωt. The graph between – induced emf and time angle for one rotation of coil will be a sine curve and the emf varying in this manner is called sinusoidal emf or alternating emf.

Question 15.
Elaborate the standard construction details of AC generator.
Answer:
Construction:
lternator consists of two major parts, namely stator and rotor. As their names suggest, stator is stationary while rotor rotates inside the stator. In any standard construction of commercial alternators, the armature winding is mounted on stator and the field magnet on rotor. The construction details of stator, rotor and various other components involved in them are given below.

(i) Stator:
The stationary part which has armature windings mounted in it is called stator. It has three components, namely stator frame, stator core and armature winding.

Stator frame:
This is the outer frame used for holding stator core and armature windings in proper position. Stator frame provides best ventilation with the help of holes provided in the frame itself.
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Stator core:
Stator core or armature core is made up of iron or steel alloy. It is a hollow cylinder and is laminated to minimize eddy current loss. The slots are cut on inner surface of the core to accommodate armature windings.
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Armature winding:
Armature winding is the coil, wound on slots provided in the armature core. One or more than one coil may be employed, depending on the type of alternator. Two types of windings are commonly used. They are (i) single-layer winding and (ii) double-layer winding. In single-layer winding, a slot is occupied by a coil as a single layer. But in double-layer winding, the coils are split into two layers such as top and bottom layers.

(ii) Rotor:
Rotor contains magnetic field windings. The magnetic poles are magnetized by DC source. The ends of field windings are connected to a pair of slip rings, attached to a common shaft about which rotor rotates. Slip rings rotate along with rotor. To maintain connection between the DC source and field windings, two brushes are used which . continuously slide over the slip rings.

There are 2 types of rotors used in alternators:

  1. salient pole rotor
  2. cylindrical pole rotor.

1. Salient pole rotor:
The word salient means projecting. This rotor has a number of projecting poles having their bases riveted to the rotor. It is mainly used in low-speed alternators.

2. Cylindrical pole rotor:
This rotor consists of a smooth solid cylinder. The slots are cut on the outer surface of the cylinder along its length. It is suitable for very high speed alternators.

The frequency of alternating emf induced is directly proportional to the rotor speed. In order to maintain the frequency constant, the rotor must run at a constant speed. These are standard construction details of alternators.

Question 16.
Explain the working of a single-phase AC generator with necessary diagram.
Answer:
Single phase AC generator: In a single phase AC generator, the armature conductors are connected in series so as to form a single circuit which generates a single-phase alternating emf and hence it is called single-phase alternator.
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The simplified version of a AC generator is discussed here. Consider a stator core consisting of 2 slots in which 2 armature conductors PQ and RS are mounted to form single-turn rectangular loop PQRS. Rotor has 2 salient poles with field windings which can be magnetized by means of DC source.

Working:
The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. The direction of magnetic field passing through the armature core. Let the field magnet be rotated in clockwise direction by the prime mover. The axis of rotation is perpendicular to the plane of the paper.

Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.

When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced cmfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming’s right hand rule. Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field.

For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Hence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is downwards B and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.
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For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B. The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP This is represented by point C.

On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature. Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency of the induced emf depends on the speed at which the field magnet rotates.
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Question 17.
How are the three different emfs generated in a three-phase AC generator? Show the graphical representation of these three emfs.
Answer:
Three-phase AC generator:
Some AC generators may have more than one coil in the armature core and each coil produces an alternating emf. In these generators, more than one emf is produced. Thus they are called poly-phase generators. If there are two alternating emfs produced in a generator, it is called two-phase generator. In some AC generators, there are three separate coils, which would give three separate emfs. Hence they are called three-phase AC generators.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-35
In the simplified construction of three-phase AC generator, the armature core has 6 slots, cut on its inner rim. Each slot is 60° away from one another. Six armature conductors are mounted in these slots. The conductors 1 and 4 are joined in series to form coil 1. The conductors 3 and 6 form coil 2 while the conductors 5 and 2 form coil 3. So, these coils are rectangular in shape and are 120° apart from one another.

The initial position of the field magnet is horizontal and field direction is perpendicular to the plane of the coil 1. As it is seen in single phase AC generator, when field magnet is rotated from that position in clockwise direction, alternating emf ε1 in coil 1 begins a cycle from origin O.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-36
The corresponding cycle for alternating emf ε2 in coil 2 starts at point A after field magnet has rotated through 120°. Therefore, the phase difference between ε1 and ε2 is 120°. Similarly, emf ε3 in coil 3 would begin its cycle at point B after 240° rotation of field magnet from initial position. Thus these emfs produced in the three phase AC generator have 120° phase difference between one another.

Question 18.
Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle:
The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-37

Construction:
In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.

The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.

Working:
If the primary’ coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coif This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils. As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil εp is almost equal and opposite to the applied voltage υp and is given by
υp = εp = -Np \(\frac {{ dΦ }_{B}}{ dt }\) …….. (1)
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil eg is given by
εs = -Ns \(\frac {{ dΦ }_{B}}{ dt }\)
where Np and Ns are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then εs = υs where υs is the voltage across secondary coil.
υs εs = -Ns \(\frac {{ dΦ }_{B}}{ dt }\) ……… (2)
From equation (1) and (2),
\(\frac {{ υ }_{s}}{{ ε }{s}}\) = \(\frac {{ N }_{s}}{{ N }{p}}\) = K …….. (3)
This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υp ip = Output power υsis
where iυp and is are the currents in the primary and secondary coil respectively. Therefore,
\(\frac {{ υ }_{s}}{{ υ }{p}}\) = \(\frac {{ N }_{s}}{{ N }{p}}\) = \(\frac {{ i }_{p}}{{ i }{s}}\)
Equation (4) is written in terms of amplitude of corresponding quantities,
\(\frac {{ V }_{s}}{{ V }{p}}\) = \(\frac {{ N }_{s}}{{ N }{p}}\) = \(\frac {{ I }_{p}}{{ I }{s}}\) = K ……. (4)

(i) If Ns > Np ( or K > 1)
∴ Vs > Vp and Is < Ip.
This is sthe case of step-up transformer in which voltage is decreased and the corresponding current is decreased.

(ii) If Ns < Np (or K < 1)
∴ Vs < Vp and Is > Ip
This is step-down transformer where voltage is decreased and the current is increased.

Question 19.
Mention the various energy losses in a transformer.
Answer:
Energy losses in a transformer: Transformers do not have any moving parts so that its efficiency is much higher than that of rotating machines like generators and motors. But there are many factors which lead to energy loss in a transformer.

(i) Core loss or Iron loss:
This loss takes place in transformer core. Hysteresis loss and eddy current loss are known as core loss or Iron loss. When transformer core is magnetized and demagnetized repeatedly by the alternating voltage applied across primary coil, hysteresis takes place due to which some energy is lost in the form of heat.

Hysteresis loss is minimized by using steel of high silicon content in making transformer core. Alternating magnetic flux in the core induces eddy currents in it. Therefore there is energy loss due to the flow of eddy current, called eddy current loss which is minimized by using very thin laminations of transformer core.

(ii) Copper loss:
Transformer windings have electrical resistance. When an electric current flows through them, some amount of energy is dissipated due to Joule heating. This energy loss is called copper loss which is minimized by using wires of larger diameter.

(iii) Flux leakage:
Flux leakage happens when the magnetic lines of primary coil arc not completely linked with secondary coil. Energy loss due to this flux leakage is minimized by winding coils one over the other.

Question 20.
Give the advantage of AC in long distance power transmission with an example.
Answer:
Advantages of AC in long distance power transmission:
Electric power is produced in a t large scale at electric power stations with the help of AC generators. These power stations are classified based on the type of fuel used as thermal, hydro electric and nuclear power stations. Most of these stations are located at remote places.

Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually consumed. This process is called power transmission. But there is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating (i2R) in the transmission lines which are hundreds of kilometer long.

This power loss can be tackled either by reducing current i or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick wires of copper or aluminium. But this increases the cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-38
Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent.

At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss.

At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically.

Illustration:
An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.

Case I:
P = 2 MW; R == 40 Ω; V = 10 kV Power,
Power, P = VI
∴ Current, I = \(\frac { P }{ V }\) = \(\frac {{ 2 × 10 }^{6}}{{ 10 × 10 }^{3}}\) 200 A
Power loss = Heat produced = I2R = (200)2 × 40 = 1.6 × 106 W
% of power loss =\(\frac {{ 1.6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) × 100% = 0.8 × 100% = 80%

Case II:
P = 2 MW; R == 40 Ω; V = 100 kV
∴ Current, I = \(\frac { P }{ V }\) = \(\frac {{ 2 × 10 }^{6}}{{ 100 × 10 }^{3}}\) 20 A
Power loss = Heat produced = I2R = (20)2 × 40 = 0.016 × 106 W
% of power loss =\(\frac {{ 0.6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) × 100% = 0.008 × 100% = 0.8%

Question 21.
Find out the phase relationship between voltage and current in a pure inductive circuit.
Answer:
AC circuit containing only an inductor:
Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source. The alternating voltage is given by the equation.
υ = Vm sin ωt …(1)
The alternating current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by
Back emf, ε -L\(\frac { di }{ dt }\)
By applying Kirchoff’s loop rule to the purely inductive circuit, we get
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-39
υ + ε = 0
Vm sin ωt = L\(\frac { di }{ dt }\)
di = L\(\frac {{ V }_{m}}{ L }\) sin ωt dt
i = \(\frac {{ V }_{m}}{ L }\) \(\int { sin } \) ωt dt = \(\frac{{ V }_{m}}{ Lω }\) (-cos ωt) + constant
The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-40
where \(\frac{{ V }_{m}}{ Lω }\) = Im, the peak value of the alternating current in the circuit. From equation (1) and (2), it is evident that current lags behind the applied voltage by \(\frac{π}{ 2 }\) in an inductive circuit.
This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by 90°.

Inductive reactance XL:
The peak value of current Im is given by Im = \(\frac{{ V }_{m}}{ Lω }\) . Let us compare this equation with Im = \(\frac{{ V }_{m}}{ R }\) from resistive circuit. The equantity ωL Plays the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance (XL). It is measured in ohm.
XL = ωL
The inductive reactance (XL) varies directly as the frequency.
XL = 2πfL …….. (3)
where ƒ is the frequency of the alternating current. For a steady current, ƒ= 0. Therefore, XL = 0. Thus an ideal inductor offers no resistance to steady DC current.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-41

Question 22.
Derive an expression for phase angle between the applied voltage and current in a series RLC circuit.
Answer:
AC circuit containing a resistor, an inductor and a capacitor in series – Series RLC
circuit:
Consider a circuit containing a resistor of resistance R, a inductor of inductance L and a capacitor of capacitance C connected across an alternating voltage source. The applied alternating voltage is given by the equation.
υ = Vm sin ωt …… (1)
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-42
Let i be the resulting circuit current in the circuit at that instant. As a result, the voltage is developed across R, L and C.
We know that voltage across R (VR) is in phase with i, voltage across L (VL) leads i by π/2 and voltage across C (VC) lags i by π/2.
The phasor diagram is drawn with current as the reference phasor. The current is represented by the phasor
\(\vec { OI } \), VR by \(\vec { OA } \) ; VL by \(\vec { OB } \) and VC by \(\vec { OC } \).
The length of these phasors are
OI = Im, OA = ImR, OB = Im,XL; OC = ImXc
The circuit is cither effectively inductive or capacitive or resistive that depends on the value of V1 or Vc Let us assume that VL > VC. so that net voltage drop across L – C combination is VL < VC which is represented by a phasor \(\vec { AD } \).
By parallelogram law, the diagonal \(\vec { OE } \) gives the resultant voltage u of VR and (VL – VC ) and its length OE is equal to Vm. Therefore,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-43
Z is called impedance of the circuit which refers to the effective opposition to the circuit current by the series RLC circuit. The voltage triangle and impedance triangle are given in the graphical figure.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-44
From phasor diagram, the phase angle between n and i is found out from the following relation
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-45
Special cases Figure: Phasor diagram for a series
(i) If XL > XC, (XL – XC) is positive and phase angle φ
is also positive. It means that the applied voltage leads the current by φ (or current lags behind voltage by φ). The circuit is inductive.
∴ υ = Vm sin ωt; i = Im sin(ωt + φ)

(ii) If XL < XC, (XL – XC) is negative and φ is also negative. Therefore current leads voltage by φ and the circuit is capacitive.
∴ υ = Vm sin ωt; i = Im sin(ωt + φ)

(iii) If XL = XC, φ is zero. Therefore current and voltage are in the same phase and the circuit is resistive.
∴ υ = Vm sin ωt; i = Im sin ωt

Question 23.
Define inductive and capacitive reactance. Give their units.
Answer:
Inductive reactance XL:
The peak value of current Im is given by Im = \(\frac{{ V }_{m}}{ Lω }\). Let us compare
this equation with Im \(\frac{{ V }_{m}}{ R }\) from resistive circuit. The quantity ωL plays the same role as the resistance R in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance (XL). It measured in ohm.
XL = ωL

Capacitive reactance XC:
The peak value of current I is given by Im = \(\frac{\mathrm{v}_{\mathrm{m}}}{1 / \mathrm{c} \omega}\). Let us compare this equation with Im = \(\frac{{ V }_{m}}{ R }\) from resistive circuit. The quantity \(\frac { 1 }{ ωC }\) plays the same role as the resistance R in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance (XC). It measured in ohm.
XC = \(\frac{ 1 }{ ωC }\).

Question 24.
Obtain an expression for average power of AC over a cycle. Discuss its special cases. Power of a circuit is defined as the rate of consumption of electric energy in that circuit.
Answer:
It is given by the product of the voltage and current. In an AC circuit, the voltage and current vary continuously with time. Let us first calculate the power at an instant and then it is averaged over a complete cycle.
The alternating voltage and alternating current in the series RLC circuit at an instant are given by
υ = Vm sin ωt and i = Im sin (ωt + 4>)
where φ is the phase angle between υ and i. The instantaneous power is then written as
P = υi = Vm Im sin ωt sin(ωt + φ)
= Vm Im sin ωt (sin ωt cos φ – cos ωt sin φ)
P = Vm Im (cos φ sin2 ωt – sin ωt cos ωt sin φ) …… (1)
Here the average of sin2 ωt over a cycle is \(\frac { 1 }{ 2 }\) and that of sin ωt cos ωt is zero. Substituting these values, we obtain average power over a cycle.
Pav = Vm Im cos φ x \(\frac { 1 }{ 2 }\) = \(\frac {{ V }_{m}}{ √2 }\) \(\frac {{ I }_{m}}{ √2 }\) cos φ
Pav = VRMS IRMS cos φ …… (2)
where VRMS IRMS is called apparent power and cos φ is power factor. The average power of an AC circuit is also known as the true power of the circuit.
Special Cases:
(i) For a purely resistive circuit, the phase angle between voltage and current is zero and cos
φ = 1.
∴ Pav = VRMS IRMS
(ii) For a purely inductive or capacitive circuit, the phase angle is ± \(\frac { π }{ 2 }\) and cos \(\left(\pm \frac{\pi}{2}\right)\) = 0
∴ Pav = 0
(iii) For series RLC circuit, the phase angle φ = tan-1 \(\left(\frac{\mathrm{x}_{\mathrm{L}}-\mathrm{x}_{\mathrm{c}}}{\mathrm{R}}\right)\)
∴ Pav = VRMS IRMS cos φ
(iv) For series RLC circuit at resonance, the phase angle is zero and cos φ = 1.
∴ Pav = VRMS IRMS

Question 25.
Show that the total energy is conserved during LC oscillations.
Answer:
Conservation of energy in LC oscillations: During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor. Although, these two forms of energy vary with time, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-46
Let us consider 3 different stages of LC oscillations and calculate the total energy of the system.

Case I:
When the charge in the capacitor, q = Qm and the current through the inductor, i = 0, the total energy is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-47
The total energy is wholly electrical.

Case II:
When charge = 0; current = Im, the total energy is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-48
The total energy is wholly electrical.

Case III:
When charge = q; current = i, the total energy is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-49
Since q = Qm cos ωt, i = \(\frac { bq }{ dt }\) = Qmω sin ωt. The negative sign in current indicates that the charge in the capacitor in the capacitor decreases with time.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-50
From above three cases, it is clear that the total energy of the system remains constant.

Question 26.
Prove that energy is conserved during electromagnetic induction.
Answer:
The mechanical energy of the spring-mass system is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-51
The energy E remains constant for varying values of x and v. Differentiating E with respect to time, we get
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-52
This is the differential equation of the oscillations of the spring-mass system. The general solution of equation (2) is of the form
x(t) = Xm cos (ωt + φ) …… (3)
where Xm is the maximum value of x(t), ω, the angular frequency and φ, the phase constant. Similarly, the electromagnetic energy of the LC system is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-53
Differentiating U with respect to time, we get
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-54
Equation (2) and (5) are proved the energy of electromagnetic induction is conserved.

Question 27.
Compare the electromagnetic oscillations of LC circuit with the mechanical oscillations of block spring system to find the expression for angular frequency of LC oscillators mathematically.
Answer:
The mechanical energy of the spring-mass system is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-51
The energy E remains constant for varying values of x and v. Differentiating E with respect to time, we get
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-52
This is the differential equation of the oscillations of the spring-mass system. The general solution of equation (2) is of the form
x(t) = Xm cos (ωt + φ) …… (3)
where Xm is the maximum value of x(t), ω, the angular frequency and φ, the phase constant. Similarly, the electromagnetic energy of the LC system is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-53
Differentiating U with respect to time, we get
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-54
Equation (2) and (5) are proved the energy of electromagnetic induction is conserved.
q(t) = Qm cos (ωt + φ) …… (6)
where Qm is the maximum value of q(t), ω, the angular frequency and φ, the phase constant.

Current in the LC circuit:
The current flowing in the LC circuit is obtained by differentiating q(t) with respect to time.
i(t) = \(\frac { dq }{ dt }\) = \(\frac { d }{ dt }\) [Qm cos (ωt + φ)] = Qm ω sin (ωt + φ) since Im = Qmω
(or)
i(t) -Im sin (ωt + φ) ……. (7)
The equation (7) clearly shows that current varies as a function of time t. In fact, it is a sinusoidally varying alternating current with angular frequency ω.

Angular frequency of LC oscillations:
By differentiating equation (6) twice, we get
\(\frac { { d }^{ 2 }q }{ dt } \) = -Qmω2 cos (ωt + φ) …….. (8)
Substituting equations (6) and (8) in equation (5),
we obtain L[-Qmω2 cos (ωt + φ)] + \(\frac { 1 }{ C }\) Qm cos (ωt + φ) = 0
Rearranging the terms, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \) …… (9)
This equation is the same as that obtained from qualitative analogy.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Numerical Problems

Question 1.
A square coil of side 30 cm with 500 turns is kept in a uniform magnetic field of 0.4 T. The plane of the coil is inclined at an angle of 30° to the field. Calculate the magnetic flux through the coil.
Solution:
Square coil of side (a) = 30 cm = 30 × 10-2m
Area of square coil (A) = a2 = (30 × 10-2)2 = 9 × 10-2 m2
Number of turns (N) = 500
Magnetic field (B) = 0.4 T
Angular between the field and coil (θ) = 90 – 30 = 60°
Magnetic flux (Φ) = NBA cos 0 = 500 × 0.4 × 9 × 10-2 × cos 60° = 18 × \(\frac { 1 }{ 2 }\)
Φ = 9 W b

Question 2.
A straight metal wire crosses a magnetic field of flux 4 mWb in a time 0.4 s. Find the magnitude of the emf induced in the wire.
Solution:
Magnetic flux (Φ) = 4 m Wb = 4 × 10-3 Wb
time (t) = 0.4 s
The magnitude of induced emf (e) = \(\frac { dΦ }{ dt }\) = \(\frac {{ 4 × 10 }^{-3}}{ 0.4 }\) 10-2
e = 10 mV

Question 3.
The magnetic flux passing through a coil perpendicular to its plane is a function of time and is given by OB = (2t3 + 4t2 + 8t + 8) Wb. If the resistance of the coil is 5 Ω, determine the induced current through the coil at a time t = 3 second.
Solution:
Magnetic flux (ΦB) = (2t3 + 8t2 + 8t + 8)Wb
Resistance of the coil (R) = 5 Ω
time (t) = 3 second
Induced current through the coil, I = \(\frac { e }{ R }\)
Induced emf, e = \(\frac {{ dΦ }_{B}}{ dt }\) = \(\frac { d }{ dt }\) ((2t3 + 4t2 + 8t + 8) = 6t2 + 8t + 8
Here time (t) = 3 second
e = 6(3)2 + 8 × 3 + 8 = 54 + 24 + 8 = 86 V
∴ Induced current through the coil, I = \(\frac { e }{ R }\) = \(\frac { 86 }{ 5 }\) = 17.2 A

Question 4.
A closely wound coil of radius 0.02 m is placed perpendicular to the magnetic field. When the magnetic field is changed from 8000 T to 2000 T in 6 s, an emf of 44 V is induced. Calculate the number of turns in the coil.
Solution:
Radius of the coil (r) = 0.02 m
Area of the coil (A) = πr² = 3.14 × (0.02)²= 1.256 × 10-3
Change in magnetic field, dB = 8000 – 2000 = 6000 T
Time, dt = 6 second
Induced emf, e = 44 V
θ = 0°
Induced emf in the coil, e = NA \(\frac { d }{ dt }\) cos θ . dt
44 = N × 1.256 × 10-3 × \(\frac { 600 }{ 6 }\) × Cos 0°
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-55
Number of turns N = 35 turns

Question 5.
A rectangular coil of area 6 cm2 having 3500 turns is kept in a uniform magnetic field of 0.4 T, Initially, the plane of the coil is perpendicular to the field and is then rotated through an angle of 180°. If the resistance of the coil is 35 Ω, find the amount of charge flowing through the coil.
Solution:
Rectangular coil of their area, A = 6 cm² = 6 x 10-4
Number of turns N = 3500 turns
Magnetic field, B = 0.4 T
Resistance of the coil, R= 35 Ω
Induced emf (e) = change in flux per second = Φ2 – Φ1
e = NAB cos 180° – NBA cos 0° = -NBA – NBA = – 2 NBA
= – 2 x 3500 x 0.4 x 6 x 10-4 – 16800 x 10-4 = – 1.68 V
Current flowing the coil, I = \(\frac { e }{ R }\) = \(\frac { -1.68 }{ 35 }\) = 0.048
Magnitude of the current, I = 48 x 10-3 A
Amount of charge flowing through the coil, q = It = 48 x 10-3 x 1 = 48 x 10-3 C

Question 6.
An induced current of 2.5 mA flows through a single conductor of resistance 100 Ω. Find out the rate at which the magnetic flux is cut by the conductor.
Solution:
Induced current, I = 2.5 mA
Resistance of conductor, R = 100 Ω
∴ The rate of change of flux, \(\frac {{ dΦ }_{B}}{ dt }\) = e
\(\frac {{ dΦ }_{B}}{ dt }\) = e = IR = 2.5 x 10-3 x 100 = 250 x 10-3 dt
\(\frac {{ dΦ }_{B}}{ dt }\) = 250 mWb s-1

Question 7.
A fan of metal blades of length 0.4 m rotates normal to a magnetic field of 4 x 10-3 T. If the induced emf between the centre and edge of the blade is 0.02 V, determine the rate of rotation of the blade.
Solution:
Length of the metal blade, l = 0.4 m
Magnetic field, B = 4 x 10-3 T
Induced emf, e = 0.02 V
Rotational area of the blade, A = πr² = 3.14 x (0.4)² = 0.5024 m²
Induced emf in rotational of the coil, e = NBA ω sin θ
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-56
= 9.95222 x 10-3 x 103
= 9.95 revolutions/second
Rate of rotational of the blade, ω = 9.95 revolutions/second

Question 8.
A bicycle wheel with metal spokes of 1 m long rotates in Earth’s magnetic field. The plane of the wheel is perpendicular to the horizontal component of Earth’s field of 4 x 10-5 T. If the emf induced across the spokes is 31.4 mV, calculate the rate of revolution of the wheel.
Solution:
Length of the metal spokes, l = 1 m
Rotational area of the spokes, A = π² = 3.14 x (1)² = 3.14 m²
Horizontal component of Earth’s field, B = 4 x 10-5 T
Induced emf, e = 31.4 mV
The rate of revolution of wheel,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-57
ω = 250 revolutions / second

Question 9.
Determine the self-inductance of 4000 turn air-core solenoid of length 2m and diameter 0. 04 m.
Solution:
Length of the air core solenoid, l = 2 m
Diameter, d = 0.04 m
Radius, r = \(\frac { d }{ 2 }\) = 0.02 m
Area of the air core solenoid, A = π2 = 3.14 x (0.02)2 = 1.256 x 10-3 m2
Number of Turns, N = 4000 turns
Self inductance, L = µ0n2 Al
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-58

Question 10.
A coil of 200 turns carries a current of 4 A. If the magnetic flux through the coil is 6 x 10-5 Wb, find the magnetic energy stored in the medium surrounding the coil.
Solution:
Number of turns of the coil, N = 200
Current, I = 4 A
Magnetic flux through the coil, Φ = 6 x 10-5 Wb
Energy stored in the coil, U = \(\frac { 1 }{ 2 }\) LI² = \(\frac { 1 }{ I2}\)
Self inductance of the coil, L = \(\frac { NΦ }{ I }\)
U =\(\frac { 1 }{ 2 }\) \(\frac { NΦ }{ I }\) x I² = \(\frac { 1 }{ 2}\) NΦI = \(\frac { 1 }{ 2}\) x 200 x 6 x 10-5 x 4
U = 2400 x 10-5 = 0.024 J (or) joules.

Question 11.
A 50 cm long solenoid has 400 turns per cm. The diameter of the solenoid is 0.04 m. Find the magnetic flux of a turn when it carries a current of 1 A.
Solution:
Length of the solenoid, l = 50 cm = 50 x 10-2 m
Number of turns per cm, N = 400
Number of turns in 50 cm, N = 400 x 50 = 20000
Diameter of the solenoid, d = 0.04 m
Radius of the solenoid, r = \(\frac { d }{ 2}\) = 0.02 m
Area of the solenoid, A = π² = 3.14 x (0.02)² = 1.256 x 10-3
Current passing through the solenoid, I = 1 A
Magnetic fluex,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-59

Question 12.
A coil of 200 turns carries a current of 0.4 A. If the magnetic flux of 4 mWb is linked with the coil, find the inductance of the coil.
Solution:
Number of turns, N = 200; Current, I = 0.4 A
Magnetic flux linked with coil, Φ = 4 mWb = 4 x 10-3 Wb
Induction of the coil, L = \(\frac { NΦ }{ I }\) = \(\frac {{ 200 × 4 × 10 }^{-3}}{ 0.4 }\) = \(\frac {{ 800 × 10 }^{-3}}{ 0.4 }\) 2 H

Question 13.
Two air core solenoids have the same length of 80 cm and same cross-sectional area 5 cm². Find the mutual inductance between them if the number of turns in the first coil is 1200 turns and that in the second coil is 400 turns.
Solution:
Length of the solenoids, l = 80 cm = 8 x 10-2 m
Cross sectional area of the solenoid, A = 5 cm2 = 5 x 10-4 m2
Number of turns in the Ist coil, N1 = 1200
Number of turns in the IInd coil, N2 = 400
Mutual inductance between the two coils,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-60

Question 14.
A long solenoid having 400 turns per cm carries a current 2A. A 100 turn coil of cross sectional area 4 cm2 is placed co-axially inside the solenoid so that the coil is in the field produced by the solenoid. Find the emf induced in the coil if the current through the solenoid reverses its direction in 0.04 sec.
Solution:
Number of turns of long solenoid per cm =\(\frac { 400 }{{10}^{ -2 }}\); N2 = 400 x 102
Number of turns inside the solenoid, N2 = 100
Cross-sectional area of the coil, A = 4 cm2 = 4 x 10-4 m2
Current through the solenoid, I = 2A; time, t = 0.04 s
Induced emf of the coil, e = -M \(\frac { dI }{ dt }\)
Mutual inductance of the coil,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-61
Induced emf of the coil,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-61-1
The current through the solenoid reverse its direction if the induced emf, e = -0.2 V

Question 15.
A 200 turn coil of radius 2 cm is placed co-axially within a long solenoid of 3 cm radius. If the turn density of the solenoid is 90 turns per cm, then calculate mutual inductance of the coil.
Solution:
Number of turns of the solenoid, N2 = 200
Radius of the solenoid, r = 2cm = 2 x 102 m
Area of the solenoid, A = πr2= 3.14 x (2 x 10-2)2 = 1.256 x 10-3 m2
Turn density of long solenoid per cm, N1 = 90 x 102
Mutual inductance of the coil,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-62
= 283956.48 x 10-8 ⇒ M = 2.84 mH

Question 16.
The solenoids S1 and S2 are wound on an iron-core of relative permeability 900. The area of their cross-section and their length are the same and are 4 cm2 and 0.04 m, respectively. If the number of turns in S1 is 200 and that in S2 is 800, calculate the mutual inductance between the coils. The current in solenoid 1 is increased from 2A to 8A in 0.04 second. Calculate the induced emf in solenoid 2.
Solution:
Relative permeability of iron core, μr = 900
Number of turns of solenoid S1, N1 = 200
Number of turns of solenoid S2, N2 = ‘800
Area of cross section, A = 4 cm2 = 4 x 10-4 m2
Length of the solenoid S1, l1 = 0.04 m
current, I =I2 – I1 = 8 – 2 = 6A
time taken, t = 0.04 second
emf induced in solenoid S2 e = -M \(\frac { dI }{ dt }\)
Mutual inductance between the two coils, M = \(\frac{\mu_{0} \mu_{r} N_{1} N_{2} A}{l}\)
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-63
M = 180864 x 10-5 = 1.81 H
Emf induced in solenoid S2, e = -M\(\frac { dI }{ dt }\) = -1.81 x \(\frac { 6 }{ 0.04 }\)
Magnitude of emf, e = 271.5 V

Question 17.
A step-down transformer connected to main supply of 220 V is made to operate 11 V, 88 W lamp. Calculate (i) Transformation ratio and (ii) Current in the primary.
Solution:
Voltage in primary coil, Vp = 220 V
Voltage in secondary coil, Vs = 11 V
Output power = 88 W
(i) To find transformation ratio, k = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) = \(\frac { 11 }{ 220 }\) = \(\frac { 1 }{ 20 }\)
(ii) Current in primary, Ip = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) x Is
So, Is = ?
Outputpower = Vs Is
⇒ 88 = 11 x Is
Is = \(\frac { 88 }{ 11 }\) = 8A
Therefore, Ip = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) x Is = \(\frac { 11 }{ 220 }\) x 8 = 0.4 A

Question 18.
A 200V/120V step-down transformer of 90% efficiency is connected to an induction stove of resistance 40 Ω. Find the current drawn by the primary of the transformer.
Solution:
Primary voltage, Vp = 200 V
Secondary voltage, Vs = 120 V
Efficiency, η = 90%
Secondary resistance, Rs = 40 Ω
Current drawn by the primary of the transformc, Ip = \(\frac {{ V }_{ s }}{{ R }_{ s }}\) x Is
Output power = Vs Is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-64
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-65

Question 19.
The 300 turn primary of a transformer has resistance 0.82 Ω and the resistance of its secondary of 1200 turns is 6.2 Ω. Find the voltage across the primary if the power output from the secondary at 1600V is 32 kW. Calculate the power losses in both coils when the transformer efficiency is 80%.
Solution:
Efficiency, η = 80% = \(\frac { 80 }{ 100 }\)
Number of turns in primary, Np = 300
Number of turns in secondary, Ns = 1200
Resistance in primary, Rp = 0.82 Ω
Resistance in secondary, Rs = 6.2 Ω
Secondary voltage, Vs = 1600 V
Output power = 32 kW
Output power = Vs Is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-66
Power loss in primary = \({ { I }_{ p }^{ 2 }{ R }_{ p } }\) = (100)² x 0.82 = 8200 = 8.2 kW
Power loss in secondary = \({ { I }_{ s }^{ 2 }{ R }_{ s } }\) = (20)² x 6.2 = 2480 = 2.48 kW

Question 20.
Calculate the instantaneous value at 60°, average value and RMS value of an alternating current whose peak value is 20 A.
Solution:
Peak value of current, Im = 20 A
Angle, θ = 60° [θ = ωt]
(i) Instantaneous value of current,
i = Im sin ωt = Im sin θ
= 20 sin 60° = 20 x \(\frac { √3 }{ 2 }\) = 10√3 = 10 x 1.732
i = 17.32 A

(ii) Average value of current,
Iav = \(\frac {{ 2I }_{m}}{ π }\) = \(\frac { 2 × 20 }{ 3.14 }\)
Iav = 12.74 A

(iii) RMS value of current,
IRMS = 0.707 Im
or \(\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}\) = 0.707 x 20
IRMS = 14. 14 A

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Conceptual Questions

Question 1.
A graph between the magnitude of the magnetic flux linked with a closed loop and time is given in the figure. Arrange the regions of the graph in ascending order of the magnitude of induced emf in the loop.
Answer:
According to electromagnetic induction, induced emf,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-67
e = \(\frac { dΦ }{ dt }\)
Ascending order of induced emf from the graphical representation is b < c < d < a.

Question 2.
Using Lenz’s law, predict the direction of induced current in conducting rings 1 and 2 when current in the wire is steadily decreasing.
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-68
According to Lenz’s law, a current will be induced in the coil which will produce a flux in the opposite direction.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-69
If the current decreases in the wire, the induced current flows in ring 1 in clockwise direction, the induced current flows in ring 2 in anti-clockwise direction.

Question 3.
A flexible metallic loop abed in the shape of a square is kept in a magnetic field with its plane perpendicular to the field. The magnetic field is directed into the paper normally. Find the direction of the induced current when the square loop is crushed into an irregular shape as shown in the figure.
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-70
The magnetic flux linked with the wire decreases due to decrease in area of the loop. The induced emf will cause current to flow in the direction. So that the wire is pulled out ward direction from all sides. According to Fleming’s left hand rule, force on wire will act outward i direction from all sides.

Question 4.
Predict the polarity of the capacitor in a closed circular loop when two bar magnets are moved as shown in the figure.
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-71
When magnet 1 is moved with its South pole towards to the coil, emf is induced in the coil as the magnetic flux through the coil changes. When seeing from the left hand side the direction of induced current appears to be clockwise. When seeing from the right hand side the direction of induced current appears to be anti-clockwise. In capacitor, plate A has positive polarity and plate B has negative polarity.

Question 5.
In series LC circuit, the voltages across L and C are 180° out of phase. Is it correct? Explain.
Answer:
In series LC circuit, the voltage across the capacitance lag the current by 90° while the voltage across the inductance lead the current by 90°. This makes the inductance and capacitance voltages 180° out of phase.

Question 6.
When does power factor of a series RLC circuit become maximum?
Answer:
For a series LCR circuit, power factor is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-72
For purely resistive, Φ = 0°, cos 0° = 1
Thus the power factor assumes the maximum value for a purely resistive circuit.

Question 7.
Draw graphs showing the distribution of charge in a capacitor and current through an inductor during LC oscillations with respect to time. Assume that the charge in the capacitor is maximum initially.
Answer:
For a capacitor, the graph between charge and time.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-73
The charge decays exponentially decreases with time.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Additional Questions solved

I Choose The Correct Answer

Question 1.
A coil of area of cross section 0.5 m2 with 10 turns is in a plane which is perpendicular to an uniform magnetic field of 0.2 Wb/m2. The flux through the coil is –
(a) 100 Wb
(b) 10 Wb
(c) 1 Wb
(d) zero
Answer:
(c) 1 Wb
Hint:
Φ = NBA cos θ
= 10 x 0.2 x 0.5 x cos 0° = 1 Wb

Question 2.
A rectangular coil of 100 turns and size 0.1 m x 0.05 m is placed perpendicular to a magnetic field of 0.1 T. If the field drops to 0.05 T in 0.05 s, the magnitude of the emf induced in the coil is-
(a) 0.5 V
(b) 0.75 V
(c) 1.0 V
(d) 1.5 V
Answer:
(a) 0.5 V
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-74
ε = 0.5 V

Question 3.
A wire of length 1 m moves with a speed of 10 ms-1 perpendicular to a magnetic field. If the emf induced in the wire is 1 V, the magnitude of the field is-
(a) 0.01 T
(b) 0.1 T
(c) 0.2 T
(d) 0.02 T
Answer:
(b) 0.1 T
Hint:
ε = Blv
⇒ B = \(\frac { ε }{ lv }\) = \(\frac { 1 }{ 1 × 10 }\) = 0.02 T

Question 4.
A coil of area 10 cm2, 10 ms-1 turns and resistance 20 Ω is placed in a magnetic field directed perpendicular to the plane of the coil and changing at the rate of 108 gauss/second. The induced current in the coil will be-
(a) 5 A
(b) 0.5 A
(c) 0.05 A
(d) 50 A
Answer:
(a) 5 A
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-75

Question 5.
A coil of cross sectional area 400 cm2 having 30 turns is making 1800 rev/min in a magnetic field of IT. The peak value of the induced emf is-
(a) 113 V
(b) 226 V
(c) 339 V
(d) 452 V
Answer:
(b) 226 V
Hint:
εm = NBA ω = 30 x 1 x 400 x 10-4 x 30 x 2π
= 226 V

Question 6.
Eddy currents are produced in a material when it is-
(a) heated
(b) placed in a time varying magnetic field
(c) placed in an electric field
(d) placed a uniform magnetic field
Answer:
(b) placed in a time varying magnetic field

Question 7.
An emf of 5 V is induced in an inductance when the current in it changes at a steady rate from 3 A to 2 A in 1 millisecond. The value of inductance is-
(a) 5 mH
(b) 5 H
(c) 5000 H
(d) zero
Answer:
(a) 5 mH

Question 8.
Faraday’s law of electromagnetic induction is related to the-
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy

Question 9.
The inductance of a coil is proportional to-
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns

Question 10.
When a direct current ‘i’ is passed through an inductance L, the energy stored is-
(a) Zero
(b) Li
(c) \(\frac { 1 }{ 2 }\) Li2
(d) \(\frac {{ L }^{ 2 }}{2i}\)
Answer:
(c) \(\frac { 1 }{ 2 }\) Li2

Question 11.
A coil of area 80 cm2 and 50 turns is rotating with 2000 revolutions per minute about an axis perpendicular to a magnetic field of 0.05 T. The maximum value of the emf developed in it is-
(a) 2000 πV
(b) \(\frac { 10π }{ 3 }\) V
(c) \(\frac { 4π }{ 3 }\)V
(d) \(\frac { 2 }{ 3 }\) V
Answer:
(c) \(\frac { 4π }{ 3 }\)V
Hint:
ε = NBA ω = 50 x 0.05 x 80 x 10-4 x \(\frac { 2π × 2000 }{ 60 }\) = \(\frac { 4π }{ 3 }\)V

Question 12.
The direction of induced current during electro magnetic induction is given by-
(a) Faraday’s law
(b) Lenz’s law
(c) Maxwell’s law
(d) Ampere’s law
Answer:
(b) Lenz’s law

Question 13.
AC power is transmitted from a power house at a high voltage as-
(a) the rate of transmission is faster at high voltages
(b) it is more economical due to less power loss
(c) power cannot be transmitted at low voltages
(d) a precaution against theft of transmission lines
Answer:
(b) it is more economical due to less power loss

Question 14.
In a step-down transformer the input voltage is 22 kV and the output voltage is 550 V. The ratio of the number of turns in the secondary to that in the primary is-
(a) 1 : 20
(b) 20 : 1
(c) 1 : 40
(d) 40 : 1
Answer:
(c) 1 : 40
Hint:
\(\frac {{ N }_{ s }}{{ N }_{ p }}\) = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) = \(\frac { 550 }{ 22000 }\) = \(\frac { 1 }{ 40 }\)

Question 15.
The self-inductance of a coil is 5 H. A current of 1 A changes to 2 A within 5 s through the coil. The value of induced emf will be-
(a) 10 V
(b) 0.1 V
(c) 1.0 V
(d) 100 V
Answer:
(c) 1.0 V
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-76

Question 16.
The low-loss transformer has 230 V applied to the primary and gives 4.6 V in the secondary. The secondary is connected to a load which draws 5 amperes of current. The current (in amperes) in the primary is-
(a) 0.1 A
(b) 1.0 A
(c) 10 A
(d) 250 A
Answer:
(a) 0.1 A
Hint:
Ip = \(\frac {{ V }_{ s }{ I }_{ s }}{{ V }_{ p }}\) = \(\frac { 4.6 × 5 }{ 230 }\) = 0.1A

Question 17.
A coil is wound on a frame of rectangular cross-section. If all the linear dimensions of the frame are increased by a factor 2 and the number of turns per unit length of the coil remains the same. Self-inductance of the coil increases by a factor of-
(a) 4
(b) 8
(c) 12
(d) 16
Answer:
(b) 8
Hint:
If all the linear dimensions are doubled, the cross-sectional are a becomes eight times. Therefore, the flux produced by a given current will become eight times. Hence, the selfinductance increases by a factor of 8.

Question 18.
If N is the number of turns in a coil, the value of self-inductance varies as-
(a) N°
(b) N
(c) N2
(d) N-2
Answer:
(c) N2
Hint:
According to self inductance of long solenoid
L = \(\frac{\mu_{0} \mathrm{N}^{2} \mathrm{A}}{l}\)
⇒ L ∝ N2

Question 19.
A magnetic field 2 x 10-2 T acts at right angles to a coil of area 100 cm2 with 50 turns. The average emf induced in the coil will be 0.1 V if it is removed from the field in time.
(a) 0.01 s
(b) 0.1 s
(c) 1 s
(d) 10 s
Answer:
(b) 0.1 s
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-77

Question 20.
Number of turns in a coil is increased from 10 to 100. Its inductance becomes-
(a) 10 times
(b) 100 times
(c) 1/10 times
(d) 25 times
Answer:
(a) 10 times

Question 21.
The north pole of a magnet is brought near a metallic ring as shown in the figure. The direction of induced current in the ring, as seen by the magnet is-
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-78
(a) anti-clockwise
(b) first anti-clockwise and then clockwise
(c) clockwise
(d) first clockwise and then anti-clockwise
Answer:
(a) anti-clockwise

Question 22.
Quantity that remains unchanged in a transformer is-
(a) voltage
(b) current
(c) frequency
(d) none of these
Answer:
(c) frequency

Question 23.
The core of a transformer is laminated to reduce.
(a) Copper loss
(b) Magnetic loss
(c) Eddy current loss
(d) Hysteresis loss
Answer:
(c) Eddy current loss

Question 24.
Which of the following has the dimension of time?
(a) LC
(b) \(\frac { R }{ L }\)
(c) \(\frac { L }{ R }\)
(d) \(\frac { C }{ L }\)
Answer:
(c) \(\frac { L }{ R }\)

Question 25.
A coil has a self-inductance of 0.04 H. The energy required to establish a steady-state current of 5 A in it is-
(a) 0.5 J
(b) 1.0 J
(c) 0.8 J
(d) 0.2 J
Answer:
(a) 0.5 J

Question 26.
Alternating current can be measured by
(a) moving coil galvanometer
(b) hot wire ammeter
(c) tangent galvanometer
(d) none of the above
Answer:
(b) hot wire ammeter

Question 27.
In an LCR circuit, the energy is dissipated in-
(a) R only
(b) R and L only
(c) R and C only
(d) R, L and C
Answer:
(a) R only

Question 28.
A 40 Ω electric heater is connected to 200 V, 50 Hz main supply. The peak value of the electric current flowing in the circuit is approximately-
(a) 2.5 A
(b) 5 A
(c) 7 A
(d) 10 A
Answer:
(c) 7 A
Hint:
I0 = \(\frac {{ V }_{0}}{ R }\) = \(\frac{200 \sqrt{2}}{40}\) 5 √2 ≈ 7A

Question 29.
The rms value of an alternating current, which when passed through a resistor produces heat three times of that produced by a direct current of 2 A in the same resistor, is-
(a) 6 A
(b) 3 A
(c) 2 A
(d) 2√3 A
Answer:
(d) 2√3 A
Hint:
\({ I }_{ rms }^{ 2 }\)R = 3(22R) (or) Irms = 2√3 A

Question 30.
An inductance, a capacitance and a resistance are connected in series across a source of alternating voltages. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of-
(a) \(\frac { π }{ 4 }\)
(b) zero
(c) π
(d) \(\frac { π }{ 2 }\)
Answer:
(b) zero

Question 31.
In an AC circuit, the rms value of the current Irms, is related to the peak current I0 as-
(a) Irms = \(\frac {{I}_{0}}{ π }\)
(b) Irms = \(\frac {{I}_{0}}{ √2 }\)
(c) Irms = √2 I0
(d) Irms = πI0
Answer:
(b) Irms = \(\frac {{I}_{0}}{ √2 }\)

Question 32.
The impedance of a circuit consists of 3 Ω resistance and 4 Ω resistance. The power factor of the circuit is
(a) 0.4
(b) 0.6
(c) 0.8
(d) 1.0
Answer:
(b) 0.6
Hint:
tan Φ = \(\frac { 4 }{ 3 }\).
Power factor = cos Φ = \(\frac { 3 }{ 5 }\) = 0.06

Question 33.
The reactance of a capacitance at 50 Hz is 5 Ω. Its reactance at 100 Hz will be-
(a) 5 Ω
(b) 10 Ω
(c) 20 Ω
(d) 2.5 Ω
Answer:
(d) 2.5 Ω.

Question 34.
In a LCR AC circuit off resonance, the current-
(a) is always in phase with the voltage
(b) always lags behind the voltage
(c) always leads the voltage
(d) may lead or lag behind the voltage
Answer:
(d) may lead or lag behind the voltage

Question 35.
The average power dissipation in a pure inductance L, through which a current I0 sin ωt is flowing is-
(a) \(\frac { 1 }{ 2 }\) L\({ I }_{ 0 }^{ 2 }\)
(b) L\({ I }_{ 0 }^{ 2 }\)
(c) 2 L\({ I }_{ 0 }^{ 2 }\)
(d) zero
Answer:
(d) zero

Question 36.
The power in an AC circuit is given by P = Vrms Irms cos Φ. The value of the power factor cos Φ in series LCR circuit at resonance is-
(a) zero
(b) 1
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ √2 }\)
Answer:
(b) 1
Hint:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-79

Question 37.
In an AC circuit with voltage V and current I, the power dissipated is-
(a) VI
(b) \(\frac { 1 }{ 2 }\) VI
(c) \(\frac { 1 }{ √2 }\) VI
(d) depends on the phase difference between I and V
Answer:
(d) depends on the phase difference between I and V

Question 38.
In an AC circuit containing only capacitance, the current-
(a) leads the voltage by 180°
(b) remains in phase with the voltage
(c) leads the voltage by 90°
(d) lags the voltage by 90°
Answer:
(c) leads the voltage by 90°

Question 39.
In a series LCR circuit R = 10 Ω and the impedance Z = 20 Ω. Then the phase difference between the current and the voltage is-
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Answer:
(c) 60°
Hint:
cos Φ = \(\frac { R }{ Z }\) = \(\frac { 10 }{ 20 }\) = \(\frac { 1 }{ 2 }\)
⇒ Φ = 60°

Question 40.
What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10 μF and ω = 1000 s-1?
(a) 1 mH
(b) 10 mH
(c) 100 mH
(d) Cannot be calculated unless R is known
Answer:
(c) 100 mH
Hint:
L = \(\frac { 1 }{ { \omega }^{ 2 }C } \) = \(\frac{1}{(1000)^{2} \times 10 \times 10^{-6}}\) = 0.1 H = 100 mH

II Fill in the Blanks

Question 1.
Electromagnetic induction is used in …………….
Answer:
transformer and AC generator

Question 2.
Lenz’s Law is in accordance with the law of …………….
Answer:
conservation of energy.

Question 3.
The self-inductance of a straight conductor is …………….
Answer:
zero

Question 4.
Transformer works on …………….
Answer:
AC only

Question 5.
The power loss is less in transmission lines when …………….
Answer:
voltage is more but current is less

Question 6.
The law that gives the direction of the induced current produced in a circuit is …………….
Answer:
Lenz’s law

Question 7.
Fleming’s right hand rule is otherwise called …………….
Answer:
generator rule

Question 8.
Unit of self-inductance is …………….
Answer:
Henry

Question 9.
The mutual induction is very large, if the two coils are wound on …………….
Answer:
soft iron core

Question 10.
When the coil is in vertical position, the angle between the normal to the plane of the coil and magnetic field is …………….
Answer:
zero

Question 11.
The emf induced by changing the orientation of the coil is ……………. in nature.
Answer:
sinusoidal

Question 12.
In a three phase AC generator, the three coils are inclined at an angle of …………….
Answer:
120°

Question 13.
The emf induced in each of the coils differ in phase by …………….
Answer:
120°

Question 14.
A device which converts high alternating voltage into low alternating voltage and vice versa is …………….
Answer:
transformer

Question 15.
For an ideal transformer efficiency η is …………….
Answer:
1

Question 16.
The alternating emf induced in the coil varies …………….
Answer:
periodically in both magnitude and direction

Question 17.
For direct current, inductive reactance is …………….
Answer:
zero

Question 18.
In an inductive circuit the average power of sinusoidal quantity of double the frequency over a complete cycle is …………….
zero

Question 19.
For direct current, the resistance offered by a capacitor is …………….
Answer:
infinity

Question 20.
In a capacitive circuit, power over a complete cycle is …………….
Answer:
zero

Question 21.
Q-factor measures the …………….in resonant circuit
Answer:
selectivity

Question 22.
Voltage drop across inductor and capacitor differ in phase by …………….
Answer:
180°

Question 23.
Angular resonant frequency (co) is …………….
Answer:
\(\frac { 1 }{ \sqrt { LC } } \)

Question 24.
A circuit will have flat resonance if its Q-value is …………….
Answer:
low

Question 25.
The average power consumed by the choke coil over a complete cycle is …………….
Answer:
zero

III Match the following

Question 1.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-80
Answer:
(i) → (c)
(ii) → (a)
(iii) → (d)
(iv) → (b)

Question 2.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-81
Answer:
(i) → (c)
(ii) → (a)
(iii) → (d)
(iv) → (b)

Question 3.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-82
Answer:
(i) → (c)
(ii) → (d)
(iii) → (b)
(iv) → (a)

Question 4.
Type of impedance Phase between voltage and current
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-83
Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 5.
Energy in two oscillatory systems: (LC oscillator and spring mass system)
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-84
Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

IV Assertion and reason

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Eddy currents is produced in any metallic conductor when flux is changed around it.
Reason: Electric potential determines the flow of charge.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
Solution:
When a metallic conductor is moved in a magnetic field, magnetic flux is “varied. It disturbs the free electrons of the metal and set up an induced emf in it. As there are no free ends of the metal i.e., it will be closed in itself so there will be induced current.

Question 2.
Assertion: Faraday’s laws are consequences of conservation of energy.
Reason: In a purely resistive AC circuit, the current lags behind the emf in phase.
Answer:
(c) If assertion is true but reason is false.
Solution:
According to Faraday’s law, the conversion of mechanical energy into electrical energy is in accordance with the law of conservation of energy. It is also clearly known that in pure resistance, the emf is in phase with the current.

Question 3.
Assertion: Inductance coil are made of copper.
Reason: Induced current is more in wire having less resistance.
Answer:
(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.
Solution:
Inductance coils made of copper will have very small ohmic resistance.

Question 4.
Assertion: An aircraft flies along the meridian, the potential at the ends of its wings will be the same.
Reason: Whenever there is a change in the magnetic flux, and emf is induced.
Answer:
(e) If assertion is false but reason is true.
Solution:
As the aircraft flies magnetic flux change through its wings due to the vertical component of the Earth’s magnetic field. Due to this, induced emf is produced across the wings of the aircraft. Therefore, the wings of the aircraft will not be at the same potential.

Question 5.
Assertion: In series LCR circuit resonance can take place.
Reason: Resonance takes place if inductance and capacitive reactances are equal and opposite.
Answer:
(a) If both assertion and reason are hue and the reason is the correct explanation of the assertion.
Solution:
At resonant frequency XL = XC
So, Impedance, Z = R (minimum)
Therefore, the current in the circuit is maximum.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Short Answer Questions

Question 1.
Define magnetic flux (ΦB).
The magnetic flux through an area A in a magnetic field is defined as the number of magnetic field lines passing through that area normally and is given by the equation,
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-85

Question 2.
Write down the drawbacks of Eddy currents.
Answer:
When eddy currents flow in the conductor, a large amount of energy is dissipated in the form of heat. The energy loss due to the flow of eddy current is inevitable but it can be reduced to a greater extent with suitable measures. The design of transformer core and electric motor armature is crucial in order to minimise the eddy current loss.

To reduce these losses, the core of the transformer is made up of thin laminas insulated from one another while for electric motor the winding is made up of a group of wires insulated from one another. The insulation used does not allow huge eddy currents to flow and hence losses are minimized.

Question 3.
Define the unit of self-inductance.
Answer:
The unit of self-inductance is henry. One henry is defined as the self-inductance of a coil in which a change in current of one ampere per second produces an opposing emf of one volt.

Question 4.
Define mutual inductance in terms of flux and current.
Answer:
The mutual inductance M21 is defined as the flux linkage of the coil 2 when 1A current flows through coil 1.
M21 = \(\frac{\mathrm{N}_{2} \mathrm{\phi}_{21}}{i_{1}}\)

Question 5.
Define mutual inductance in terms of emf and current.
Answer:
Mutual inductance M21 is also defined as the opposing emf induced is the coil 2 when the rate of change of current through the coil 1 is 1 As-1.
M12 = \(\frac{-\varepsilon_{1}}{d i_{2} / d t}\)

Question 6.
List out the advantages of three phase alternator.
Answer:
Three-phase system has many advantages over single-phase system, which is as follows:
(i) For a given dimension of the generator, three-phase machine produces higher power output than a single-phase machine.
(ii) For the same capacity, three-phase alternator is smaller in size when compared to single phase alternator.
(iii) Three-phase transmission system is cheaper. A relatively thinner wire is sufficient for transmission of three-phase power.

Question 7.
Mentions the differences betw een a step up and step down transformer.
Answer:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-86

Question 8.
Define efficiency of transformer.
Answer:
The efficiency p of a transformer is defined as the ratio of the useful output power to the input power. Thus
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-87
Transformers are highly efficient devices having their efficiency in the range of 96 – 99%. Various energy losses in a transformer will not allow them to be 100% efficient.

Question 9.
What is meant by sinusoidal alternating voltage?
Answer:
If the waveform of alternating voltage is a sine wave, then it is known as sinusoidal alternating voltage, which is given by the relation.
υ = Vm sin ωt

Question 10.
An inductor blocks AC but it allows DC. Why? and How?
Answer:
An inductor L is a closely wound helical coil. The steady DC current flowing through L produces uniform magnetic field around it and the magnetic flux linked remains constant. Therefore there is no self-induction and self-induced emf (back emf). Since inductor behaves like a resistor, DC flows through an inductor.

The AC flowing through L produces time-varying magnetic field which in turn induces self- induced emf (back emf). This back emf, according to Lenz’s law, opposes any change in the current. Since AC varies both in magnitude and direction, its flow is opposed in L. For an ideal inductor of zero ohmic resistance, the back emf is equal and opposite to the applied emf. Therefore L blocks AC.

Question 11.
A capacitor blocks DC but allows AC. Explain.
Answer:
Capacitive reactance, XC = \(\frac { 1 }{ ωC }\) = \(\frac { 1 }{ 2πƒc }\)
where, ƒ = 0, XC = ∞
where, ƒ is the frequency of the ac supply. In a dc circuit ƒ = 0. Hence the capacitive reactance has infinite value for dc and a finite value for ac. In other words, a capacitor serves as a block for dc and offers an easy path to ac.

Question 12.
Why dc ammeter cannot read ac?
Answer:
A dc ammeter cannot read ac because, the average value of ac is zero over a complete cycle.

Question 13.
Write down the applications of series RLC resonant circuit.
Answer:
RLC circuits have many applications like filter circuits, oscillators and voltage multipliers, etc. An important use of series RLC resonant circuits is in the tuning circuits of radio and TV systems. The signals from many broadcasting stations at different frequencies are available in the air. To receive the signal of a particular station, tuning is done.

Question 14.
What is meant by ‘Wattful current’?
Answer:
The component of current (Irms cos Φ) which is in phase with the voltage is called active component. The power consumed by this current = Vrms Irms cos Φ. So that it is also known as ‘Wattful’ current.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Long Answer Questions

Question 1.
Derive an expression for Mutual Inductance between two long co-axial solenoids.
Answer:
Mutual inductance between two long co-axial solenoids:
Consider two long co-axial solenoids of same length 1. The length of these solenoids is large when compared to their radii so that the magnetic field produced inside the solenoids is uniform and the fringing effect at the ends may be ignored. Let A1 and A2 be the area of cross section of the solenoids with A1 being greater than A2. The turn density of these solenoids are n1 and n2 respectively.
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-88
Let i1 be the current flowing through solenoid 1, then the magnetic field produced inside it is
B1 = μ0n1i1.
As the field lines of \(\vec {{ B }_{1}} \) are passing through the area bounded by solenoid 2, the magnetic flux is linked with each turn of solenoid 2 due to solenoid 1 and is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-89
since θ = 0°
The flux linkage of solenoid 2 with total turns N2 is
N2Φ21 = (n2l)(μ0 n1 i1)
since N2 = n2l
N2Φ21 = (μ0 n1 n2 A2l)i1 ….. (1)
From equation of mutual induction
N2Φ21 = M21 i1 …… (2)
Comparing the equations (1) and (2),
M21 = μ0 n1 n2 A2l ….. (3)
This gives the expression for mutual inductance M21 of the solenoid 2 with respect to solenoid 1. Similarly, we can find mutual inductance M21 of solenoid 1 with respect to solenoid 2 as given below.
The magnetic field produced by the solenoid 2 when carrying a current i2 is
B2 = μ0 n2 i2
This magnetic field B2 is uniform inside the solenoid 2 but outside the solenoid 2, it is almost zero. Therefore for solenoid 1, the area A2 is the effective area over which the magnetic field B2 is present; not area A2 Then the magnetic flux Φ12 linked with each turn of solenoid 1 due to solenoid 2 is
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-90
The flux linkage of solenoid 1 with total turns N1 is
[Since N1 = n1l]
[Since N1 Φ12 = M12i2]
N1 Φ12 = (n1l) (μ0 n2 i2) A2
N1 Φ12 = (μ0 n1 n2 A2l) i2
M12i2 = (μ0 n1 n2 A2l) i2
Therefore, we get
∴ M12 = μ0 n1 n2 A2l ……. (4)
From equation (3) and (4), we can write
M12 = M21 = M ……. (5)
In general, the mutual inductance between two long co-axial solenoids is given by
M= μ0 n1 n2 A2l ……. (6)
If a dielectric medium of relative permeability’ pr is present inside the solenoids, then
M = μn1 n2 A2l
or M = μ0 μr n1 n2 A2l

Question 2.
How will you define the unit of mutual-inductance?
Answer:
Unit of mutual inductance:
The unit of mutual inductance is also henry (H).
If iA= 1 A and N2 Φ21 = 1 Wb turns, then M21 = 1 H.
Therefore, the mutual inductance between two coils is said to be one henry if a current of 1A in coil 1 produces unit flux linkage in coil 2.
If \(\frac {{ di }_{1}}{ 2 }\) = 1 As-1 and ε2 = -1V, theen M21 = 1H.
Therefore, the mutual inductance between two coils is one henry if a current changing at the rate of lAs-1 in coil 1 induces an opposing emf of IV in coil 2.

Question 3.
Find out the phase relationship between voltage and current in a pure resister circuit.
Answer:
AC circuit containing pure resistor:
Consider a circuit containing a pure resistor of resistance R connected across an alternating voltage source. The instantaneous value of the alternating voltage is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-91
υ = Vm sin ωt ….. (1)
An alternating current i flowing in the circuit due to this voltage, develops a potential drop across R and is given by
VR = iR ……. (2)
Kirchoff’s loop rule states that the algebraic sum of potential differences in a closed circuit is zero. For this resistive circuit,
υ – VR = 0
From equation (1) and (2),
Vm sin ωt = iR
⇒ i = \(\frac {{ V }_{m}}{ R }\) sin ωt
i = Im sin ωt …… (3)
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-92
where V\(\frac {{ V }_{m}}{ R }\) = Im the peak value of alternating current in the circuit. From equations (1) and and (3), it is clear that the applied voltage and the current are in phase with each other in a resistive circuit. It means that they reach their maxima and minima simultaneously. This is indicated in the phasor diagram. The wave diagram also depicts that current is in phase with the applied voltage.

Question 4.
Find out the phase relationship between voltage and current in a pure capacitor circuit.
Answer:
AC circuit containing only a capacitor:
Consider a circuit containing a capacitor of capacitance C connected across an alternating voltage source.
The alternating voltage is given by
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-93
Vm sin ωt …… (1)
Let q be the instantaneous charge on the capacitor. The emf across the capacitor at that instant is \(\frac { q }{ C }\). According to Kirchoff’s loop rule,.
υ = \(\frac { q }{ C }\) = 0
⇒ CVm sin ωt
By the definition of current,
i = \(\frac { dq }{ dt }\) = \(\frac { d }{ dt }\) CVm \(\frac { d }{ dt }\) (sin ωt)
= CVm sin ωt
or i = \(\frac{\frac{\mathrm{V}_{m}}{1}}{\frac{\mathrm{l}}{\mathrm{C} \omega}}\) sin \(\left(\omega t+\frac{\pi}{2}\right)\)
i = Im sin \(\left(\omega t+\frac{\pi}{2}\right)\) ….. (2)
where \(\frac{\frac{\mathrm{V}_{m}}{1}}{\frac{\mathrm{l}}{\mathrm{C} \omega}}\) = Im, the peak value of the alternating current. From equation (1) and (2), it is clear that current leads the applied voltage by π/2 in a capacitive circuit. The wave diagram for a capacitive circuit also shows that the current leads the applied voltage by 90°.

Question 5.
What are LC oscillation? and explain the generation of LC oscillation.
Answer:
Whenever energy is given to a circuit containing a pure inductor of inductance L and a capacitor of capacitance C, the energy oscillates back and forth between the magnetic field of the inductor and the electric field of the capacitor. Thus the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations.

Generation of LC oscillations:
Let us assume that the capacitor is fully charged with maximum charge Qm at the initial stage. So that the energy stored in the capacitor is maximum and is given by UEm = \(\frac{\mathrm{Q}_{\mathrm{m}}^{2}}{2 \mathrm{C}}\) As there is no current in the inductor, the energy stored in it is zero i.e., UB = 0. Therefore, the total energy is wholly electrical.

The capacitor now begins to discharge through the inductor that establishes current i in clockwise direction. This current produces a magnetic field around the inductor and the energy stored in the inductor is given by UB = \(\frac {{ Li }_{ 2 }}{ 2 }\). As the charge in the capacitor decreases, the energy stored in it also decreases and is given by UE = \(\frac {{ q }_{ 2 }}{ 2C }\). Thus there is a transfer of some part of energy from the capacitor to the inductor. At that instant, the total energy is the sum of electrical and magnetic energies.

When the charges in the capacitor are exhausted, its energy becomes zero i.e., UE = 0. The energy is fully transferred to the magnetic field of the inductor and its energy is maximum. This maximum energy is given by UB = \(\frac{\mathrm{LI}_{\mathrm{m}}^{2}}{2}\) where Im is the maximum current flowing in the circuit. The total energy is wholly magnetic.

Even though the charge in the capacitor is zero, the current will continue to flow in the same direction because the inductor will not allow it to stop immediately. The current is made to flow with decreasing magnitude by the collapsing magnetic field of the inductor. As a result of this, the capacitor begins to charge in the opposite direction. A part of the energy is transferred from the inductor back to the capacitor. The total energy is the sum of the electrical and magnetic energies.

When the current in the circuit reduces to zero, the capacitor becomes frilly charged in the opposite direction. The energy stored in the capacitor becomes maximum. Since the current is zero, the energy stored in the inductor is zero. The total energy is wholly electrical. The state of the circuit is similar to the initial state but the difference is that the capacitor is charged in opposite direction. The capacitor then starts to discharge through the inductor with anti-clockwise current. The total energy is the sum of the electrical and magnetic energies.

As already explained, the processes are repeated in opposite direction. Finally, the circuit returns to the initial state. Thus, when the circuit goes through these stages, an alternating current flows in the circuit. As this process is repeated again and again, the electrical oscillations of definite frequency are generated. These are known as LC oscillations. In the ideal LC circuit, there is no loss of energy. Therefore, the oscillations will continue indefinitely. Such oscillations are called undamped oscillations.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Numerical Problems

Question 1.
A coil has 2000 turns and area 70 cm2. The magnetic field perpendicular to the plane of the coil is 0.3 Wb/m2. The coil takes 0.1 s to rotate through 180°. Then what is the value of induced emf?
Solution:
Magnitude of change in flux,
|∆Φ | = |NBA (cos 180° – cos 0°
= |NBA(-1 – 1)| = |-2 NBA| = |2 NBA|
Where,
N = 2000
B = 0.3 Wb/m2
A = 70 x 10-4 m2
t = 0.1 sec
Induced emf, ε = \(\frac { \left| \Delta \phi \right| }{ \Delta t } \) = \(\frac { 2NBA }{ ∆t }\) = \(\frac {{ 2 × 2000 × 0.3 × 70 × 10 }^{-4}}{ 0.1 }\)
ε = 84 V

Question 2.
A rectangular loop of sides 8 cm and 2 cm is lying in a uniform magnetic field of magnitude 0.5 T with its plane normal to the field. The field is now gradually reduced at the rate of 0.02 T/s. If the resistance of the loop is 1.6 Ω, then find the power dissipated by the loop as heat.
Solution:
Induced emf, |ε| = \(\frac { dΦ }{ dt } \) = A \(\frac { dB }{ dt } \) = 8 × 2 × 10-4 × 0.02
ε = 3.2 × 10-5 V
Induced current, I = \(\frac { ε }{ R } \) = 2 × 10-5 A
Power loss = I2R = 4 × 10-10 × 1.6 = 6.4 × 10-10 W

Question 3.
A current of 2 A flowing through a coil of 100 turns gives rise to a magnetic flux of 5 x 10-5 Wb per turn. What is the magnetic energy associated with the coil?
Solution:
Self inductance of coil, L = \(\frac { NΦ }{ I } \) = \(\frac {{ 100 × 5 × 10 }^{-3}}{ 2 } \)
= 2.5 × 10-3 H
Magnetic energy associated with inductance,
U = \(\frac { 1 }{ 2 }\) LI2 = \(\frac { 1 }{ 2 }\) × 2.5 × 10-3 × (2)2
= \(\frac { 1 }{ 2 }\) × 2.5 × 10-3 × 4 = 5 × 10-3 J

Question 4.
A transformer is used to light a 140 W, 24 V bulb from a 240 V AC mains. The current in the main cable is 0.7 A. Find the efficiency of the transformer.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-95
η = \(\frac { 140 }{ 240 × 0.7 }\) × 100 = 83.3%

Question 5.
In an ideal step up transformer the turns ratio is 1 : 10. A resistance of 200 ohm connected across the secondary is drawing a current of 0.5 A. What are the primary voltage and current?
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-96
Primary current, Ip = 5 A
Promary voltage, Ep = 10 V

Question 6.
A capacitor of capacitance 2 μF is connected in a tank circuit oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA, then find the voltage across the capacitor.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-97

Question 7.
An ideal inductor takes a current of 10 A when connected to a 125 V, 50 Hz AC supply. A pure resistor across the same source takes 12.5 A. If the two are connected in series across a 100 √2 V, 40 Hz supply, then calculate the current through the circuit.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-98

Question 8.
An LCR series circuit containing a resistance of 120 Ω. has angular resonance frequency 4 x 105 rad s-1. At resonance the voltages across resistance and inductance are 60 V and 40 V, respectively. Find the values of L and C.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-99

Question 9.
A coil of inductive reactance 31 Ω has a resistance of 8 Ω. It is placed in series with a capacitor of capacitance reactance 25 Ω. The combination is connected to an ac source of 110 volt. Find the power factor of the circuit.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-100
Power faactor = 0.8

Question 10.
The power factor of an RL circuit is \(\frac { 1 }{ √2 }\). If the frequency of AC is doubled, what will be the power factor?
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-101

Question 11.
The instantaneous value of alternating current and voltage are given as i = \(\frac { 1 }{ √2 }\) sin (100 πt) A and e = \(\frac { 1 }{ √2 }\) sin(100 πt + \(\frac { π }{ 3 }\)) volt. Find the average power in watts consumed in the circuit.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current-102

Common Errors and its Rectifications:

Common Errors:

  1. Students sometimes may confuse the peak current and instantaneous value of current and emf.
  2. They may confuse in the area of R, L and C with AC. The relation between current and induced emf.

Rectifications:

  1. Instantaneous current, i = I0 sin tot Peak current, I0 = √2 Irms Instantaneous emf, e = E0 sin cor Peak emf, E0 = √2 Erms
  2. In Inductor: current is \(\frac { π }{ 2 }\) rad less than that of emf.
    In Resistor: current and emf are same phase.
    In Capacitor: current is \(\frac { π }{ 2 }\) rad greater than that of emf.

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Samacheer Kalvi 12th English Solutions Poem Chapter 3 All the World’s a Stage

Students can Download English Poem 3 All the World’s a Stage Questions and Answers, Summary, Activity, Notes, Samacheer Kalvi 12th English Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th English Solutions Poem Chapter 3 All the World’s a Stage

Warm Up

This is Life Cycle of butterfly.

All The World's A Stage Poem Questions And Answers Samacheer Kalvi 12th English Solutions Poem Chapter 3

All The World’s A Stage Poem Questions And Answers Question 1.
Discuss with your partner the different stages in the grow th of man from a new born to an adult.
Answer:
An infant pukes on the mother’s arms. As he is unable to articulate his needs, he keeps on crying like a kitten. Then he goes to school giving up his freedom. He is made to learn things he doesn’t want to learn. Then he becomes an adult hopelessly in love. He wastes his purple youth writing love letters or songs admiring the beauty of his love. Some join army or police force to serve the nation. At the peak of adulthood, they are quite touchy about honour and believe it to be more important than life itself.

Samacheer Kalvi 12th English All the World’s a Stage Textual Questions

1. Fill in the blanks using the words given in the box to complete the summary of the poem.

attention treble reluctantly
actors maturity reputation
serious faculties composing
enter promises dependent

Shakespeare considers the whole world a stage where men and women are only (1) _____ They (2) _____ the stage when they are borm and exit when they die. Every man, during his life time; plays seven roles based on age. In the first act, as an infant, he is wholly (3) _____ on the mother or a nurse. Later, emerging as a school child, he slings his bag over his shoulder and creeps most (4) _____ to school. His next act is that of a lover, busy (5) _____ ballads for his beloved and yearns for her (6) _____ In the fourth stage, he is aggressive and ambitious and seeks (7) _____ in all that he does. He (8) _____ solemnly to guard his country and becomes a soldier. As he grows older, with (9) _____ and wisdom, he becomes a fair judge. During this stage, he is firm and (10) _____ In the sixth act, he is seen with loose pantaloons and spectacles. His manly voice changes into a childish (11) _____ The last scene of all is his second childhood. Slowly, he loses his (12) _____ of sight, hearing, smell and taste and exits from the roles of his life.
Answer:

  1. actors
  2. enter
  3. dependent
  4. reluctantly
  5. composing
  6. attention
  7. reputation
  8. promises
  9. maturity
  10. serious
  11. treble
  12. faculties

2. From your understanding of the poem, answer the following questions briefly in a sentence or two.

12th English All The World’s A Stage Paragraph Question (a)
What is the world compared to?
Answer:
The world is compared to a stage.

12th English Poem All The World’s A Stage Question (b)
“And they have their exits and their entrances” – What do the words ‘exits’ and ‘entrances’ mean?
Answer:
‘Entrances’ means life. ‘Exits means death.

All The World’s A Stage Poem Appreciation Questions And Answers Question (c)
What is the first stage of a human’s life?
Answer:
The first stage of human life is “infant”. The babe on nurse’s arms pukes and mewls.

All The World’s A Stage Questions And Answers Question (d)
Describe the second stage of life as depicted by Shakespeare.
Answer:
The second stage is school boy. The boy goes to school with a heavy heart like a snail.

12th English Unit 3 Poem Question (e)
How does a man play a lover’s role?
Answer:
As a lover, man sings serenades seeking the attention of his lady love.

12th English All The World’s A Stage Question (f)
Bring out the features of the fourth stage of a man as described by the poet.
Answer:
In the fourth stage, man becomes aggressive and ambitious and seeks glory in all his pursuits. He is ready to enter the mouth of cannon for a moment of glory.

All The World’s A Stage Question And Answers Question (g)
When does a man become a judge? How?
Answer:
In the fifth stage, man grows mature and wise. He becomes an impartial judge. He is firm and serious about his opinions.

All The World’s A Stage Poem Questions And Answers Pdf Question (h)
Which stage of man’s life is associated with the ‘shrunk shank’?
Answer:
In the sixth stage, man becomes thin and weak. His fashionable dresses of youthful days have now become too lose to use for his shrunk shank (i.e.) legs that have become very lean with age.

All The World’s A Stage Question Answer Question (i)
Why is the last stage called second childhood?
Answer:
The last stage is called the second childhood. The old man slowly loses all his senses. He requirs the support of a nurse or wife to do anything. In this stage, he departs from the world.

3. Explain the following lines briefly with reference to the context.

All The World’s A Stage Question And Answer Question (a)
“They have their exits and their entrances;
And one man in his time plays many parts”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says this while hinting at the beginning and the end of life. The poet divides man’s life into seven stages. The first stage symbolises birth and the last stage death. So, he uses the words “entrances and exits”.

Question Answer Of All The World’s A Stage Question (b)
‘‘Jealous in honour, sudden and quick in quarrel,
Seeking the bubble reputation”.
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says these words while describing the fourth stage when the young man becomes a soldier and runs after short-lived glory. He has inflated sense of honour and ready to insist on duels to settle matters touching his honour. He does not realise that the reputation he seeks is short-lived like a bubble.

All The World’s A Stage Summary Question (c)
“Is second childishness and mere oblivion;
Sans teeth, sans eyes, sans taste, sans everything.”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says this while man gets ready to leave this world (i.e.) the last stage of his life on this lonely planet. In this stage, man becomes totally forgetful. He loses his teeth, eyesight and taste. He loses all his senses of perception. Like a baby, he can’t do anything on his own. So, the poet calls this stage “second childhood” when the old man behaves in a childish manner.

All The World’s A Stage Poem Paragraph Additional Questions

Explain the following lines briefly with reference to the context.

All The World’s A Stage English Workshop Answers Question (a)
“His acts being seven ages. At first the infant,
Mewling and puking in the nurse’s arms;”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William . Shakespeare.
Context and Explanation: The poet says these words while describing the first seven stages of life on the stage (i.e.) earth. The first stage/Act is infancy. The babe vomits on the arms of the nurse and cries like a kitten.

All The World’s A Stage Reference To Context Question (b)
“All the world’s a stage,
And all the men and women merely players;”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says these words while philosophising and classifying stages of life. The poet compares the world to a stage. All men and women are simply actors playing different roles on the different stages of life.

All The World’s A Stage Poem Questions Question (c)
“Then the whining school-boy, with his satchel
And shining morning face, creeping like snail’
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context’and Explanation: The poet says these words while describing the second stage of life. During boyhood, the school boy goes to school reluctantly in snail speed with a heavy heart. In ‘Romeo and Juliet’, Shakespeare compares a school boy going to school like a lover going away from his lady love with a heavy heart.

All The World’s A Stage Pictures Question (d)
“Sighing like furnace, with a woeful ballad
Made to his mistress’ eyebrow.”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage’ written by William Shakespeare.
Context and Explanation: The poet says these words while describing the third stage of life when he becomes a lover. At this stage, he yearns for the attention of his lady love. He composes ballads expressing his agony caused by unrequitted love. He sings songs praising the beauty of his mistress trying to win her heart.

All The World’s A Stage Poem In Tamil Question (e)
“Seeking the bubble reputation
Even in the cannon’s mouth.”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words while describing the fourth stage of life. In this stage, youngman becomes a soldier. He is quick to anger and attaches great importance to honour. He is ready to lay down his life for the fleeting bubble of reputation.

All The World’s A Stage Poem Reference To Context Question (f)
“…And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances;”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William ‘ Shakespeare.
Context and Explanation: The poet says these words while describing the fifth stage of life. At this stage, he behaves like a judge pronouncing his decisive opinions with the modem instances. He quotes wise maxims from his own life experiences to influence other people. He is fond of eating delicacies unmindful of the protruding belly size.

Question (g)
“And so he plays his part The sixth age shifts
Into the lean and slipper ’dpantaloon,”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words while describing the impact of ageing on the physical appearance. In the sixth stage, he becomes old, thin and unsteady.

Question (h)
“Hisyouthful hose, well said, a world too wide
For his shrunk shank; and his big manly voice,”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words to describe the unsuitability of one’s own dress as one advances in years. As the young man turns old, his legs become thin and his trousers become very loose giving easy access to legs but tough to wear as the waistline has also thinned. His manly voice has become feeble. When he speaks, it looks like a child piping up his dreams.

Question (i)
“…Last scene of all,
That ends this strange eventful history,”
Answer:
Reference: These lines are from the poem ‘All the world’s a stage” written by William Shakespeare.
Context and Explanation: The poet says these words while describing the preparedness of the old man in the last stage of life to exit from this lonely planet. The poet beautifully says the “eventful history” (i.e.) life which was spiced up with many interesting things is now coming to a dramatic close. The eternal jewel of life, ‘the soul’, is going to depart the body which had kept it imprisoned for long. The soul celebrates the joy of freedom in death.

Appreciate The Poem

4. Read the poem once again carefully and identify the figure of speech that has been used in each of the following lines from the poem.

“All the world’s a stage,
And all the men and women merely players;
They have their exits and their entrances;
And one man in his time plays many parts,
His acts being seven ages. At first the infant,

Mewling and puking in the nurse’s arms;
Then the whining school-boy, with his satchel
And shining morning face, creeping like snail
Unwillingly to school. And then the lover,
Sighing like furnace, with a woeful ballad
Made to his mistress’ eyebrow. Then a soldier,
Full of strange oaths, and bearded like the pard,
Jealous in honour, sudden and quick in quarrel,

Seeking the bubble reputation
Even in the cannon’s mouth. And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances;
And so he plays his part. The sixth age shifts
Into the lean and slipper’d pantaloon,
12th English All The World's A Stage Paragraph Solutions Poem Chapter 3 Samacheer Kalvi
With spectacles on nose and pouch on side;
His youthful hose, well sav’d, a world too wide
For his shrunk shank; and his big manly voice,
Turning again toward childish treble, pipes
And whistles in his sound. Last scene of all,
That ends this strange eventful history,
Is second childishness and mere oblivion;
Sans teeth, sans eyes, sans taste, sans everything.”

Question (a)
“All the world’s a stage”
Answer:
Metaphor

Question (b)
“And all the men and women merely players”
Metaphor

Question (c)
“And shining morning face, creeping like snail’
Simile

Question (d)
“Full of strange oaths, and bearded like the pard,”
Answer:
Simile

Question (e)
“Seeking the bubble reputation”
Answer:
Metaphor

Question (f)
“Hisyouthful hose, well sav’d, a world too wide”
Answer:
Alliteration

Question (g)
“and his big manly voice, turning again toward childish treble”
Answer:
Metaphor

5.Pick out the words in ‘alliteration’ in the following lines,

Question (a)
“and all the men and women merely players”
Answer:
and all the men and women merely players

Question (b)
“And one man in his time plays many parts”
Answer:
And one man in his time plays many parts

Question (c)
“Jealous in honour, sudden and quick in quarrel ”
Answer:
Jealous in honour, sudden and quick in quarrel.

6. Read the given lines and answer the questions that follow.

(a) “Then the whining school-boy, with his satchel
And shining morning face, creeping like snail
Unwillingly to school ”

Question (i)
Which stage of life is being referred to here by the poet?
Answer:
Boyhood is referred to here.

Question (ii)
What are the characteristics of this stage?
Answer:
Innocence, joy and care-free life are the characteristics of this stage in life.

Question (iii)
How does the boy go to school?
Answer:
The boy goes to school unwillingly. He is slow like a snail.

Question (iv)
Which figure of speech has been employed in the second line?
Answer:
Simile is employed in the second line.

(b) “Then a soldier,
full of strange oaths, and bearded like the pard,
Jealous in honour, sudden and quick in quarrel,
Seeking the bubble reputation Even in the cannon’s mouth.

Question (i)
What is the soldier ready to do?
Answer:
The soldier is ready to lay down his life.

Question (ii)
Explain ‘bubble reputation’.
Answer:
Reputation is a transitory thing. It doesn’t even last a minute like the life of a bubble.

Question (iii)
What are the distinguishing features of this stage?
In this stage, the youthful soldier attaches great value to honour. He is quick to temper and challenges people for fight for the sake of honour. He often swears to assert his valour.

(c) “And then the justice,
In fair round belly with good capon lin’d,
With eyes severe and beard of formal cut,
Full of wise saws and modern instances;’’’’

Question (i)
Whom does justice refer to?
Answer:
Justice refers to man in his fifth stage when he becomes critical of everyone else’s opinion in life.

Question (ii)
Describe his appearance.
Answer:
He has a pot belly and is fond of eating delicacies.

Question (iii)
How does he behave with the people around him?
Answer:
His eyes are severe. He often gives advice to people.

Question (iv)
What does he do to show his wisdom?
Answer:
To show of his wisdom, he often quotes modem examples and words of wisdom.

Additional Questions

Read the given lines and answer the questions that follow.

(a) “All the world’s a stage
And all the men and women merely players:
They have their exits and their entrances;
And one man in his time plays many parts,
His acts being seven ages.”

Question (i)
What are all the men and women of this world?
Answer:
The men and women of the world are just like players on the stage of life.

Question (ii)
Explain: ‘They have their exits and their entrances’.
Answer:
They take birth and enter the world. They die and depart from the world.

Question (iii)
How many parts does every man enact and play?
Answer:
Every man enacts and plays seven different roles in life.

Question (iv)
Why is this world compared to a stage?
Answer:
This world is like a big stage where men and women are ever busy in playing their respective roles.

(b) “At first the infant,
Mewling and puking in the nurse’s arms.
Then the whining schoolboy, with his satchel
And shining morning face, creeping like snail
Unwillingly to school.”

Question (i)
What does man do in the first stage of life?
Answer:
In the first stage of life man plays the role of an infant. He is always crying and vomiting in the nurse’s arms.

Question (ii)
Does the schoolboy show eagerness to go to school?
Answer:
No, the schoolboy doesn’t show any interest in going to school. Rather he is unwilling to go there.

Question (iii)
How does the schoolboy walk up to his school?
Answer:
He is inching slowly and unwillingly like a snail towards his school.

Question (iv)
Explain, ‘Mewling and pucking’.
Answer:
It means crying and vomiting.

(c) “And then the lover,
Sighing like furnace, with a woeful ballad
Made to his mistress’ eyebrow.”

Question (i)
What is the third stage of life?
The third stage of man’s life is that of a lover.

Question (ii)
What is the poetic device used in the second line?
Answer:
‘Simile’ is used as a poetic device in the second line.

Question (iii)
What does the lover do for his mistress?
Answer:
The lover is ahvays sighing and longing for his beloved. He writes a sad ballad describing the eyebrow of his mistress.

Question (iv)
Explain, ‘sighing like furnace’.
Answer:
It means moaning, breathing deeply and sadly like a fire place.

(d) “Then a soldier.
Full of strange oaths, and bearded like the pard,
Jealous in honour, sudden and quick in quarrel,
Seeking the bubble reputation.
Even in the cannon’s mouth.”

Question (i)
Describe the two traits of a soldier.
Answer:
A soldier is always ready to swear and is full of oaths. He is ever ready to compete for honour and glory.

Question (ii)
What is the poetic device used in : ‘bearded like a pard’?
Answer:
The poet uses a simile for comparison.

Question (iii)
Why does the soldier risk his life and what for?
Answer:
The soldier risks his life a momentary reputation and is ready even to enter the cannon’s mouth.

Question (iv)
How is the soldier bearded?
Answer:
He is bearded like a pard or a leopard.

(e) “The sixth age shifts
Into the lean and slippered pantaloon,
With spectacles on nose and pouch on side,
His youthful hose, well saved, a world too wide For his shrunk shank; and his big manly voice,
Turning again toward childish treble, pipes
And whistles in his sound.’’

Question (i)
What is a ‘lean and slippered pantaloon’?
Answer:
It means a thin old man wearing slippers and loose trousers.

Question (ii)
What does the phrase ‘a world too wide’ here mean?
Answer:
The stockings he bought in his youth have become too loose for his shrunk and thin legs.

Question (iii)
How does the ‘mainly voice’ turn into ‘childish’ in the sixth stage of life?
Answer:
His manly voice turns into childish trebles and whistles when he speaks as he has no teeth in his mouth.

Question (iv)
What is the sixth stage of man’s life?
Answer:
In the sixth stage of life man plays the role of a ‘lean and slippered pantaloon’.

(f) “Last scene of all,
That ends this strange eventful history,
Is second childishness and mere oblivion,
Sans teeth, sans eyes, sans taste, sans everything.”

Question (i)
What is the last scene of man’s life?
Answer:
The last scene that ends man’s eventual life is a ‘second-childishness’. In this stage he appears and behaves like a child.

Question (ii)
Why is the last stage of man has been called a ‘second childishness’?
Answer:
The last stage of man’s life has been called a ‘second childishness’ as man’s appearance and activities in this stage are quite similar to those of a child.

Question (iii)
How is the last stage of man’s life a ‘mere oblivion’?
Answer:
The last stage of life is a ‘mere oblivion’ as old age is another stage of forgetfulness.

Question (iv)
Explain ‘eventful history’.
Answer:
It means the life-long history of man full of interesting incidents and experiences.

7. Complete the table based on your understanding of the poem.

Stage Characteristic
crying
judge
soldier
unhappy
second childhood
whining
old man

Answer:

Stage Characteristic
Baby (first stage) crying
judge Firm and serious
soldier Aggressive and Ambitious
Lover unhappy
second childhood Loses senses
Boyhood (school) whining
old man Wise and judges others

8. Based on your understanding of the poem, answer the following questions in about 100 – 150 words each. You may add your own ideas if required, to present and justify your point of view.

Question (а)
Describe the various stages of a man’s life picturised in the poem “All the World’s a stage.”
Answer:
Shakespeare has beautifully portrayed this world as a huge open theatre where in all humans play seven acts/ages. In the first act, he is a helpless infant puking on the nurse’s arms mewling like a kitten. In the second stage, he is the grumbling/whining school student. He moves to school like a snal/unwllingly with his slate and bag. In the third Act, he is a lover sighing and yearning for the attention of his lady love.

He composes romantic ballads complaining his love that he needs a better deal. In the fourth Act, he becomes a quick-tempered soldier, aggressive and ambitious, ready to stake his life for the sake of bubble reputation. As he matures, he becomes a wise judge of contemporary life quoting wise maxims to endorse his opinion. He is firm and serious. In the sixth act, his stout legs become thin making his trousers of youth unsuitable. Thin and lean legs easily travel through them but are unable to stay due to a slimmed waist. His bass voice has become treble like that of a child. In the last act, he is sans teeth, sanys eyes, sans taste and sans everything (i.e.) loses all senses. He departs the world.

Question (b)
Shakespeare has skill fully brought out the parallels between the life of man and actors on stage. Elaborate this statement with reference to the poem.
Answer:
Shakespeare has beautifully compared the growth of humans by stages with his emergent role during that stage. In the first stage man plays the role of an infant. As an infant, he does represent characterisation of mewling and puking. In the second Act, he does the role of a school boy with the characteristics of unwillingness to go to schools and innocence shining in his face. In the third Act, he performs the role of a lover head over heels in love with a beautiful lady. He composes woeful romantic ballads and sings serenades to impress his love. In the fourth act, he plays the impressive role of a short-tempered, honour pursuing soldier.

He is ready to put his mouth in the Cannon’s mouth for conquering the bubble like honour in order to defend the territory of his country. In the fifth Act, he performs the role of a mature and fair judge criticising the ways of the world often spicing up his conversations with wise remarks and wit. His pot belly and well-cut beard shows the social status he enjoys in life. In the sixth act, he is old. He performs the role of a thin old man wearing ill-fitting loose garments with a changed treble in his voice. He is bespectacled and slow in walking. In the final act, he becomes a total invalid losing all senses of hearing, taste and sight. Then the performer leaves the stage (i.e.) the lonely planet.

Speaking Activity

Shakespeare describes the characteristics of the various stages of man. You are in the second stage of life. What do you think of your roles and responsibilities at this stage? Discuss with your partner and share your ideas with the class.
Answer:
At school age, imagination takes wings. Inquisitiveness is common among my peers. Parents, society and teachers want us only to study. But we need to explore the world around us. At home, it is our responsibility to keep our things in order. We need to assist the perennial worker, we mean, our moms in completing their domestic chores. Occasionally, we shall take care of siblings too not as a work but as a duty towards a family member who will be a life long companion to us.

Listening Activity

Listen to the poem and fill in the blanks with appropriate words and phrases. If required listen to the poem again.

The World Is Too Much with Us
The world is too much with us; late and soon, Getting and spending, we lay waste our powers; Little we see in Nature that is ours;We have given our hearts away, a sordid boon! This Sea that bares her bosom to the moon, The winds that will be howling at all hours,And are up- gathered now like sleeping flowers,

For this, for everything, we are out of tune; It moves us not. – Great God! I’d rather be A Pagan suckled in a creed outworn; So might I, standing on this pleasant lea, Have glimpses that would make me less forlorn; Have sight of Proteus rising from the sea; Or hear old Triton blow his wreathed horn.

The World Is Too Much with Us:
The world is too much with us; late and soon
Getting and spending, we lay waste our powers
Little we see in (1) ______ that is ours;
We have given (2) ______ away, a sordid boon!
This Sea that bares her bosom (3) ______
(4) ______ that will be howling at all hours,
And are up-gathered now like (5) ______
,For this, for everything, we are (6) ______ ;
It (7) ______ . us not. Great God! I’d rather be
A Pagan suckled in a creed outworn;
So might I, standing on this pleasant lea
Have glimpses that would make me less forlorn;
Have sight of Proteus rising (8) ______
Or hear old Triton blow his wreathed horn.
Answers:

  1. Nature
  2. our hearts
  3. to the moon
  4. The winds
  5. sleeping flowers
  6. out of tune
  7. moves
  8. from the sea

All the World’s a Stage About The Poet

12th English Poem All The World's A Stage Solutions Poem Chapter 3 Samacheer Kalvi

William Shakespeare (1564-1616) was a prolific writer during the Elizabethan and Jacobean ages of British theatre (sometimes called the English Renaissance). Shakespeare’s plays are perhaps his most enduring legacy. Shakespeare’s poems remain popular to this day. Shakespeare’s rich and diverse works have spawned countless adaptations across multiple genres and cultures. His writings have been compiled in various iterations of The Complete Works of William Shakespeare. William Shakespeare continues to be one of the most important literary7 figures of the English language.

All the World’s a Stage Summary in english

Introduction
‘All the world’s a stage’ is an extract from the play ‘As you like it’, a romantic comedy by Shakespeare.

A metaphor defining world
Shakespeare claims this world as a stage in a theatre. All men and women are only actors. The stage has both exits and entrances. Similarly, men and women take birth and enter the world. They live their lives and go out of it when they die. Every man plays seven emergent roles and lives through seven stages of life.

Infancy and boyhood
With the birth of an infant begins the first stage of man’s life. The infant cries and vomits on the arms of his nurse. Then he grows into a school-going boy. He is unwilling to go to the school. He moves towards school at a snail’s speed.

Thirst for love and glory
In the third stage, man plays the role of a lover. He sighs like a fumace.He keeps on writing ballads praising the beauty of the eyes of his beloved. The fourth stage is that of a soldier. He keeps a beard like that of a leopard. He always runs after honour and fame. He is ready even to enter a cannon’s mouth just for momentary glory and bubble of reputation.

Wisdom and failing health
In the fifth stage, man plays the role of a justice. He is fond of eating chicken and develops a fat round belly. He is full of wise sayings and modem instances. He is a man of wisdom and knowledge. In the sixth stage, man becomes weak and thin in body. He wears slippers, spectacles and clothes that he bought when he was young. These pants and stockings have become loose for his shrunk and thin legs.

Second childhood
The seventh stage is the ‘second childhood’. In this stage, man becomes very old and starts behaving like a child. He is left with no teeth and becomes weak in eyesight. Actually, he loses taste and becomes a victim of forgetfulness. The poet describes this helpless state as “Sans teeth, sans eyes, sans taste and sans everything” nicely. Then the man departs from this world.

Conclusion
Shakespeare condenses the life of man beautifully and portrays it well. The revisit of childhood in old age proves his profound understanding of human life.

All the World’s a Stage Summary in Tamil

முன்னுரை
‘All the World’s a stage’ (‘உலகம் ஒரு நாடக மேடை’) என்ற கவிதை சேக்ஸ்பியரின் ‘As you like it’, என்ற | நகைச்சுவை கலந்த கற்பனை கதையின் ஒரு சாரம் ஆகும்.

உலகம் ஒரு நாடக மேடை:
சேக்ஸ்பியர் உலகத்தை ஒருநாடக மேடையாகக் கருதுகிறார். அதில் அனைத்து ஆணும், பெண்ணும் நடிகர்களே. இந்த நாடக மேடையின் உள்ளே வரவும் வெளியே செல்லவும் வழிகள் உள்ளன. அதே போல தான் மனிதன் பிறந்து இந்த உலகத்துக்கு வருகிறான். அவனது வாழ்நாளை வாழ்ந்துவிட்டு வெளியே போய்விடுகிறான். ஒவ்வொரு மனிதனும் ஏழு கதாபாத்திரங்களாக வாழ்க்கை மேடையில் நடிக்கிறான்.

குழந்தை பருவமும், விடலைப் பருவமும்:
குழந்தை பருவமே மனிதனின் முதல் பாகம் ஆகும். வாந்தியும், அழுகையுமாக முதல் பாகம் செவிலிப் பெண் தோளில் இருக்கிறான். பிறகு பள்ளிப் பருவம் அடைகிறான். பள்ளிக்கூடம் போக மனமில்லாது இருக்கிறான். பள்ளிக்கூடத்தை நோக்கி நத்தை போல் நகர்கிறான்.

காதல், புகழ் என ஈர்ப்புக்குள்ளாகிறான்:
மூன்றாம் பாகத்தில் காதலனாக கதாபாத்திரம் ஏற்கிறான். எரியும் அடுப்பைப் போன்று குமுறுகிறான். தன் காதலியின் கண்களைக் குறித்து கவிதை மழை பொழிகிறான். நான்காம் பாகத்தில் சிப்பாய் வேடம் ஏற்கிறான். சிறுத்தை போன்று மீசையை வளர்த்துக் கொள்கிறான். பேர் மற்றும் புகழின் பின்னால் ஓடுகிறான். தற்காலிக பேருக்கும், புகழுக்கும் ஆசைப்பட்டு பீரங்கி | வாயினுள் நுழையவும் தயாராக இருக்கிறான்.

அறிவு முதிர்ச்சியும், குன்றும் ஆரோக்கியமும்:
ஐந்தாம் பாகத்தில் தானே ஒரு நீதிபதி ஆகிறான். கோழி மாமிசத்தின் பால் ஆவல் கொண்டு அதை உண்டு பெரிய தொப்பையுடன் தோன்றுகிறான். அறிவு முதிர்ச்சியுடனும், புதுப் பொலிவுடனும் தோன்றுகிறான். அறிவும் ஆற்றலுமுடையவனாய்த் திகழ்கிறான். ஆறாம் பாகத்தில் உடல் வலுவிழந்து சோர்வடைகிறான், ஒல்லி வடிவமாய், காலில் செருப்புமாய் ஒரு சிரிப்பு நடிகனைப் போல் தோற்றமளிக்கிறான். இளம் வயதில் அணிந்த கண்ணாடியும், துணிகளும், செருப்பும் அணிந்து கொள்கிறான். சுருங்கிய தோல்களுக்கும், ஒல்லியான கால்களுக்கும் இந்த உடையும், செருப்பும் தொள தொளவென காணப்படுகின்றன.

இரண்டாம் குழந்தை பருவம்:
ஏழாம் பருவம் இரண்டாவது குழந்தை பருவம் எனலாம். இந்த பாகத்தில் மிகவும் வயது முதிர்ந்த ஒரு குழந்தையின் இயலாமைத் தனத்தை செயல்பாட்டில் காட்டுகிறான். பற்களை இழந்து, கண் பார்வைக் குன்றிப் போகிறான். குழந்தையின் குரல் போல் மாறி, குரல் ஒரு விசில் சத்தமாய் மாறுகிறது. இது கடைசி அத்தியாயம் எனலாம். அவனின் அதிசயமான பரப்பரப்பூட்டும் நிகழ்வுகள் நிறைந்த வரலாறானது ஒரு முடிவுக்கு வருகிறது. தன் இரண்டாம் குழந்தைப் பருவத்தில் பிறரைச் சார்ந்து வாழும் நிலையை அடைகிறான். பற்களை இழந்து, கண் பார்வையை இழந்து, நாவின் | சுவை இழந்து, பின் அனைத்தையும் இழக்கிறான். உலகத்தை விட்டு வெளியேறுகிறான்.

முடிவுரை:
சேக்ஸ்பியர் மனித வாழ்க்கையை சுருக்கி அழகாக அதை வர்ணித்து இருக்கிறார். வயோதிகத்தில் திரும்பும் குழந்தைத் தனம் என்பது அவர் மனித வாழ்க்கையை அவர் ஆழ்ந்து அறிந்து கொண்டதை உணர்த்துகிறது.

All the World’s a Stage Glossary

Textual:

All The World's A Stage Poem Appreciation Questions And Answers Samacheer Kalvi 12th English Solutions Poem Chapter 3

Additional:

All The World's A Stage Questions And Answers Samacheer Kalvi 12th English Solutions Poem Chapter 3

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Samacheer Kalvi 12th Physics Current Electricity Multiple Choice Questions   

Current Electricity Class 12 Problems With Solutions State Board Question 1.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?
Current Electricity Class 12 Problems With Solutions State Board Chapter 2 Samacheer Kalvi
(a) 2 ohm
(b) 4 ohm
(c) 8 ohm
(d) 1 ohm
Answer:
(a) 2 ohm

Current Electricity Class 12 State Board Question 2.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is-
Current Electricity Class 12 State Board Solutions Chapter 2 Samacheer Kalvi
(a) π Ω
(b) \(\frac { π }{ 2 }\) Ω
(c) 2π Ω
(d) \(\frac { π }{ 4 }\) Ω
Answer:
(b) \(\frac { π }{ 2 }\) Ω

12th Physics Chapter 2 Book Back Answers Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is
(a) 400 W
(b) 2 W
(c) 480 W
(d) 240 W
Answer:
(c) 480 W

12th Physics Samacheer Kalvi Question 4.
A carbon resistor of (47 ± 4.7) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be
(a) Yellow – Green – Violet – Gold
(b) Yellow – Violet – Orange – Silver
(c) Violet – Yellow – Orange – Silver
(d) Green – Orange – Violet – Gold
Answer:
(b) Yellow – Violet – Orange – Silver

Samacheer Kalvi 12th Physics Question 5.
What is the value of resistance of the following resistor?
12th Physics Chapter 2 Book Back Answers Current Electricity Samacheer Kalvi
(a) 100 k Ω
(b) 10 k Ω M
(c) 1 k Ω
(d) 1000 k Ω
Answer:
(a) 100 k Ω

Class 12 Physics Samacheer Kalvi Question 6.
Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose RA = 3 RB, then what is the ratio of radius of wire A to that of B?
(a) 3
(b) √3
(c) \(\frac { 1 }{ √3 }\)
(d) \(\frac { 1 }{ 3 }\)
Answer:
(c) \(\frac { 1 }{ √3 }\)

Current Electricity Short Notes Question 7.
A wire connected to a power supply of 230 V has power dissipation P1 Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P2. The ratio \(\frac {{ p }_{2}}{{ p }_{1}}\) is.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Samacheer Kalvi Guru 12th Physics Question 8.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be
(a) R
(b) 2R
(c) \(\frac { R }{ 4 }\)
(d) \(\frac { R }{ 2 }\)
Answer:
(c) \(\frac { R }{ 4 }\)

Samacheer Kalvi 12th Physics Solutions Question 9.
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW are connected. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be (IIT-JEE 2014)
(a) 14 A
(b) 8 A
(c) 10 A
(d) 12 A
Answer:
(d) 12 A

Samacheerkalvi.Guru 12th Physics Question 10.
There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?
12th Physics Samacheer Kalvi Solutions Chapter 2 Current Electricity
(a) 1.5 Ω
(b) 2.5 Ω
(c) 3.5 Ω
(d) 4.5 Ω
Answer:
(c) 3.5 Ω

Samacheer Kalvi Physics Question 11.
What is the current out of the battery?
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity
(а) 1 A
(b) 2 A
(c) 3 A
(d) 4 A
Answer:
(а) 1 A

12th Physics Solutions Samacheer Kalvi Question 12.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1 Ω. The resistance of the wire will be 2 Ω. at
(a) 1154 K
(ft) 1100 K
(c) 1400 K
(d) 1127 K
Answer:
(d) 1127 K

Samacheer Kalvi 12th Physics Solution Book Question 13.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10Ω is
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Samacheer Kalvi Guru Physics Question 14.
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
(a) each of them increases
(b) each of them decreases
(c) copper increases and germanium decreases
(d) copper decreases and germanium increases
Answer:
(d) copper decreases and germanium increases

Physics Class 12 Samacheer Kalvi Question 15.
In Joule’s heating law, when I and t are constant, if the H is taken along the y axis and I2 along the x axis, the graph is
(a) straight line
(b) parabola
(c) circle
(d) ellipse
Answer:
(a) straight line

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Samacheer Kalvi.Guru 12th Physics Question 1.
Why current is a scalar?
Answer:
The current I is defined as the scalar product of current density and area vector in which the charges cross.
I = \(\vec { j } \) . \(\vec { A } \)
The dot product of two vector quantity is a scalar form. Hence, current is called as a scalar quantity.

12th Samacheer Physics Solutions Question 2.
Distinguish between drift velocity and mobility.
Answer:
Class 12 Physics Samacheer Kalvi Solutions Chapter 2 Current Electricity

Physics Solution Class 12 Samacheer Kalvi Question 3.
State microscopic form of Ohm’s law.
Answer:
Current density J at a point in a conductor is the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.
Current Electricity Short Notes Samacheer Kalvi 12th Physics Solutions Chapter 2

Current Electricity Pdf Question 4.
State macroscopic form of Ohm’s law.
Answer:
The macroscopic form of Ohm’s Law relates voltage, current and resistance. Ohm’s Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object’s resistance.
V = IR.

Current Electricity Class 12 Question 5.
What are ohmic and non-ohmic devices?
Answer:
Materials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm’s law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm’s law are said to be non-ohmic.

Samacheer Kalvi Class 12 Physics Solutions Question 6.
Define electrical resistivity.
Answer:
Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

Question 7.
Define temperature coefficient of resistance.
Answer:
It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T0
Samacheer Kalvi Guru 12th Physics Solutions Chapter 2 Current Electricity

Question 8.
What is superconductivity?
Answer:
The ability of certain metals, their compounds and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.

Question 9.
What is electric power and electric energy?
Answer:
1. Electric power:
It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
P = \(\frac { W }{ t }\) = VI = I2R = \(\frac {{ V }_{2}}{ R }\)

2. Electric energy:
It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Pt = VIt joule = I2Rt joule.

Question 10.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
J = \(\frac { 1 }{ A }\)
The S.I unit of current density.
\(\frac { A }{{ m }^{2}}\)
Or
Am2

Question 11.
Derive the expression for power P = VI in electrical circuit.
Answer:
The electrical power P is the rate at which the electrical potential energy is delivered,
P = \(\frac { dU }{ dt }\) = \(\frac { d }{ dt }\) (V.dQ) = V\(\frac { dQ }{ dt }\)
Since the electric current I = \(\frac { dQ }{ dt }\)
So the equation can be rewritten as P = VI.

Question 12.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power P is the rate at which the electrical potential energy is delivered,
P = \(\frac { dU }{ dt }\) = \(\frac { 1 }{ dt }\) (V.dQ) = V.\(\frac { dQ }{ dt }\)
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
P = VI
The electric power delivered to the resistance R is expressed in other forms.
P = VI = I(IR) = I2R
P = IV = \((\frac { V }{ R })\) V = \(\frac {{ V }^{2}}{ R }\).

Question 13.
State Kirchhoff’s current rule.
Answer:
It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Question 14.
State Kirchhoff’s voltage rule.
Answer:
It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

Question 15.
State the principle of potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 16.
What do you mean by internal resistance of a cell?
Answer:
The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. An ideal battery has zero internal resistance and the potential difference across the battery equals to its emf. But a real battery is made of electrodes and electrolyte, there is resistance to the flow of charges within the battery. A freshly prepared cell has low internal resistance and it increases with ageing.

Question 17.
State Joule’s law of heating.
Answer:
It states that the heat developed in an electrical circuit due to the flow of current varies directly as:

  1. the square of the current
  2. the resistance of the circuit and
  3. the time of flow.
    H = I2R?

Question 18.
What is Seebeck effect?
Answer:
Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an emf (potential difference) is developed.

Question 19.
What is Thomson effect?
Answer:
Thomson showed that if two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result the potential difference is created between these points. Thomson effect is also reversible.

Question 20.
What is Peltier effect?
Answer:
When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect.

Question 21.
State the applications of Seebeck effect.
Answer:
Applications of Seebeck effect:

  1. Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.
  2. This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
  3. Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.

Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Describe the microscopic model of current and obtain genera! form of Ohm’s Law.
Answer:
Microscopic model of current: Consider a conductor with area of cross-section A and an electric field E applied from right to left. Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity \(\vec { v } \)d.
The drift velocity of the electrons = vd
The electrons move through a distance dx within a small interval of dt
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity
vd = \(\frac { dx }{ dt }\); dx = vddt ….. (1)
Since A is the area of cross section of the conductor, the electrons available in the volume or length dx is
= volume x number per unit volume
= A dx × n …… (2)
Substituting for dx from equation (1) in (2)
= (A vd dt)n
Total charge in volume element dQ = (charge) x (number of electrons in the volume element)
dQ= (e)(A vd dt)n
Hence the current, I = \(\frac { dQ }{ dt }\) = \(\frac{n e A v_{d} d t}{d t}\)
I = ne A vd …….. (3)
Current denshy (J):
The current density (J) is defined as the current per unit area of cross section of the conductor
J = \(\frac { I }{ A }\)
The S.I. unit of current density,\(\frac { A }{{ m }^{2}}\) (or) Am-2
J = \(\frac {{ neA v }_{d}}{ A }\) (from equation 3)
J = nevd …….. (4)
The above expression is valid only when the direction of the current is perpendicular to the area A. In general, the current density is a vector quantity and it is given by
\(\vec { J } \) = ne\(\vec { v } \)d
Substituting i from equation \(\vec { v } \)d = \(\frac { -eτ }{ m }\) \(\vec { E } \)
\(\vec { J } \) = –\(\frac{n \cdot e^{2} \tau}{m}\)\(\vec { E } \) …… (5)
\(\vec { J } \) = -σ\(\vec { E } \)
But conventionally, we take the direction of (conventional) current density as the direction of electric field. So the above equation becomes
\(\vec { J } \) = σ\(\vec { E } \) …….. (6)
where σ = \(\frac{n \cdot e^{2} \tau}{m}\) is called conductivity.
The equation 6 is called microscopic form of ohm’s law.

Question 2.
Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.
Samacheerkalvi.Guru 12th Physics Solutions Chapter 2 Current Electricity
When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as V = El
As we know, the magnitude of current density
J = σE = σ\(\frac { V }{ l }\) ……. (1)
But J = \(\frac { I }{ A }\), so we write the equation as
\(\frac { I }{ A }\) σ\(\frac { V }{ l }\)
By rearranging the above equation, we get
V = I \(\left( \frac { l }{ \sigma A } \right) \) ……… (2)
The quantity \(\frac { l }{ \sigma A } \)is called resistance of the conductor and it is denoted as R. Note that the oA resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR.

Question 3.
Explain the equivalent resistance of a series and parallel resistor network.
Answer:
1. Resistors in series:
When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors R1, R2 and R3 connected in series.
Samacheer Kalvi Physics 12th Solutions Chapter 2 Current Electricity
The amount of charge passing through resistor R1 must also pass through resistors R2 and R3 since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V1, V2 and V3 be the potential difference (voltage) across each of the resistors R1, R2 and R3 respectively, then we can write V1 = IR1, V2 = IR2 and V3= IR3. But the total voltage V is equal to the sum of voltages across each resistor.
V = V1 + V2 + V3
= IR1 + IR2 + IR3 ….. (1)
V = I(R1 + R2 +R3)
V = I.RS …… (2)
where Rs = R1 + R2 R3 ……. (3)
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).
12th Physics Solutions Samacheer Kalvi Chapter 2 Current Electricity
Note:
The value of equivalent resistance in series connection will be greater than each individual resistance.

2. Resistors in parallel:
Resistors are in parallel when they are connected across the same potential difference as shown in figer.
Samacheer Kalvi 12th Physics Solution Book Chapter 2 Current Electricity
In this case, the total current I that leaves the battery in split into three separate paths. Let I1, I2 and I3 be the current through the resistors R1, R2 and R3 respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I1 + I2 + I3 ……. (1)
Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have
I1 = \(\frac { V }{{ R }_{1}}\) I2 = \(\frac { V }{{ R }_{2}}\),I1 = \(\frac { V }{{ R }_{3}}\)
Substituting these values in equation (1), we get
I = \(\frac { V }{{ R }_{1}}\) + \(\frac { V }{{ R }_{2}}\) +\(\frac { V }{{ R }_{3}}\) = V\(\left[ \frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \right] \)
I = \(\frac { V }{{ R }_{p}}\)
\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
Here RP is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).
Samacheer Kalvi Guru Physics 12th Solutions Chapter 2 Current Electricity
Note:
The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 4.
Explain the determination of the internal resistance of a cell using voltmeter.
Answer:
Determination of internal resistance:
The emf of cell ξ is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the cell. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).
Physics Class 12 Samacheer Kalvi Solutions Chapter 2 Current Electricity
The potential drop across the resistor R is
V= IR …… (1)
Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ξ. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.
Samacheer Kalvi.Guru 12th Physics Solutions Chapter 2 Current Electricity
V = ξ – Ir
Ir = ξ – V …… (2)
Dividing equation (2) by equation (1), we get
\(\frac { Ir }{IR}\) = \(\frac { ξ – V }{V}\)
r = |\(\frac { ξ – V }{V}\)| R …… (3)
Since ξ, V and R are known, internal resistance r can be determined.

Question 5.
State and explain Kirchhoff’s rules.
Answer:
Kirchhoff’s first rule (current rule or junction rule):
Statement: It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.
12th Samacheer Physics Solutions Chapter 2 Current Electricity

Explanation:
All charges that enter a given junction in a circuit must leave that junction since charge cannot build up or disappear at a junction. Current entering the junction is taken as positive and current leaving the junction is taken as negative.
Applying this law to the junction A,
I1 + I2 – I3 – I4 – I5 = o
Or
I1 + I2 = + I3 I4 + I5
Kirchhoff’s second rule (voltage rule or loop rule):
Statement: It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system. (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).

Explanation:
The product of current and resistance is taken as positive when the direction of the current is followed. Suppose if the direction of current is opposite to the direction of the loop, then product of current and voltage across the resistor is negative. It is shown in Fig. (a) and (b). The emf is considered positive when proceeding from the negative to the positive terminal of the cell. It is shown in Fig. (c) and (d).
Physics Solution Class 12 Samacheer Kalvi Chapter 2 Current Electricity
Kirchhoff voltage rule has to be applied only when all currents in the circuit reach a steady state condition (the current in various branches are constant).

Question 6.
Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:
An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The – bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is IG and its resistance is G.
Current Electricity Pdf Samacheer Kalvi 12th Physics Solutions Chapter 2
Applying KirchhofFs current rule to junction B,
I1 – IG – I3 = 0 …….. (1)
Applying Kirchhoff’s current rule to junction D,
I2 – IG – I4 = 0 …….. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I1P + IGG – I2R = 0 …….. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I1P + I3Q – I4S – I2R = 0 …….. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (IG = 0). Substituting IG = 0 in equation, (1), (2) and (3), we get
I1 = I3 …….. (5)
I2 = I4 …….. (6)
I1P = I2R …….. (7)
Substituting the equation (5) and (6) in equation (4)
I1P + I1Q – I2R = 0
I1(P + Q) = I2 (R + S) …….. (8)
Dividing equation (8) by equation (7), we get
\(\frac { P + Q }{ P }\) = \(\frac { R + S }{ R }\)
1 + \(\frac { Q }{ P }\) = 1 + \(\frac { S }{ R }\)
⇒ \(\frac { Q }{ P }\) = \(\frac { S }{ R }\)
\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) …….. (9)
This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

Question 7.
Explain the determination of unknown resistance using meter bridge.
Answer:
The meter bridge is another form of Wheatstone’s bridge. It consists of a uniform manganin wire AB of one meter length. This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G1 and G2 An unknown resistance P is connected in G1 and a standard resistance Q is connected in G2.

A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR). The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.
Current Electricity Class 12 Samacheer Kalvi Physics Solutions Chapter 2
The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the point be J. The lengths AJ and JB of the bridge wire now replace the resistance R and S of the Wheatstone’s bridge. Then
\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) = \(\frac { R’.AJ }{ R’.JB }\) …….. (1)
where R’ is the resistance per unit length of wire
\(\frac { P }{ Q }\) = \(\frac { AJ }{ JB }\) = \(\frac {{ l }_{1}}{ { l }_{2} }\) …….. (2)
P = Q \(\frac {{ l }_{1}}{ { l }_{2} }\) ……… (3)
The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.
To find the specific resistance of the material of the wire in the coil P, the radius r and length l of the wire is measured. The specific resistance or resistivity r can be calculated using the relation
Resistance = ρ-\(\frac { l }{ A }\)
By rearranging the above equation, we get
ρ = Resistance x \(\frac { A }{ l }\) …….. (4)
If P is the unknown resistance, equation (4) becomes
ρ = P\(\frac {{ πr }^{2}}{ l }\).

Question 8.
How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer:
To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ1 and ξ2 to be compared are connected to the terminals M1, N1 and M2, N2 of the DPDT switch.
Samacheer Kalvi Class 12 Physics Solutions Chapter 2 Current Electricity
The positive terminals of Bt, ξ1 and ξ2 should be connected to the same end C. The DPDT switch is pressed towards M1, N1 so that cell ξ1 is included in the secondary circuit and the balancing length l1 is found by adjusting the jockey for zero deflection, Then the second cells ξ2 is included in the circuit and the balancing length l2 is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have.
ξ1 = Irl1 …… (1)
ξ2 = Irl2 ……. (2)
By dividing (1) by (2)
\(\frac {{ ξ }^{1}}{ { ξ }^{2} }\) = \(\frac {{ l }^{1}}{ { l }^{2} }\) ……. (3)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
The following graphs represent the current versus voltage and voltage versus current for the six conductors A,B,C,D,E and F. Which conductor has least resistance and which has maximum resistance?
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-21
According to ohm’s law, V = IR
Resistance of conductor, R = \(\frac { V }{ I }\)

Graph-I:
Conductor A, I = 4 A and V = 2 V
R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 4 }\) = 0.5 Ω
Conductor B, I = 3 A and V = 4 V
R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 3 }\) = 1.33 Ω
Conductor C, I = 2 A and V = 5 V
R= \(\frac { V }{ I }\) = \(\frac { 5 }{ 2 }\) = 2.5 Q Ω.

Graph-II:
Conductor D, I = 2 A and V = 4 V
R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 2 }\) = 2 Ω d
Conductor E, I = 4A and V = 3 V
R = \(\frac { V }{ I }\) = \(\frac { 3 }{ 4 }\) = 1.75 Ω
Conductor F, I = 5A and V = 2 V
R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 5 }\) = 0.4 Ω
Conductor F has least resistance, RF = 0.4 Ω,
Conductor C has maximum resistance, RC = 2.5 G.

Question 2.
Lightning is very good example of natural current. In typical lightning, there is 109 J energy transfer across the potential difference of 5 x 107 V during a time interval of 0.2 s.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-22
Using this information, estimate the following quantities (a) total amount of charge transferred between cloud and ground (b) the current in the lightning bolt (c) the power delivered in 0.2 s.
Solution:
During the lightning energy, E = 109 J
Potential energy, V = 5 x 107 V
Time interval, t = 0.2 s
(a) Amount of charge transferred between cloud and ground,
q = It

(b) Current in the lighting bolt, E = VIt
I = \(\frac { E }{ Vt }\) = \(\frac{10^{9}}{5 \times 10^{7} \times 0.2}\) = 1 × 109 × 1-7
I = 1 × 102
I = 100 A
∴ q = It = 100 × 0.2
q = 20 C.

(c) Power delivered, E = VIt
P = VI = 5 × 107 × 100 = 500 × 107
I = 5 × 109 W
P = 5 GW.

Question 3.
A copper wire of 106 m2 area of cross section, carries a current of 2 A. If the number of electrons per cubic meter is 8 x 1028, calculate the current density and average drift velocity.
Solution:
Cross-sections area of copper wire, A = 106 m2
I = 2 A
Number of electron, n = 8 x 1028
Current density, J = \(\frac { 1 }{ A }\) = \(\frac { 2 }{{ 10 }^{-6}}\)
J = 2 × 106 Am-2
Average drift velocity, Vd = \(\frac { 1 }{ neA }\)
e is the charge of electron = 1.6 × 10-9 C
Vd = \(\frac{2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}\) = \(\frac { 1 }{{ 64. × 10 }{3}}\)
Vd = 0.15625 × 10-3
Vd = 15.6 × 10-5 ms-1

Question 4.
The resistance of a nichrome wire at 0 °C is 10 Ω. If its temperature coefficient of resistance is 0.004/°C, find its resistance at boiling point of water. Comment on the result.
Solution:
Resistance of a nichrome wire at 0°C, R0 = 10 Ω
Temperature co-efficient of resistance, α = 0.004/°C
Resistance at boiling point of water, RT = ?
Temperature of boiling point of water, T = 100 °C
RT=R0 ( 1 + αT) = 10[1 + (0.004 x 100)]
RT= 10(1 +0.4) = 10 x 1.4
RT = 14 Ω
As the temperature increases the resistance of the wire also increases.

Question 5.
The rod given in the figure is made up of two different materials.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-23
Both have square cross sections of 3 mm side. The resistivity of the first material is 4 x 10-3 Ω.m and it is 25 cm long while second material has resistivity of 5 x 10-3 Ω.m and is of 70 cm long. What is the resistivity of rod between its ends?
Solution:
Square cross section of side, a = 3 mm = 3 x 10-3 m
Cross section of side, A = a2 = 9 x 106 m
First material:
Resistivity of the material, ρ1 = 4 x 10-3 Ωm
length, l1 = 25 cm = 25 x 10-2 m
Resistance of the lord, R1 = \(\frac{\rho_{l} l_{l}}{A}\) = \(\frac{4 \times 10^{-3} \times 25 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{100 \times 10^{-5} \times 10^{6}}{9}\)
R1 = 11.11 x 101

Second material:
Resistivity of the material, ρ2 = 5 x 10-3 Ωm
length, l2 = 70 cm = 70 x 10-2 m
Resistance of the rod, R2 = \(\frac{\rho_{2} l_{2}}{A}\) = \(\frac{5 \times 10^{-3} \times 70 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{350 \times 10^{-5} \times 10^{6}}{9}\)
R2 = 38.88 x 101
R2 = 389 Ω
Total resistance between the ends f the rods
R = R1 + R2 = 111 + 389
= 500 Ω

Question 6.
Three identical lamps each having a resistance R are connected to the battery of emf as shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-24
Suddenly the switch S is closed, (a) Calculate the current in the circuit when S is open and closed (b)
What happens to the intensities of the bulbs A,B and C. (c) Calculate the voltage across the three bulbs when S is open and closed (d) Calculate the power delivered to the circuit when S is opened and closed (e) Does the power delivered to the circuit decreases, increases or remain same?
Solution:
Resistance of the identical lamp = R
Emf of the battery = ξ
According to Ohm’s Law, ξ = IR
(a) Current:
When Switch is open— The current in the circuit. Total resistance of the bulb,
Rs = R1 + R2 + R3
R1 = R2 = R3 = R
Rs = R + R + R = 3R
∴ Current, I = \(\frac { ξ }{{ R }_{s}}\)
⇒ I0 = \(\frac { ξ }{ 3R }\)
Switch is closed— The current in the circuit. Total resistance of the bulb,
Rs = R + R = 2R
Current I = \(\frac { ξ }{{ R }_{s}}\)
Ic = \(\frac { ξ }{ 2R }\).

(b) Intensity:
When switch is open — All the bulbs glow with equal intensity.
When switch is closed — The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current pass through it.

(c) Voltage across three bulbs:
When switch is open — Voltage across bulb A, VA = I0 R = \(\frac { ξ }{ 3R }\) x R = \(\frac { ξ }{ 3 }\)
similarly:
Voltage across bulb B, VB = \(\frac { ξ }{ 3 }\)
Voltage across bulb C, VC = \(\frac { ξ }{ 3 }\)
When switch is closed— Voltage across bulb A, VA = IcR = \(\frac { ξ }{ 2R }\) x \(\frac { ξ }{ 2 }\)
similarly:
Voltage across bulb B, VB = IcR \(\frac { ξ }{ 2 }\)
Voltage across bulb C, VC = 0

(d) Power delivered to the circuit,
When switch is opened — Power P, = VI
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-25
When switch is closed — Power P, = VI
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-26

(e) Total power delivered to circuit increases.

Question 7.
The current through an element is shown in the figure. Determine the total charge that pass through the element at
(a) t = 0 s
(b) t = 2 s
(c) t = 5 s
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-27
Solution :
Rate of flow of charge is called current, I = \(\frac { dq }{ dt }\)
Total charge pass through element, dq = Idt
(a) At t= 0 s, I = 10 A
dq = Idt= 10 x 0 = 0 C.

(b) At t = 2 s, I = 5 A
dq = Idt = 5 x 2 = 10 C.

(c) At t = 5 s, I = 0
dq = Idt = 0 x 5 = 0 C.

Question 8.
An electronics hobbyist is building a radio which requires 150 Ω in her circuit, but she has only 220 Ω, 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Solution:
Required effective resistance = 150 Ω
Given resistors of resistance, R1 = 220 Ω, R2 = 79 Ω, R3 = 92 Ω
Parallel combination R1 and R2
\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{{ R }_{1}}\) + \(\frac { 1 }{{ R }_{2}}\) = \(\frac { 1 }{ 220 }\) + \(\frac { 1 }{ 79 }\) = \(\frac { 79 + 220 }{ 220 × 79 }\)
Rp = 58 Ω
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-28
Parallel combination Rp and R3
Rs = Rp + R3 = 58 + 92
Rs = 150 Ω
Parallel combination of 220 Ω and 79 Ω in series with 92 Ω.

Question 9.
A cell supplies a current of 0.9 A through a 2 Ω resistor and a current of 0.3 A through a 7 Ω resistor. Calculate the internal resistance of the cell.
Solution:
Current from the cell, I1 = 0.9 A
Resistor, R1 = 2 Ω
Current from the cell, I2 = 0.3 A
Resistor, R2 7 Ω
Internal resistance of the cell, r = ?
Current in the circuit I1 = \(\frac { ξ }{{ r + R }_{1}}\)
ξ = I1 (r + R1) …… (1)
Current in the circuit, I2 = \(\frac { ξ }{{ r + R }_{2}}\) …… (2)
Equating equation (1) and (2),
I1r + I1R1 = I2R2 + I2r
(I1 – I2)r = I2R2 – I1R1
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-29
r = 0.5 Ω.

Question 10.
Calculate the currents in the following circuit.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-30
Solution:
Applying Kirchoff’s 1st Law at junction B
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-31
I1 – I1 – I3 = 0
I3 = I1 – I2 …… (1)
Applying Kirchoff’s IInd Law at junction in ABEFA
100 I3 + 100 I1 = 15
100 (I3 + I1) = 15
100 I1 – 100 I2 + 100 I1 =15
200 I1 – 100 I2 = 15 …… (2)
Applying Kirchoff’s IInd Law at junction in BCDED
-100I2 + 100 I3 = 9
-100 I2+ 100(I1 – I2) = 9
100I1 – 200 I2 = 9
Solving equating (2) and (3)
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-32
Substitute I1 values in equ (2)
200(0.07) – 100 I2 = 15
14 – 100 I2 = 15
– 100 I2 = 15 – 14
I2 = \(\frac { -1 }{ 100 }\)
I2 = -0.01A
Substitute I1 and I2 value in equ (1), we get
I3 = I1 – I2 = 0.07 – (-0.01)
I3 = 0.08 A.

Question 11
A potentiometer wire has a length of 4 m and resistance of 20 Ω. It is connected in series with resistance of 2980 Ω and a cell of emf 4 V. Calculate the potential along the wire.
Solution:
Length of the potential wire, l = 4 m
Resistance of the wire, r = 20 Ω
Resistance connected series with potentiometer wire, R = 2980 Ω
Emf of the cell, ξ = 4 V
Effective resistance, Rs = r + R = 20 + 2980 = 3000 Ω
Current flowing through the wire, I = \(\frac { ξ }{{ r + R }_{s}}\) = \(\frac { 4 }{ 3000 }\)
I = 1.33 x 10-3 A
Potential drop across the wire, V = Ir
= 1.33 x 10-33 x 20
V = 26.6 x 10-3 V
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-33
= 6.65 x 10-3
Potential garadient = 0.66 x 10-2 Vm-1

Question 12.
Determine the current flowing through the galvanometer (G) as shown in the figure.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-34
Solution:
Current flowing through the circuit, I = 2A
Applying Kirchoff’s Ist law at junction P, I = I1 + I2
I2 = I – I1 …(1)
Applying Kirchoff’s Ind law at junction PQSP
5 I1 + 10 Ig – 15 I2 = 0
5 I1 + 10 Ig -15(I – I1) = 0
20 I1 + 10 Ig = 15 I
20 I1 + 10 Ig = 15 x 2
÷ by 10 21 I1 + Ig = 3 … (2)
Applying Kirchoff’s IInd law at junction QRSQ
10(I1 – Ig) – 20(I – I1 – Ig) – 10 Ig = 0
10 I1 – 10g – 20(I – I1 – Ig) – 10 Ig = 0
10 I1 – 10g – 20 I + 20 I1 -20Ig – 10 Ig = 0
30 I 1 – 40 Ig = 20 I
÷ by 10 ⇒ 3 I1 – 4 Ig = 20 I
3 I1 – 4 Ig = 2 I
3 I1 – 4 Ig = 2 x 2 = 4
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-35

Question 13.
Two cells each of 5 V are connected in series across a 8 Ω resistor and three parallel resistors of 4 Ω, 6 SI and 12 Ω. Draw a circuit diagram for the above arrangement. Calculate (i) the current drawn from the cell (ii) current through each resistor.
Solution:
V1 = 5 V; V2 = 5 V
R1 = 8 Ω; R2 = 4 Ω; R3 = 6 Ω; R4 = 12 Ω
Three resistors R2, R3 and R4 are connected parallel combination
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-36
Resistors R1 and Rp are connected in series combination
Rs = R1 +Rp = 8 + 2 = 10
Rs = 10Ω
Total voltage connected series to the circuit
V = V1 + V2
= 5 + 5 = 10
V = 10 V.
(i) Current through the circuit, I = \(\frac { V }{{ R }_{s}}\) = \(\frac { 10 }{ 10 }\)
I = 1 A
Potential drop across the parallel combination,
V’ = IRp = 1 x 2
V’ 2 V

(ii) Current in 4 Ω resistor, I = \(\frac { V’ }{{ R }_{2}}\) = \(\frac { 2 }{ 4 }\) = 0.5 A
Current in 6 Ω resistor, I = \(\frac { V’ }{{ R }_{3}}\) = \(\frac { 2 }{ 6 }\) = 0.33 A
Current in 12 Ω resistor, I = \(\frac { V’ }{{ R }_{4}}\) = \(\frac { 2 }{ 12 }\) = 0.17 A

Question 14.
Four light bulbs P, Q, R, S are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-37
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-38

Question 15.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
Solution:
Emf of the cell1, ξ1 = 1.25 V
Balancing length of the cell, l1 = 35 cm = 35 x 10-2 m
Balancing length after interchanged, l2 = 63 cm = 63 x 10-2 m
Emf of the cell1, ξ2 = ?
The ration of emf’s, \(\frac {{ ξ }_{1}}{{ ξ }_{2}}\) = \(\frac {{ l }_{1}}{{ l }_{2}}\)
The ration of emf’s, ξ2 = ξ1 = \(\left( \frac { { l }_{ 2 } }{ { l }_{ 1 } } \right) \)
= 1. 25 x \(\left( \frac { { 63×10 }^{ -2 } }{ { 35×10 }^{ -2 } } \right) \) = 12.5 x 1.8
ξ2 = 2.25 V.

Samacheer Kalvi 12th Physics Current Electricity Additional Questions Solved

I. Choose the Correct Answer

Question 1.
When current I flows through a wire, the drift velocity of the electrons is v. When current 21 flows through another wire of the same material having double the length and area of cross-section, the drift velocity of the electrons will be-
(a) \(\frac { v }{ 4 }\)
(b) \(\frac { v }{ 2 }\)
(c) v
(d) 2 v
Answer:
(c) v
Hint:
Vd = \(\frac { 1 }{ nAe }\); v’d = \(\frac { 2I }{ (2A)ne }\) = vd

Question 2.
A copper wire of length 2 m and area of cross-section 1.7 x 10-6 m2 has a resistance of 2 x 10-2 Ω. The resistivity of copper is
(a) 1.7 x 10-8 Ωm
(b) 1.9 x -8 Ωm
(c) 2.1 x 10-7 Ωm
(d) 2.3 x 10-7 Ωm
Answer:
(a) 1.7 x 10-8 Ωm
Hint:
Resistivity, ρ = \(\frac { RA }{ l }\) = \(\frac{2 \times 10^{-2} \times 1.7 \times 10^{-6}}{2}\) = 1.7 x 10-8 Ωm.

Question 3.
If the length of a wire is doubled and its cross-section is also doubled, then its resistance will
(a) become 4 times
(b) become 1 / 4
(c) becomes 2 times
(d) remain unchanged
Answer:
(d) remain unchanged

Question 4.
A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3 V and a resistance of 10 Ω. The potential gradient along the wire in volt per meter is
(a) 6.02
(b) 0.1
(c) 0.2
(d) 1.2
Answer:
(c) 0.2
Hint:
Potential difference across the wire = \(\frac { 20 }{ 3 }\) x 3 = 2 V
Potential gradient = \(\frac { v }{ l }\) = \(\frac { 2 }{ 10 }\) = 0.2 V/m

Question 5.
The resistivity of a wire
(a) varies with its length
(b) varies with its mass
(c) varies with its cross-section
(d) does not depend on its length, cross-section and mass.
Answer:
(d) does not depend on its length, cross-section and mass.

Question 6.
The electric intensity E, current density and conductivity a are related as
(a) j = σE
(b) j = \(\frac { E }{ σ }\)
(c) JE = s
(d) j = σ2E
Answer:
(a) j = σE

Question 7.
For which of the following dependences of drift velocity Vd on electric field E, is Ohm’s law obeyed?
(a) vd ∝ E
(b) vd ∝ E2
(c) vd ∝ √E
(d) vd = constant
Answer:
(a) vd ∝ E
Hint:
vd = \(\frac { 1 }{ nAe }\) = \(\frac { j }{ ne }\) = \(\left( \frac { σ }{ ne } \right) \)E ⇒ vd ∝ E.

Question 8.
A cell has an emf of 1.5 V. When short circuited, it gives a current of 3A. The internal resistance of the cell is .
(a) 0.5 Ω
(b) 2.0 Ω
(c) 4.5 Ω
(d) \(\frac { 1 }{ 4.5 }\) Ω
Answer:
(a) 0.5 Ω
Hint:
r = \(\frac { ξ }{ I }\) = \(\frac { 1.5 }{ 3 }\) = 0.5 Ω.

Question 9.
The resistance, each of 1 Ω, are joined in parallel. Three such combinations are put in series. The resultant resistance is
(a) 9 Ω
(b) 3 Ω
(c) 1 Ω
(d) \(\frac { 1 }{ 3 }\) Ω
Answer:
(c) 1 Ω
Hint:
Rp = 1 + 1 + 1 = 3 Ω;
\(\frac { 1 }{{ R }_{s}}\) = \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 3 }{ 3 }\) = 1
⇒ Rs = 1 Ω.

Question 10.
Constantan is used for making standard resistance because it has
(a) high resistivity
(b) low resistivity
(c) negligible temperature coefficient of resistance
(d) high melting point
Answer:
(c) negligible temperature coefficient of resistance

Question 11.
Kirchhoff’s two laws for electrical circuits are magnifestations of the conservation of
(a) charge only
(b) both energy and momentum
(c) energy only
(d) both charge and energy
Answer:
(d) both charge and energy

Question 12.
The resistance R0 and Rt of a metallic wire at temperature 0° C and t° C are related as (a is the temperature co-efficient of resistance).
(a) Rt = R0(1 + αt)
(b) Rt = R0(1 – αt)
(c) Rt = R0(1 + αt)2
(d) Rt = R0(1 – αt) 2
Answer:
(a) Rt = R0(1 + αt)

Question 13.
A cell of emf 2 V and internal resistance 0.1 Ω is connected with a resistance of 3.9 Ω. The voltage across the cell terminals will be
(a) 0.5 V
(b) 1.9 V
(c) 1.95 V
(d) 2 V
Answer:
(c) 1.95 V
Hint:
V = \(\frac { ER }{ R + r }\) = \(\frac { 2 x 3.9 }{ 3.9 + 0.1 }\) = 1.95 V.

Question 14.
A flow of 107 electrons per second in a conduction wire constitutes a current of
(a) 1.6 x 10-26 A
(b) 1.6 x 1012 A
(c) 1.6 x 10-12 A
(d) 1.6 x 1026 A
Answer:
(c) 1.6 x 10-12 A
Hint:
I = \(\frac { Q }{ t }\) = \(\frac{10^{7} \times 1.6 \times 10^{-19}}{1}\) = 1.6 x 1012 A.

Question 15.
Sensitivity of a potentiometer can be increased by
(a) increasing the emf of the cell
(b) increasing the length of the wire
(c) decreasing the length of the wire
(d) none of the above
Answer:
(b) increasing the length of the wire

Question 16.
Potential gradient is defined as
(a) fall of potential per unit length of the wire.
(b) fall of potential per unit area of the wire.
(c) fall of potential between two ends of the wire.
(d) none of the above.
Answer:
(a) fall of potential per unit length of the wire.

Question 17.
n equal resistors are first connected in series and then in parallel. The ratio of the equivalent resistance in two cases is
(a) n
(b) \(\frac { 1 }{{ n }^{2}}\)
(c) n2
(d) \(\frac { 1 }{ n }\)
Answer:
(c) n2
Hint:
Required ratio = \(\frac{n \mathrm{R}}{\left(\frac{\mathrm{R}}{n}\right)}\) = n(c) n2

Question 18.
A galvanometer is converted into an ammeter when we connect a
(a) high resistance in series
(b) high resistance in parallel
(c) low resistance in series
(d) low resistance in parallel
Answer:
(d) low resistance in parallel

Question 19.
The reciprocal of resistance is
(a) conductance
(b) resistivity
(c) conductivity
(d) none of the above
Answer:
(a) conductance

Question 20.
A student has 10 resistors, each of resistance r. The minimum resistance that can be obtained by him using these resistors is
(a) 10r
(b) \(\frac { r }{ 10 }\)
(c) \(\frac { r }{ 100 }\)
(d) \(\frac { r }{ 5 }\)
Answer:
(b) \(\frac { r }{ 10 }\)

Question 21.
The drift velocity of electrons in a wire of radius r is proportional to
(a) r
(b) r2
(c) r3
(d) none of the above
Answer:
(d) none of the above

Question 22.
Kirchhoff’s first law, i.e. ∑I = 0 at a junction deals with conservation of
(a) charge
(b) energy
(c) momentum
Answer:
(a) charge

Question 23.
The resistance of a material increases with temperature. It is a
(a) metal
(b) insulator
(c) semiconductor
(d) semi-metal
Answer:
(a) metal

Question 24.
Five cells, each of emf E, are joined in parallel. The total emf of the combination is
(a) 5E
(b) \(\frac { E }{ 5 }\)
(c) E
(d) \(\frac { 5E }{ 2 }\)
Answer:
(c) E

Question 25.
A carbon resistance has colour bands in order yellow, brown, red. Its resistance is
(a) 41 Ω
(b) 41 x 102
(c) 41 x 103
(d) 4.2 Ω
Answer:
(b) 41 x 102

Question 26.
The conductivity of a superconductor is
(a) infinite
(b) very large
(c) very small
(d) zero
Answer:
(a) infinite

Question 27.
The resistance of an ideal voltmeter is
(a) zero
(b) very high
(c) very low
(d) infinite
Answer:
(d) infinite

Question 28.
Carriers of electric current in superconductors are
(a) electrons
(b) photons
(c) holes
Answer:
(c) holes

Question 29.
Potentiometer measures potential more accurately because
(a) It measure potential in the open circuit.
(b) It uses sensitive galvanometer for null detection.
(c) It uses high resistance potentiometer wire.
(d) It measures potential in the closed circuit.
Answer:
(a) It measure potential in the open circuit.

Question 30.
Electromotive force is most closely related to
(a) electric field
(b) magnetic field
(c) potential difference
(d) mechanical force
Answer:
(c) potential difference

Question 31.
The capacitance of a pure capacitor is 1 farad. In DC circuit, the effective resistance will be
(a) zero
(b) infinite
(c) 1 Ω
(d) 0.5 Ω
Answer:
(b) infinite

Question 32.
The resistance of an ideal ammeter is
(a) zero
(b) small
(c) high
(d) infinite
Answer:
(a) zero

Question 33.
A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as a voltmeter of range 10 V, the resistance that must be connected in series with it is
(a) 999 Ω
(b) 1000 Ω
(c) 9 Ω
(d) 99 Ω
Answer:
(a) 999 Ω
Hint:
R = \(\frac { V }{{ I }_{g}}\)-Rg = \(\frac { 10 }{{ 10 × 10 }^{-3}}\)-1 = 999 Ω.

Question 34.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is. 0.5 A. The terminal voltage of the battery when the circuit is closed is
(a) 10 V
(b) zero
(c) 8.5 V
(d) 1.5 V
Answer:
(c) 8.5 V
Hint:
V = ξ – Ir = 10 – (0.5 x 3) = 8.5 V.

Question 35.
Good resistance coils are made of
(a) copper
(b) manganin
(c) iron
(d) aluminium
Answer:
(b) manganin

Question 36.
A wire of resistance R is stretched to three times its original length. The new resistance is
(a) 3R
(b) 9R
(c) R/3
(d) R/9
Answer:
(b) 9R

Question 37.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to
(a) n2R
(b) \(\frac { R }{{n}^{ 2 }}\)
(c) \(\frac { R }{ n }\)
(d) nR
Answer:
(a) n2R
Hint:
Resistance in parallel combination, R = \(\frac { r }{ n }\) ⇒ r = Rn
Resistance in series combination, R’ = nr = n2R

Question 38.
When a wire of uniform cross-section, having resistance R, is bent into a complete circle, the resistance between any two of diametrically opposite points will be
(a) \(\frac { R }{ 8 }\)
(b) \(\frac { R }{ 2 }\)
(c) 4R
(d) \(\frac { R }{ 4 }\)
Answer:
(d) \(\frac { R }{ 4 }\)
Hint:
It becomes two resistors each of (d) \(\frac { R }{ 2 }\), connected in parallel.

Question 39.
A steady current is set up in a metallic wire of non uniform cross-section. How is the rate of flow K of electrons related to the area of cross-section A?
(a) K is independent of A
(b) K ∝ A
(c) K ∝ A-1
(d) K ∝ A2
Answer:
(c) K ∝ A-1

Question 40.
Ohm’s Law is not obeyed by
(a) electrolytes
(b) discharge tubes
(c) vacuum tubes
(d) all of these
Answer:
(d) all of these

Question 41.
Which of the following has negative temperature coefficient of resistance?
(a) Copper
(b) Aluminium
(c) Germanium
(d) Iron
Answer:
(c) Germanium

II. Fill in the blanks

Question 1.
The material through which electric charge can flow easily is ……………….
Answer:
Copper.

Question 2.
A toaster operating at 240 V has a resistance of 120 Ω. The power is ……………….
Answer:
480 W.

Question 3.
In the case of insulators, as the temperature decreases, resistivity ……………….
Answer:
Increases.

Question 4.
When n resistors of equal resistance (R) are connected in series, the effective resistance is ……………….
Answer:
nR.

Question 5.
The net flow of charge at any point in the conductor is ……………….
Answer:
Zero.

Question 6.
The flow of free electrons in a conductor constitutes ……………….
Answer:
Electric current.

Question 7.
The rate of flow of charge through any wire is called ……………….
Answer:
Current.

Question 8.
The drift velocity acquired per unit electric field is the ……………….
Answer:
Mobility.

Question 9.
The reciprocal of resistance is ……………….
Answer:
Conductance.

Question 10.
The unit of specific resistance is ……………….
Answer:
Ohm meter.

Question 11.
The reciprocal of electrical resistivity is called ……………….
Answer:
Electrical conductivity.

Question 12.
With increase in temperature the resistivity of metals ……………….
Answer:
Increases.

Question 13.
The resistivity of insulators is of the order of ……………….
Answer:
108 1014 Ωm.

Question 14.
The resistivity of semiconductors is of the order of ……………….
Answer:
10-2 -102 Ωm.

Question 15.
The materials which conduct electricity at zero resistance are called ……………….
Answer:
Superconductors.

Question 16.
Conductors turn into superconductors at ……………….
Answer:
Low temperatures.

Question 17.
The resistance of superconductors is ……………….
Answer:
Zero.

Question 18.
The phenomenon of superconductivity was discovered by ……………….
Answer:
Kammerlingh onnes.

Question 19.
Mercury becomes a superconductor at ……………….
Answer:
4.2 K.

Question 20.
With increase of temperature, resistance of conductors ……………….
Answer:
increases

Question 21.
In insulators and semiconductors, as temperature increases, resistance ……………….
Answer:
Decreases.

Question 22.
A material with a negative temperature coefficient is called a ……………….
Answer:
Thermistor.

Question 23.
The temperature coefficient for alloys is ……………….
Answer:
Low.

Question 24.
The electric current in an external circuit flows from the ……………….
Answer:
Positive to negative terminal.

Question 25.
In the electrolyte of the cell, current flows from ……………….
Answer:
Negative to positive terminal.

Question 26.
A freshly prepared cell has ………………. internal resistance.
Answer:
Low.

Question 27.
Kirchhoff’s first law is ……………….
Answer:
Current law.

Question 28.
The current law states that the algebraic sum of the currents meeting at any junction in a circuit is ……………….
Answer:
Zero.

Question 29.
Current law is a consequence of conservation of ……………….
Answer:
Charges.

Question 30.
Kirchhoff’s second law is ……………….
Answer:
Voltage law.

Question 31.
Kirchhoff’s second law is a consequence of conservation of ……………….
Answer:
Energy.

Question 32.
Wheatstone bridge is an application of ……………….
Answer:
Kirchhoff’s Law.

Question 33
………………. is a form of Wheatstone’s bridge.
Answer:
Metre bridge.

Question 34.
The temperature coefficient of manganin wire is ……………….
Answer:
Low.

Question 35
………………. is an instrument to measure potential difference.
Answer:
Potentiometer.

Question 36.
Unit of electrical energy is ……………….
Answer:
Joule.

Question 37.
An instrument to measure electrical power consumed is ……………….
Answer:
Watt meter.

Question 38.
………………. first introduced the electrochemical battery
Answer:
Volta.

Question 39.
Charging is a process of reproducing ……………….
Answer:
Active materials.

III. Match the following

Question 1.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-39
Answer:
(i) → (b)
(ii) → (a)
(iii) → (d)
(iv) →(c)

Question 2.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-40
Answer:
(i) → (b)
(ii) → (c)
(iii) → (d)
(iv) → (a)

Question 3.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-41
Answer:
(i) → (d)
(ii) → (c)
(iii) → (b)
(iv) → (a)

Question 4.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-42
Answer:
(i) → (b)
(ii)→ (d)
(iii) → (a)
(iv) → (c)

IV.Assertion and reason type

(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Fuse wire must have high resistance and low melting point.
Reason: Fuse is used for small current flow only
Answer:
(c) If assertion is true but reason is false.

Question 2.
Assertion: In practical application, power rating of resistance is not important.
Reason: Property of resistance remains same even at high temperature
Answer:
(d) If the assertion and reason both are false.

Question 3.
Assertion: Electric appliances with metallic body e.g. heaters, presses, etc, have three pin connections, whereas an electric bulb has two pins.
Reason: Three pin connection reduce heating of connecting cables.
Answer:
(c) If assertion is true but reason is false.

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Question 1.
Define current?
Answer:
Current is defined as a net charge Q passes through any cross section of a conductor in time t
then, I = \(\frac { Q }{ t }\).

Question 2.
Define instantaneous current?
Answer:
The instantaneous current I is defined as the limit of the average current, as ∆t → 0.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-43

Question 3.
What is resistance? Give its unit?
Answer:
The resistance is the ratio of potential difference across the given conductor to the current passing through the conductor V.
R = \(\frac { V }{ I }\).

Question 4.
What is meant by transition temperature?
Answer:
The resistance of certain materials become zero below certain temperature Tc. This temperature is known as critical temperature or transition temperature.

Question 5.
What is Joule’s heating effect?
Answer:
When current flows through a resistor, some of the electrical energy delivered to the resistor is converted into heat energy and it is dissipated. This heating effect of current is known as Joule’s heating effect.

Question 6.
What is meant by thermoelectric effect?
Answer:
Conversion of temperature differences into electrical voltage and vice versa is known as thermoelectric effect.

Question 7.
What is a thermopile? On what principle does it work?
Answer:
Thermopile is a device used to detect thermal radiation. It works on the principle of seebeck effect.

Question 8.
What is a thermistor?
Answer:
A material with a negative temperature coefficient is called a thermistor.
Eg:

  1. Insulator
  2. Semiconductor.

Question 9.
State principle of potentiometer?
Answer:
The principle of potentiometer states that the emf of the cell is directly proportional to its balancing length.
ξ ∝ l
ξ = Irl.
Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Explain the concept of colour code for carbon resistors.
Answer:
Color code for Carbon resistors:
Carbon resistors consists of a ceramic core, on which a thin layer of crystalline carbon is deposited. These resistors are inexpensive, stable and compact in size. Color rings are used to indicate the value of the resistance according to the rules.

Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth color, silver or gold, shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%. For the resistor, the first digit = 5 (green), the second digit = 6 (blue), decimal multiplier = 103 (orange) and tolerance = 5% (gold). The value of resistance = 56 x 103 Q or 56 kΩ with the tolerance value 5%.

Question 2.
Explain in details of temperature dependence of resistivity.
Answer:
Temperature dependence of resistivity:
The resistivity of a material is dependent on temperature. It is experimentally found that for a wide range of temperatures, the resistivity of a conductor increases with increase in temperature according to the expression,
ρT = ρ0 [1 + α(T -T0)] ……. (1)
where ρT is the resistivity of a conductor at T0C, ρ0 is the resistivity of the conductor at some reference temperature To (usually at 20°C) and a is the temperature coefficient of resistivity. It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity atT0.
From the equation (1), we can write
ρT – ρ0 = αρ0 (T -T0)
∴ α = \(\frac{\rho_{\mathrm{T}}-\rho_{0}}{\rho_{0}\left(\mathrm{T}-\mathrm{T}_{0}\right)}\) = \(\frac{\Delta p}{\rho_{0} \Delta T}\)
where ∆ρ = ρT – ρ0 is change in resistivity for a change in temperature ∆T = T – T0. Its unit is per °C.

1. α of conductors:
For conductors a is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases. Even though, the resistivity of conductors like metals varies linearly for wide range of temperatures, there also exists a nonlinear region at very low temperatures. The resistivity approaches some finite value as the temperature approaches absolute zero. As the resistance is directly proportional to resistivity of the material, we can also write the resistance of a conductor at temperature T °C as
RT -R0 = [1 + α(T -T0)] ……. (2)
The temperature coefficient can be also be obtained from the equation (2), +
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-44
where ∆R = RT -R0 is change in resistance during the change in temperature ∆T = T – T0

2. α of semiconductors:
For semiconductors, the resistivity decreases with increase in temperature. As the temperature increases, more electrons will be liberated from their atoms (Refer unit 9 for conduction in semi conductors). Hence the current increases and therefore the resistivity decreases. A semiconductor with a negative temperature coefficient of resistance is called a thermistor.
We can understand the temperature dependence of resistivity in the following way. The electrical conductivity, σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\). As the resistivity is inverse of σ, it can be written as,
σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\) …… (4)
The resistivity of materials is

  1. inversely proportional to the number density (n) of the electrons
  2. inversely proportional to the average time between the collisions (τ).

In metals, if the temperature increases, the average time between the collision (τ) decreases and n is independent of temperature. In semiconductors when temperature increases, n increases and τ decreases, but increase in n is dominant than decreasing x, so that overall resistivity decreases.

Question 3.
Explain the effective internal resistance of cells connected in series combination. Compare the results to the external resistance.
Answer:
Cells in series Several cells can be connected to form a battery. In series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of second cell is connected to the positive terminal of the third cell and so on. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-45
Suppose n cells, each of emf ξ volts and internal resistance r ohms are connected in series with an external resistance R.
The total emf of the battery = nξ
The total resistance in the circuit = nr + R
By Ohm’s law, the current in the circuit is
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-46
Case (a) If r << R, then,
I = \(\frac { nξ }{ R }\)nI1 ≈ nI1 ……. (2)
where, I, is the current due to a single cell \(\left(\mathrm{I}_{1}=\frac{\xi}{\mathrm{R}}\right)\)
Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If r >> R, I = \(\frac { nξ }{ nr }\) ≈ \(\frac { ξ }{ R }\) …….. (3)
It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells. Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.

Question 4.
Explain the effective internal resistance of cells connected in parallel combination. Compare the results to the external resistance.
Answer:
Cells in parallel: In parallel connection all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
Let n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B. Let ξ, be the emf and r the internal resistance of each cell.

The equivalent internal resistance of the battery is \(\frac { 1 }{{ r }_{eq}}\) = \(\frac { 1 }{ r }\) + \(\frac { 1 }{ r }\) + ….. \(\frac { 1 }{ r }\) (n terms) = \(\frac { n }{ r }\).
So reg = \(\frac { r }{ n }\) and the total resistance in the circuit = R + \(\frac { r }{ n }\). The total emf is the potential difference between the points A and B, which is equal to ξ. The current in the circuit is given by
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-47
Case (a) If r << R, then,
I = \(\frac { nξ }{ R }\) = nI1 …….. (2)
where II is the current due to a single cell and is equal to \(\frac { ξ }{ R }\) when R is negligible. Thus, the
current through the external resistance due to the whole battery is n times the current due to a single cell.

Case (b) If r << R. I = \(\frac { ξ }{ R }\) …… (3)
The above equation implies that current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connect cells in parallel when the external resistance is very small compared to the internal resistance of the cells.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
Show that one ampere is equivalent to a flow of 6.25 x 1018 elementary charges per second.
Solution:
Here I = 1 A, t= 1s, e = 1.6 x 10-19 C
As I = \(\frac { q }{ t }\) = \(\frac { ne }{ t }\)
Number of electrons, n = \(\frac { It }{ e }\) = \(\frac { 1 × 1 }{{ 1.6 × 10 }^{19}}\) = 6.25 × 1018

Question 2.
Calculate the resistivity of a material of a wire 10 m long. 0.4 mm in diameter and having a resistance of 2.0 Ω.
Solution:
Here l = 10 cm, r = 0.2 mm = 0.2 x 10-3 m, R = 2 Ω
\(\left[ r\quad =\quad \frac { d }{ 2 } \right] \)
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-48
= 2.513 x 10-8 Ω.

Question 3.
A wire of 10 ohm resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?
Solution:
(i) Resistivity p remains unchanged because it is the property of the material of the wire.
(ii) In both cases, volume of wire is same, so
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-49

Question 4.
A copper wire has a resistance of 10 Ω. and an area of cross-section 1 mm2. A potential difference of 10 V exists across the wire. Calculate the drift speed of electrons if the number of electrons per cubic metre in copper is 8 x 1028 electrons.
Solution:
Here, R = 10 Ω, A = 1 mm2 = 10-6 m2, V = 10 V, n = 8 x 1028 electrons/m3
Now,
I = enAvd
∴ \(\frac { V }{ R }\) = enAvd (or) vd = \(\frac { V }{ enAR }\)
= \(\frac{10}{1.6 \times 10^{-19} \times 8 \times 10^{28} \times 10^{-6} \times 10}\) = 0.078 x 10-3 ms-1
= 0.078 x ms-1

Question 5.
(i) At what temperature would the resistance of a copper conductor be double its resistance at 0°C.
(ii) Does this temperature hold for all copper conductors regardless of shape and size? Given a for Cu = 3.9 x 10-3 °C-1.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-50
Thus the resistance of copper conductor becomes double at 256 °C.
(ii) Since a does not depend on size and shape of the conductor. So the above result holds for all copper conductors.

Question 6.
Find the value of current I in the circuit shown in figure.
Solution:
Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity-51
In the circuit, the resistance of arm ACB (30 + 30 = 60 Ω) is the parallel with the resistance of arm AB (= 30 Ω).
Hence, the effective resistance of the circuit is
R = \(\frac { 30 × 60 }{ 30 + 60 }\) = 20 Ω
Current, I = \(\frac { V }{ R }\) = \(\frac { 2 }{ 20 }\) = 0.1 A.

Common Errors and Its Rectifications:

Common Errors:

  1. Sometimes students think that charge and current are same.
  2. In doing calculation part students can’t give the importance to mention the units.
  3. They may confuse the parallel and series network of the resistance.

Rectifications:

  1. Charge q = ne Current I = q/t
  2. Unit is very importance to the every physical quantities.
  3. If the resistors are series, their resultant is sum of the all the reciprocal of individual resistance. If the resistors are parallel their resultant is sum of the individual resistance.

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Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

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Samacheer Kalvi 12th Chemistry p-Block Elements – II TextBook Evalution

I. Choose the correct answer.

12th Chemistry Chapter 3 Book Back Answers Question 1.
In which of the following, NH3 is not used?
(a) Nessler’s reagent
(b) Reagent for the analysis of IV group basic radical
(c) Reagent for the analysis of III group basic radical
(d) Tollen’s reagent
Answer:
(a) Nessler’s reagent

Samacheer Kalvi Guru 12th Chemistry Question 2.
Which is time regarding nitrogen?
(a) least electronegative element
(b) has low ionisation enthalpy than oxygen
(c) d-orbitals available
(d) ability to form pπ – pπ bonds with itself
Answer:
(d) ability to form pπ – pπ bonds with itself

12th Chemistry 3rd Lesson Book Back Answers Question 3.
An element belongs to group 15 and 3 rd period of the periodic table, its electronic configuration would be …………
(a) 1s22s22p4
(b) 1s22s22p3
(c) 1s22s22p63s23p2
(d) 1s22s22p63s23p3
Answer:
(d) 1s22s22p63s23p3

Samacheerkalvi.Guru 12th Chemistry Question 4.
Solid (A) reacts with strong aqueous NaOH liberating a foul smelling gas(B) which spontaneously bum in air giving smoky rings. A and B are respectively …………
(a) P4(red) and PH3
(b) P4(white) and PH3
(c) S8 and H2S
(d) P4(white) and H2S
Answer:
(b) P4(white) and PH3

Samacheer Kalvi Guru Class 12 Chemistry Question 5.
In the brown ring test, brown colour of the ring is due to …………
(a) a mixture of NO and NO2
(b) Nitroso ferrous sulphate
(c) Ferrous nitrate
(d) Ferric nitrate
Answer:
(b) Nitroso ferrous sulphate

Samacheer Kalvi 12th Chemistry Question 6.
On hydrolysis, PCl3 gives …………
(a) H3PO3
(b) PH3
(c) H3PO4
(d) POOL
Answer:
(a) H3PO3

Chemistry Class 12 Samacheer Kalvi Question 7.
P4O6 reacts with cold water to give …………
(a) H3PO3
(b) H4P2O7
(c) HPO3
(d) H3PO4
Answer:
(a) H3PO3

Samacheer Kalvi Class 12 Chemistry Solutions Question 8.
The basicity of pyrophosphorous acid ( H4P2O5) is …………
(a) 4
(b) 2
(c) 3
(d) 5
Answer:
(b) 2

Samacheer Kalvi Guru Chemistry Question 9.
The molarity of given orthophosphoric acid solution is 2M. its normality is …………
(a) 6N
(b) 4N
(c) 2N
(d) none of these
Answer:
(a) 6N

Chapter 3 Chemistry Class 12 Notes Question 10.
Assertion – bond dissociation energy of fluorine is greater than chlorine gas
Reason – chlorine has more electronic repulsion than fluorine
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d) Both assertion and reason are false. The converse is true.

Samacheer Kalvi 12th Chemistry Solutions Question 11.
Among the following, which is the strongest oxidizing agent?
(a) Cl2
(b) F2
(c) Br2
(d) I2
Answer:
(b) F2

Class 12 Chemistry Samacheer Kalvi Question 12.
The correct order of the thermal stability of hydrogen halide is …………
(a) HI > HBr > HCl > HF
(b) HF > HCl > HBr > HI
(c) HCl > HF > HBr > HI
(d) HI > HCl > HF > HBr
Answer:
(b) HF > HCl > HBr > HI

Samacheer Kalvi 12 Chemistry Solutions Question 13.
Which one of the following compounds is not formed?
(a) XeOF4
(b) XeO3
(c) XeF2
(d) NeF2
Answer:
(d) NeF2

Class 12 Chemistry Chapter 3 Notes Pdf Download Question 14.
Most easily liquefiable gas is …………
(a) Ar
(b) Ne
(c) He
(d) Kr
Answer:
(c) He

12th Chemistry Chapter 7 Book Back Answers Question 15.
XeF6 on complete hydrolysis produces …………
(a) XeOF4
(b) XeO2F4
(c) XeO3
(d) XeO2
Answer:
(c) XeO3

Question 16.
On oxidation with iodine, sulphite ion is transformed to …………
(a) S4\({ O }_{ 6 }^{ 2- }\)
(b) S2\({ O }_{ 6 }^{ 2- }\)
(c) S\({ O }_{ 4 }^{ 2- }\)
(d) S\({ O }_{ 3 }^{ 2- }\)
Answer:
(c) S\({ O }_{ 4 }^{ 2- }\)

Question 17.
Which of the following is strongest acid among all?
(a) HI
(b) HF
(c) HBr
(d) HCl
Answer:
(a) HI

Question 18.
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
(a) Br2 > I2 > F2 > Cl2
(b) F2 > Cl2 > Br2 > I2
(c) I2 > Br2 > Cl2 > F2
(d) Cl2 > Br2 > F2 > I2
Answer:
(d) Cl2 > Br2 > F2 > I2

Question 19.
Among the following the correct order of acidity is …………
(a) HClO2 < HCIO < HClO3 < HClO4
(b) HClO4 < HClO2 < HCIO < HClO3
(c) HClO3 < HClO4 < HClO2 < HCIO
(d) HCIO < HClO2 < HClO3 < HClO4
Answer:
(d) HCIO < HClO2 < HClO3 < HClO4

Question 20.
When copper is heated with cone HNO3 it produces …………
(a) CU(NO3)2 , NO and NO2
(b) Cu(NO3)2 and N2O
(c) CU(NO3)2 and NO2
(d) Cu(NO3)2 and NO
Answer:
(c) CU(NO3)2 and NO2

II. Answer the following questions:

Question 1.
What is inert pair effect?
Answer:
In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

Question 2.
Chalcogens belongs to p-block. Give reason.
Answer:
Chalcogens are ore forming elements. Most of the ores are oxides and sulphides, therefore oxygen, sulphur and other group 16 elements are called Chalcogens. In O, S, Se, Te and Po last electron enters to p-orbital. Therefore Chalcogens belongs to p-block.

Question 3.
Explain why fluorine always exhibit an oxidation state of -1?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibit an oxidation state of-1 and others in halogen family shows +1, +3, +5 and +7 oxidation states.

Question 4.
Give the oxidation state of halogen in the following.

  1. OF2
  2. O2F2
  3. Cl2O3
  4. I2O4

Answer:
1. OF2
+2 + 2(x) = 0
+2 = -2x
2x = -2
x =-1

2. O2F2
2(+1) + 2x = 0
2x = – 2
x = – 1

3. Cl2O3
2(x) + 3(-2) =0
2x = +6
x = +3

4. I2O4
2(x) + 4(-2) =0
2x = +8
x = +4

Question 5.
What are interhalogen compounds? Give examples.
Answer:
Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide (F2O) and difluorine dioxide (F2O2).

Question 6.
Why fluorine is more reactive than other halogens?
Answer:
Fluorine is the most reactive element among halogen. This is due to the minimum value of F – F bond dissociation energy. Hence fluorine is more reactive than other halogens.

Question 7.
Give the uses of helium.
Answer:

  1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
  2. Helium is used to provide inert atmosphere in electric arc welding of metals
  3. Helium has lowest boiling point hence used in cryogenics (low temperature science).
  4. It is much less denser than air and hence used for filling air balloons

Question 8.
What is the hybridisation of iodine in IF7? Give its structure.
Answer:
Hybridisation of iodine in IF7 is sp3d3 Structure of IF7 is pentagonal bipyramidal.
12th Chemistry Chapter 3 Book Back Answers P-Block Elements - II Samacheer Kalvi

Question 9.
Give the balanced equation for the reaction between chlorine with cold NaOH and hot NaOH.
Answer:
1. Reaction between chlorine with cold NaOH:
Cl2+ H2O → HCl + HOCl
HCl + NaOH → NaCl + H2O
HOCl + NaOH → NaOCl + H2O
Overall reaction
Samacheer Kalvi Guru 12th Chemistry Solutions Chapter 3 P-Block Elements - II
Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite.

2. Reaction between chlorine with hot NaOH:
Cl2 + H2O → HCl + HOCl
HCl + NaOH → NaCl + H2O
HOCl + NaOH → NaOCl + H2O
3NaOCl → NaClO3+ 2NaCl
Overall reaction
12th Chemistry 3rd Lesson Book Back Answers P-Block Elements - II Samacheer Kalvi
Chlorine reacts with hot NaOH to give sodium chlorate and sodium chloride.

Question 10.
How will you prepare chlorine in the laboratory?
Answer:
1. Chlorine is prepared by the action of cone, sulphuric acid on chlorides in presence of manganese dioxide.
4NaCl + MnO2 + 4H2SO4 → Cl2 + MnCl2 + 4NaHSO4 + 2H2O

2. It can also be prepared by oxidising hydrochloric acid using various oxidising agents such as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
PbO2 + 4HCl → PbCl2 + 2H2O + Cl2
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
2KMnO4 + 16HCl → 2KCl + 2MnCl + 8H2O + 5Cl2
K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2

3. When bleaching powder is treated with mineral acids chlorine is liberated
CaOCl2 + 2HCl → CaCl2 + H2O + Cl2
CaOCl2 + H2SO4 → CaSO4 + H2O + Cl2

Question 11.
Give the uses of sulphuric acid.
Answer:

  1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc.
  2. It is used as a drying agent and also used in the preparation of pigments, explosives etc.

Question 12.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono ( H2SO4. H2O ) and di ( H2SO4. 2H2O ) hydrates and the reaction is exothermic. The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.
Samacheerkalvi.Guru 12th Chemistry Solutions Chapter 3 P-Block Elements - II

Question 13.
Write the reason for the anamolous behaviour of Nitrogen.
Answer:
1. Due to its small size, high electro negativity, high ionisation enthalpy and absence of d-orbitals.

2. N, has a unique ability to form pπ – pπ multiple bond whereas the heavier members of this group (15) do not form pπ – pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ – pπ bond due to the absence of d – orbitals whereas other elements can.

Question 14.
Write the molecular formula and structural formula for the following molecules.
(a) Nitric acid
(b) dinitrogen pentoxide
(c) phosphoric acid
(d) phosphine
Answer:
Samacheer Kalvi Guru Class 12 Chemistry Solutions Chapter 3 P-Block Elements - II

Question 15.
Give the uses of argon.
Answer:
Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

Question 16.
Write the valence shell electronic configuration of group-15 elements.
Answer:
General electronic configuration of group 15 elements are ns2np3.

  1. Nitrogen – [He] 2s2 2p3
  2. Phosphorous – [Ne] 3s2 3p3
  3. Arsenic – [Ar] 3d10 4s2 4p3
  4. Antimony – [Kr] 4d10 5s2 5p3
  5. Bismuth – [Ne] 4f14 5s10 6s2 6p3

Question 17.
Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
1. Phosphine reacts with halogens to give phosphorous penta halides.
PH3 + 4Cl2 → PCl5 + 3HCl

2. Phosphine forms coordination compound with lewis acids such as boron trichloride.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 P-Block Elements - II

3. Phosphine precipitates some metal from their salt solutions.
3AgNO3 + PH3 → Ag3P + 3HNO3

Question 18.
Give a reaction between nitric acid and a basic oxide.
Answer:
Nitric acid reacts with bases and basic oxides to form salts and water.

  1. ZnO + 2HNO3 → Zn(NO3)2 + H2O
  2. 3FeO + 10HNO3 → 3Fe(NO3)3 + NO + 5H2O

Question 19.
What happens when PCl5 is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.
PCl5 \(\underrightarrow { \triangle }\) PCl3 + Cl2

Question 20.
Suggest a reason why HF is a weak acid, whereas binary acids of the all other halogens are strong acids.
Answer:
The hydrogen halides are extremely soluble in water due to the ionisation.
X + H2O → H3O+ + X
( X = F, Cl, Br or I )

Solutions of hydrogen halides are therefore acidic and known as hydrohalic acids. Hydrochloric, hydrobromic and hydroiodic acids are almost completely ionised and are therefore strong acids but HF is a weak acid. For HF,

  • HF + H2O \(\rightleftharpoons\) H3O+ + F
  • HF + F → \({ HF }_{ 2 }^{ – }\)

At high concentration, the equilibrium involves the removal of fluoride ions is important. Since it affects the dissociation of hydrogen fluoride, therefore it is a weak acid.

Question 21.
Deduce the oxidation number of oxygen in hypofluorous acid – HOF.
Answer:
In case of O – F bond is HOF, fluorine is most electronegative element. So its oxidation number is -1. Thereby oxidation number of O is +1. Similarly in case of O – H bond is HOF. O is highly electronegative than H. So its oxidation number is -1 and oxidation number of H is +1. So, Net oxidation of oxygen is – 1 + 1 = 0.
Chemistry Class 12 Samacheer Kalvi Solutions Chapter 3 P-Block Elements - II

Question 22.
What type of hybridisation occur in

  1. BrF5
  2. BrF3

Answer:
1. BrF5
BrF5 is a AX5 type. Therefore is has sp3d2 hybridisation. Hence, BrF5 molecule has square pyramidal shape.

2. BrF3
BrF3 is a AX3 type. Therefore it has sp3d hybridisation. Hence, BrF3 molecule has T-shape.

Question 23.
Complete the following reactions.
Answer:
Samacheer Kalvi Class 12 Chemistry Solutions Chapter 3 P-Block Elements - II

Samacheer Kalvi 12th Chemistry p-Block Elements – II Evaluate yourself

Question 1.
Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.
Answer:
Samacheer Kalvi Guru Chemistry Solutions Chapter 3 P-Block Elements - II

Samacheer Kalvi 12th Chemistry p-Block Elements – II Additional Questions

Samacheer Kalvi 12th Chemistry p-Block Elements – II 1 Mark Questions and Answers

I. Choose the best answer.

Question 1.
About 78% of earth atmosphere contains,…………
(a) P
(b) As
(c) N
(d) Bi
Answer:
(c) N

Question 2.
Which one of the following is not a pnictogens?
(a) Nitrogen
(b) Oxygen
(c) Phosphorous
(d) Antimony
Answer:
(b) Oxygen

Question 3.
Which one of the following shows isotopes?
(a) Nitrogen
(b) Arsenic
(c) Antimony
(d) Bismuth
Answer:
(a) Nitrogen

Question 4.
Nitrogen gas in atmosphere is separated industrially from liquid air by …………
(a) simple distillation
(b) Fractional distillation
(c) Sublimation
(d) Distillation under reduced pressure
Answer:
(b) Fractional distillation

Question 5.
Bond order for nitrogen molecule is …………
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(c) 3

Question 6.
Nitrogen gas is …………
(a) Inert
(b) Noble
(c) More reactive
(d) Less reactive
Answer:
(a) Inert

Question 7.
Which one of the following is used in cryosurgery?
(a) Liq N2
(b) Liq NH3
(c) Liq Na
(d) Liq H2
Answer:
(a) Liq N2

Question 8.
The dielectric constant of ammonia is (K) …………
(a) 10-30
(b) 10-14
(c) 1030
(d) 1014
Answer:
(a) 10-30

Question 9.
When ammonia reacts with copper sulphate solution to give complex, the colour of complex is …………
(a) violet
(b) deep blue
(c) blue
(d) Red
Answer:
(b) deep blue

Question 10.
H – N – H bond angle in NH3 is …………
(a) 109° 28’
(b) 107° 28’
(c) 104°
(d) 107°
Answer:
(d) 107°

Question 11.
Shape of ammonia is …………
(a) Planar
(b) Square planar
(c) Pyramidal
(d) Square pyramidal
Answer:
(c) Pyramidal

Question 12.
Nitric acid prepared in large scales using …………
(a) Ostwald’s process
(b) Haber’s process
(c) Contact process
(d) Deacon’s process
Answer:
(a) Ostwald’s process

Question 13.
Benzene undergoes nitration reaction to form nitrobenzene in this reaction takes place due to the formation of …………
(a) Hydronium ion
(b) Hydride ion
(c) Nitronium ion
(d) Nitrasonium ion
Answer:
(c) Nitronium ion

Question 14.
Oxidation state of N m FINO3 is…………
(a) ±2
(b) +3
(c) +4
(d) +5
Answer:
(d) +5

Question 15.
Compound used in photography is …………
(a) AgNO3
(b) AgBr
(c) AgCl
(d) AgI
Answer:
(a) AgNO3

Question 16.
Sodium nitrate
(a) Photography
(b) Firearms
(c) Royal water Sosurgerv …………
(d) Cryosurgery
Answer:
(b) Firearms

Question 17.
Nitrogen sesquoxide colour is …………
(a) colourless
(b) Brown
(c) Blue
(d) Red
Answer:
(c) Blue

Question 18.
White phosphorous is also called as …………
(a) Red phosphorous
(b) Black phosphorous
(c) Scarlet phosphorous
(d) Yellow phosphorous
Answer:
(d) Yellow phosphorous

Question 19.
White (Yellow) phosphorous glows in the dark due to oxidation which is called …………
(a) phosphorescence
(b) phosphorus
(c) Fluorescence
(d) Liminoscence
Answer:
(a) phosphorescence

Question 20.
Yellow phosphorous reacts with alkali on boiling in an inert atmosphere liberates …………
(a) Phosphorous acid
(b) Phosphoric acid
(c) Phosphine
(d) Pyrophosphoric acid
Answer:
(c) Phosphine

Question 21.
Consider the following statements.
(i) phosphine is the most important hydride of phosphorous
(ii) phosphine is a poisonous gas with rotten egg smell.
(iii) phosphine is a powerful reducing agent

Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Ansswer:
(c) (i) and (iii)

Question 22.
When phosphine is heated with air it bums to gives …………
(a) Orthophosphoric acid
(b) Metaphosphoric acid
(c) Pyrophosphoric acid
(d) Phosphoroustrioxide
Answer:
(b) Metaphosphoric acid

Question 23.
Hybridisation of P in phosphine is …………
(a) sp3d
(b) sp3d2
(c) sp3d3
(d) sp3
Answer:
(d) sp3

Question 24.
Compounds used in Holme’s signal are …………
(a) Phosphine + Acetylene
(b) H3PO3+H3PO3
(c) Calcium carbide + calcium phosphide
(d) Calcium carbonate + calcium phosphate
Answer:
(c) Calcium carbide + calcium phosphide

Question 25.
Chalgogens are also called as …………
(a) Ore forming elements
(b) Group-16 elements
(c) group 17 elements
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 26.
Element present in the volcanic ashes is …………
(a) Oxygen
(b) Sulphur
(c) Selenium
(d) Tellurium
Answer:
(b) Sulphur

Question 27.
The decomposition of potassium chlorate speed up in the presence of …………
(a) MnO2
(h) Mn3O4
(c) MnSO4
(d) KMnO4
Answer:
(a) MnO2

Question 28.
Pure ozone is …………
(a) yellow gas
(b) blue gas
(c) Pale blue gas
(d) bright blue gas
Answer:
(c) Pale blue gas

Question 29.
Shape of ozone …………
(a) V-shape
(b) Linear shape
(c) bent shape
(d) spherical shape
Answer:
(c) bent shape

Question 30.
The rate of decompositon of ozone drops sharply in …………
(a) acidic medium
(b) alkaline medium
(c) neutral medium
(d) Ether medium
Answer:
(b) alkaline medium

Question 31.
Which one of the following used as fuel in rockets?
(a) Liq O2
(b) Liq CO2
(c) Liq N2
(d) Liq He – O2
Answer:
(a) Liq O2

Question 32.
Find out crystalline allotrophic form of sulphur?
(a) γ – sulphur
(b) λ – sulphur
(c) α – sulphur
(d) milk of sulphur
Answer:
(c) α – sulphur

Question 33.
Consider the following statements
(i) α – sulphur is the only thermodynamically stable allotrophic form.
(ii) At 140 ° C the mono clinic sulphur melts to form mobile pale yellow liquid called γ – sulphur
(iii) Monoclinic sulphur is stable between 96°C-119°C and slowly changes into λ- sulphur

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii)
Answer:
(d) (ii) and (iii)

Question 34.
Sulphur di oxide, how many times heavier than air?
(a) 2 times
(b) 2.5 times
(c) 2.2 times
(d) 2.3 times
Answer:
(c) 2.2 times

Question 35.
Which one of following has temporary bleaching action?
(a) Chlorine
(b) SO3
(c) H3SO4
(d) SO2
Answer:
(d) SO2

Question 36.
Sulphuric acid can be manufactured by …………
(a) Ostwald’s process
(b) Lead chamber process
(c) Deacon’s process
(d) Haber’s process
Answer:
(b) Lead chamber process

Question 37.
Sulphuric acid is manufactured by contact process, catalyst used in contact process is …………
(a) V2O5
(b) TiCl4
(c) Fe
(d) Mo
Answer:
(c) V2O5

Question 38.
Benzene reacts with sulphuric acid to gives …………
(a) sulphate
(b) sulphide
(c) sulphonic acid
(d) sulphite
Answer:
(c) sulphonic acid

Question 39.
Reagent used to detect sulphate ion is …………
(a) BaCl2
(b) BaSO3
(c) (CH,COO),Pb
(d) both (a) and (c)
Answwer:
(d) both (a) and (c)

Question 40.
Deacon’s process is used to manufacture …………
(a) Cl2
(b) F2
(C) Br
(d) I2
Answer:
(a) Cl2

Question 41.
Catalyst used in Deacon’s process is …………
(a) CuCl2
(b) Cu2Cl2
(c) CuBr
(d) Cu2Br2
Answer:
(b) Cu2Cl2

Question 42.
C10H16 + 8C12 \(\underrightarrow { \triangle }\) A. Identify A?
(a) Methane
(b) Ethane
(c) Carbon
(d) Propane
Answer:
(c) Carbon

Question 43.
Passing chlorine gas through dry slaked lime to produce …………
(a) CaOCl
(b) CaOCl2
(c) CaO
(d) CaCl2
Answer:
(b) CaOCl2

Question 44.
Which one of the following is used for purification of drinking water?
(a) SO3
(b) SO2
(c) Br2 / H2O
(d) Cl2
Answer:
(d) Cl2

Question 45.
Which one of the following is a weak acid?
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer:
(a) HF

Question 46.
Reagent not stored in glass bottles?
(a) HCI
(b) HBr
(c) HF
(d) HI
Answer:
(c) HF

Question 47.
More reactive element is
(a) Fluorine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer:
(a) Fluorine

Question 48.
The correct order of the acidity of hydrohalic acids?
(a) HF > HCI > HBr > HI
(b) HCI >HF >HBr >HI
(c) HBr > HCI >HF > HI
(d) HI > HBr > HCI > HF
Answer:
(d) HI > HBr > HCI > HF

Question 49.
Consider the following statements
(i) In interhalogen compounds, the central atom will be the smaller one.
(ii) It can be formed only between two halogen and not more than two halogens.
(iii) They are strong reducing agents.

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iii) only
Answer:
(c) (i) and (iii)

Question 50.
Shape of ClF3 is …………
(a) Linear
(b) T-shape
(c) Pyrimidal
(d) Square planar
Answer:
(b) T-shape

Question 51.
Which one of the following is more acidic?
(a) HOCl
(b) HCIO2
(c) HClO3
(d) HClO4
Answer:
(d) HClO4

Question 52.
When XeF6 reacts with 2.5 M NaOH gives …………
(a) Na4XeO6
(b) Na2XeO3
(c) XeO2F2
(d) XeO3
Answer:
(a) Na4XeO6

Question 53.
Shape of XeF6 is …………
(a) Octahedron
(b) Distorted octahedron
(c) Pyramidal
(d) Tetrahedron
Answer:
(b) Distorted octahedron

Question 54.
Which one of the following can penetrate through dense fog?
(a) He
(b) Ne
(c) Kr
(d) Rn
Answer:
(c) Kr

Question 55.
Find out radioactive element?
(a) He
(b) Rn
(c) Xe
(d) Ar
Answer:
(b) Rn

II. Fill in the blanks:

  1. The 11th most abundant element is ………….
  2. …………. is the principle gas of atmosphere.
  3. Nitrogen is chemically ………….
  4. …………. process for the synthesis of ammonia.
  5. …………. is used for the manufacture of calcium cyanamide
  6. …………. is a pungent smelling gas.
  7. Ammonia acts as a …………. agent.
  8. With excess of chlorine, ammonia reacts to give …………. an explosive substance.
  9. When excess ammonia is added to aqueous solution of copper sulphate …………. colour compound is formed.
  10. Pure nitric acid becomes …………. on standing.
  11. …………. is used in gunpower for firearms.
  12. The decomposition of ammonium nitrate gives ………….
  13. White phosphorous is colourless but becomes pale yellow due to formation of a …………. upon standing.
  14. The white phosphorous can be changed into …………. by heated it to 420°C in the absence of air and light.
  15. …………. reacts with alkali on boiling in an inert atmosphere liberating phosphine.
  16. …………. is used in the match boxes.
  17. Phosphine is …………. smelling gas.
  18. Phosphine has a …………. shape.
  19. …………. is used for producing smoke screen.
  20. When phosphorous trichloridie is hydrolysed with cold water it gives ………….
  21. Elements belonging group 16 are called ………….
  22. Under ordinary condition oxygen exists as a …………. gas.
  23. Allotrophic form of oxygen is …………. and ………….
  24. Pure ozone is …………. gas.
  25. …………. is used in welding purpose.
  26. Monoclinic sulphur is stable between 96° and 119°C and slowly changes into ………….
  27. …………. gas is found in volcanic eruptions.
  28. A large amount of …………. gas is released into atmosphere from plants using coal and oil and copper melting plants.
  29. Sulphurdioxide gas has …………. odour.
  30. Sulphurdioxide can be used for …………. and …………. in agriculture.
  31. In SO3 S-atom undergoes …………. hybridisation.
  32. In SO3 a double bond arises between S and O is due to …………. overlapping.
  33. High boiling point and viscosity of sulphuric acid is due to ………….
  34. …………. is used as a drying agent.
  35. The main source of fluorine is ………….
  36. The main source of chlorine is ………….
  37. Chlorine is a …………. gas.
  38. Chlorine is soluble in water and its solution is referred as ………….
  39. …………. is produced by passing chlorine gas through dry slaked lime.
  40. …………. is used in extraction of gold and platinum.
  41. …………. is used for extraction of glue from bone.
  42. At room temperature, hydrogen halides are gases but …………. can be readily liquefied.
  43. Liberation of iodine which gives a …………. colouration with starch.
  44. Each halogen combines with other halogens to form a series of compounds called ………….
  45. Structure of AX7 type is ………….
  46. Oxidation state of Cl in HClO4 is ………….
  47. Noble gases have the …………. ionisation energy.
  48. Xenon reacts with PtF6 and gave an ………….
  49. Kr and fluorine gases are irradiated with SbF it forms ………….
  50. Shape of XeOF4 is ………….
  51. Helium used for filling air ………….
  52. …………. is used in fluorescent bulbs.
  53. …………. is used in high speed electronic flash bulbs.
  54. Radon is a source of …………. rays.
  55. …………. is formed by the hydrolysis of urea.
  56. …………. element subtimes at 889 K.
  57. Yellow phosphorus has a characteristics …………. smell.

Answers:

  1. phosporous
  2. Nitrogen
  3. Inert
  4. Haber’s
  5. Nitrogen
  6. Ammonia
  7. Reducing
  8. Nitrogen trichloride
  9. Deep blue
  10. Yellow
  11. Nitric acid / NaNO
  12. Nitrous oxide
  13. Layer of red phosphorous
  14. Red phosphorous
  15. White phosphorous
  16. Red phosporous
  17. Rotten fish
  18. Pyramidal
  19. Phosphine
  20. phosporous acid
  21. Chalgogens
  22. diatomic
  23. dioxygen and ozone
  24. pale blue
  25. Oxyacetylene
  26. Rhombic sulphur
  27. Sulphurdioxide
  28. SO2
  29. Suffocating
  30. disinfecting crops and plants
  31. sp2
  32. pπ – dπ
  33. Hydrogen bonding
  34. Sulphuric acid
  35. Fluorite
  36. Sodium chloride
  37. Green yellow
  38. Chlorine water
  39. Bleaching powder
  40. Chlorine
  41. Hydrochloric acid
  42. hydrogen fluroide
  43. blue-black
  44. Inter-halogen compounds
  45. pentagonal bipyramidal
  46. +7
  47. largest
  48. orange yellow solid [XePtF6]
  49. KrF2. 2SbF3
  50. Square pyramidal
  51. Balloons
  52. Krypton
  53. Xenon
  54. Gamma
  55. Ammonia
  56. Arsenic
  57. Garlic

II. Match the following:

Question 1.
(i) Haber’s process – (a) HNO3
(ii) Deacon’s process – (b) Ammonia
(iii) Contact process – (c) Chlorine
(iv) Ostwald’s process – (d) H2SO4
Answer:
(i) b
(ii) c
(iii) d
(iv) a

Question 2.
(i) Nitric acid – (a) Purification of bone black
(ii) HCl – (b) Photography
(iii) White (yellow) phosphorous – (c) Rotten fish smell
(iv) Phosphine – (d) Phosphorescence
Answer:
(i) b
(ii) a
(iii) d
(iv) c

Question 3.
(i) Nitrogen sesquoxide – (a) H2N2O2
(ii) Nitrous oxide – (b) H4N2O4
(iii) Hyponitrous acid – (c) N2O
(iv) Hydronitrous acid – (d) N2O3
Answer:
(i) d
(ii) c
(iii) a
(iv) b

Question 4.
(i) N2O – (a) +5
(ii) N2O4 – (b) +3
(iii) N2O5 – (c) +1
(iv) N2O3 – (d) +4
Answer:
(i) c
(ii) d
(iii) a
(iv) b

Question 5.
(i) White phosphorous – (a) Volcanic eruptions
(ii) Red phosphorous – (b) Yellow phosphorous
(iii) Phosphine – (c) Match boxes
(iv) SO2 – (d) smoke screen
Answer:
(i) b
(ii) c
(iii) d
(iv) a

Question 6.
(i) ammonia – (a) suffocating odour
(ii) SO2 – (b) Rotten fish smell
(iii) PH3 – (c) Greenish yellow gas
(iv) Cl2 – (d) pungent smelling gas
Answer:
(i) d
(ii) a
(iii) b
(iv) c

Question 7.
(i) XeF4 – (a) sp3
(ii) XeOF2 – (b) sp3d2
(iii) XeO3 – (c) sp3d3
(iv) XeF – (d) sp3d
Answer:
(i) b
(ii) d
(iii) a
(iv) c

Question 8.
(i) Helium – (a) flash bulbs
(ii) Neon – (b) radioactive
(iii) Krypion – (c) air balloons
(iv) Radon – (d) Brilliant red
Answer:
(i) c
(ii) d
(iii) a
(iv) b

IV. Assertion and Reason

Question 1.
Assertion (A) – Xenon is used in high speed electronic flash bulbs used by photographers.
Reason (R) – Xenon emits an intense light in discharge tubes instantly.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 2.
Assertion (A) – Noble gases have the largest ionisation energy compared to any other elements.
Reason (R) – Noble gases have incomplete filled orbital.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(c) A is correct but R is wrong

Question 3.
Assertion (A) – Hydrogen iodide decomposes at 400<sup>0</sup>C while hydrogen fluoride and hydrogen chloride are stable at this temperature.
Reason (R) – Thermal stability of hydrogen halides decreases from fluoride to iodide.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 4.
Assertion (A) – The bleaching of chlorine is temporary.
Reason (R) – Chlorine oxidises ferrous salts to ferric salts.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(d) A is wrong but R is correct

Question 5.
Assertion (A) – Sulphuric acid is highly reactive.
Reason (R) – Sulphuric acid can act as strong acid and an oxidising agent.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct but doesn’t explains A

Question 6.
Assertion (A) – Sulphuric acid is a high boiling point and viscous liquid.
Reason (R) – This is due to the association of molecules together through hydrogen bonding.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 7.
Assertion (A) – Monoclinic sulphur is less stable than rhomobic sulphur.
Reason (R) – Monoclinic sulphur is stable between 96°C – 119°C and slowly changes into rhombic sulphur.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 8.
Assertion (A) – Nitrogen gas is chemically inert.
Reason (R) – Nitrogen has low bonding energy.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(c) A is correct but R is wrong

V. Find the odd one out
Question 1.
(a) NO
(b) HNO3
(C) NO2
(d) N2O
Answer:
(b) HNO3
Hint: HNO3 is acid and other are oxides.

Question 2.
(a) Nitrous acid
(b) Nitric acid
(c) Hyponitrous acid
(d) Pemitrous acid
Answer:
(d) Pemitrous acid
Hint: Pemitrous acid contains peroxide linkage others doesn’t have peroxide linkage.

Question 3.
(a) White phosphorous
(b) Red phosphorous
(c) phosphorous pentaoxide
(d) black phosphorous
Answer:
(c) phosphorous pentaoxide
Hint: P2O5 is a compound of phosphorous and others are allotropic form of phosphorous.

Question 4.
(a) PH3
(b) HPO3
(c) H3PO3
(d) H3PO4
Answer:
(a) PH3
Hint: PH3 is hydrides of phosphorous and others are oxo acids of phosphorous.

Question 5.
(a) He
(b) Ne
(c) Ar
(d) Xe
Answer:
(d) Xe
Hint: Xe forms several chemical compounds than others.

VI. Find out the correct pair.

Question 1.
(a) Helium – filament bulbs
(b) Krypton – prevent bonds
(c) Xenon – Lasers
(d) Radon – flash bulbs
Answer:
(c) Xenon – Lasers

Question 2.
(a) Ra – gamma rays
(b) Xe – cancer growth
(c) Ne – balloons
(d) Kr – advertisement bulb
Answer:
(a) Ra – gamma rays

Question 3.
(a) ClF3 – Linear
(b) BrF5 – T Shaped
(c) IF4 – square pyrimidal
(d) BrF5 – square pyrimidal
Answer:
(d) BrF5 – square pyrimidal

Question 4.
(a) XeOF2 – sp3
(b) XeF6 – sp3d3
(c) XeF4 – sp3
(d) XeOF4 – sp3d
Answer:
(b) XeF6 – sp3d3

Question 5.
(a) OF2 = -1
(b) Cl4O4 = -1
(c) I2O4 = -1
(d) I2O9 = -1
Answer:
(a) OF2 = -1

Question 6.
(a) Royal water – Dissolving gold
(b) Chlorine – Entraction of glue from bone
(c) HCl- Extraction of gold
(d) Chlorine – Temporary bleaching
Answer:
(a) Royal water – Dissolving gold

Question 7.
(a) O – 2s22p3
(b) S – 3s23p4
(c) Se – 4d105s25p4
(d) Te – 3d104s24p4
Answer:
(b) S – 3s23p4

VI.Find out the incorrect pair.

Question 1.
(a) Nitrogen gas – inert
(b) Ammonia – pungent smelling gas
(c) Nitric acid – oxidizing agent
(d) Phosphine – rotten egg smell
Answer:
(d) Phosphine – rotten egg smell

Question 2.
(a) Liquid nitrogen – biological preservation
(b) Nitric acid – photography
(c) white phosphorous – yellow phosphorous
(d) phosphorous – welding
Answer:
(d) phosphorous – welding

Question 3.
(a) N2O = +1
(b) N2O = +2
(c) N2O3 = +5
(d) NO2 = +4
Answer:
(c) N2O3 = +5

Question 4.
(a) Hvponitrous acid – N2O
(b) Nitrous acid – HNO2
(c) pernitric acid – HNO4
(d) pernitrous acid – HOONO
Answer:
(a) Hvponitrous acid – N2O

Question 5.
(a) PH3 – Holme’s signal
(b) O2 – welding
(c) H2SO4 – Disinfecting crops
(d) SO3 – Bleaching hair
Answer:
(c) H2SO4 – Disinfecting crops

Question 6.
(a) H2SO4 drying agent
(b) Chlorine – Deacon’s process
(c) HCI – purification of bone black
(d) Helium – flash bulbs
Answer:
(d) Helium – flash bulbs

Question 7.
(a) ICl – Linear
(b) CIF3 – T shape
(c) IF5 – pentagonal bipyramidal
(d) IF7 – pentagonal bipyramidal
Answer:
(c) IF5 – pentagonal bipyramidal

Question 8.
(a) HOCl = +2
(b) HOCl = +3
(e) HOCI = +5
(d) HOCl4 = +7
Answer:
(a) HOCl = +2

Question 9.
(a) XeF4 – sp3d3
(b) XeF6 – sp3d3
(e) XeOF4 – sp3d2
(d) XeO3 – sp3d
Answer:
(d) XeO3 – sp3d

Question 10.
(a) He – cryogenics
(b) Ne – advertisement
(c) Kr – flurescent bulbs
(d) Ra – Lasers.
Answer:
(d) Ra – Lasers.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 2 Mark Questions and Answers

Question  1.
What are pnictogens?
Answer:
The group – 15 elements like nitrogen, phosphorous Arsenic, Antimony and Bismuth are collectively called as pnictogens. Their general outer electronic configuration is ns2np3.

Question 2.
What happen when sodium azide undergoes thermal decomposition?
Answer:
Pure nitrogen gas can be obtained by the thermal decomposition of sodium azide about 575K
2NaN3 \(\underrightarrow { 575K }\) 2Na + 3N2

Question 3.
How will you prepare ammonia from nitrogen? and mention the name of process?
Answer:
Nitrogen directly reacts with hydrogen gives ammonia. This reaction is favoured by high pressures and at optimum temperature in the presence of iron catalyst,
N2 + 3H3 \(\rightleftharpoons\) 2NH3
This process is called as Haber’s process.

Question 4.
Why nitrogen gas is chemically inert?
Answer:
The chemically inert character of nitrogen is largely due to high bonding energy of the molecules 225 cal mol-1 .Interestingly the triply bonded species is notable for its less reactivity in comparison with other iso-electronic triply bonded systems such as -C \(\equiv\) C – C \(\equiv\) Q, X- C \(\equiv\) N, etc.

Question 5.
Mention the uses of nitrogen?
Answer:

  1. Nitrogen is used for the manufacture of ammonia, nitric acid and calcium cyanamide etc.
  2. Liquid nitrogen is used for producing low temperature required in cryosurgery, and so in biological preservation

Question 6.
Explain the action of heat on ammonia.
Answer:
Above 500°C ammonia decomposes into its elements. The decomposition may be accelerated by metallic catalysts like Nickel, Iron. Almost complete dissociation occurs on continuous sparking.
2NH3 \(\underrightarrow { { 500 }^{ 0 }C } \) N2 + 3H2

Question 7.
Prove ammonia act as a reducing agent?
Answer:
Ammonia reduces the metal oxides to metal when passed over heated metallic oxide.
3PbO + 2NH3 → 3Pb + N2 + 3H2O

Question 8.
What happen when copper sulphate reacts with ammonia?
Answer:
When excess ammonia is added to aqueous solution copper sulphate a deep blue colour Complex [Cu(NH3)4 ]2+ is formed.
Chapter 3 Chemistry Class 12 Notes P-Block Elements - II Samacheer Kalvi

Question 9.
Pure nitric acid is colourless, on standing it becomes yellow. Justify your answer.
Answer:
Nitric acid decomposes on exposure to sunlight or on being heated, into nitrogen dioxide, water and oxygen.
4HNO4 → 4NO2 + 2H2O + O2
Due to this reaction pure acid or its concentrated solution becomes yellow on standing.

Question 10.
Write the products formed in the reaction of nitric acid with dilute and concentrated with magnesium.
Answer:
1. Magnesium with cone. HNO3:
4Mg + 10HNO3 → 4Mg(NO3)2 + NH4NO3 + 3H2O
Magnesium reacts with concentraated nitric acid to gives both magnesium nitrate and ammonium nitrate.

2. Magnesium with dil HNO3:
Magnesium reacts with dilute nitric acid to gives both magnesium nitrate and nitrous oxide.
4Mg + 10HNO3 → 4 Mg(NO3)2 + N2O + 5H2O

Question 11.
Give the uses of nitric acid.
Answer:

  1. Nitric acid is used as a oxidising agent and in the preparation of aquaregia.
  2. 4. Salts of nitric acid are used in photography (AgNO3) and gunpowder for firearms. (NaNO3) .

Question 12.
How will you prepare nitric oxide from sodium nitrite?
Answer:
When sodium nitrite reacts with ferrous sulphate in the presence of sulphuric acid to gives nitric oxide.
2NaNO2 + 2FeSO4 + 3H2S → Fe2(SO4)3 + 2NaHSO4 + 2H2O + 2NO

Question 13.
How will you prepare nitrogen pentoxide?
Answer:
Nitric acid reacts with phosphorous pentaoxide to give nitrogen pentoxide.
2HNO3 + P2O5 → N2O5 + 2HPO3

Question 14.
Draw the structure of

  1. N2O
  2. N2O3

Answer:
1. N2O (Nitrous oxide)
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 P-Block Elements - II

2. N2O3 (Nitrogen sesquoxide)
Class 12 Chemistry Samacheer Kalvi Chapter 3 P-Block Elements - II

Question 15.
Draw- the structure of

  1. N2O4
  2. N2O3

Answer:
1. N2O4 (Nitrogen tetraoxide)
Samacheer Kalvi 12 Chemistry Solutions Chapter 3 P-Block Elements - II
2. N2O5
Class 12 Chemistry Chapter 3 Notes Pdf Download P-Block Elements - II Samacheer Kalvi

Question 16.
Draw the structure of

  1. Hyponitrous acid
  2. Hydronitrous acid.

Answer:
(a) Hypon trous acid – H2N2O2
HO – N = H – OH

(b) Hydronitrous acid:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-16

Question 17.
Mention the allotropic forms of phosphorous?
Answer:
The most common allotropic forms of phosphorous

  1. white phosphorous
  2. Red phosphorous
  3. Black phosphorous

Question 18.
Why white phosphorous is also known as yellow phosphorous?
Answer:
The freshly prepared white phosphorus is colourless but becomes pale yellow due to formation of a layer of red phosphorus upon standing. Hence it is also known as yellow phosphorus.

Question 19.
What is phosphorescence?
Answer:
White (yellow) phosphorous glows in the dark due to oxidation which is called phosphorescence.

Question 20.
Why white phosphorous undergoes spontaneous combustion in air?
Answer:
White phosphorous ignition temperature is very low and hence it undergoes spontaneous combustion in air at room temperature and during in the combusition process, white phosphorous produces P2O5.

Question 21.
Draw the structure of

  1. white phosphorous
  2. red phosphorous

Answer:
1. White phosphorous
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-17

2. Red phosphorous
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-18

Question 22.
How will you convert red phosphorous into P2O3 and P2Q5?
Answer:
Red phosphorus reacts with oxygen on heating to give phosphorus trioxide or phosphorus pentoxide.
P4 + 3O2 \(\underrightarrow { \triangle }\) P4O6(phosphorous trioxide)
P4 + 5O2 \(\underrightarrow { \triangle }\) P4Q10 (phosphorous pentaoxide)

Question 23.
How will you prepare orthophosphoric acid from phosphorous?
Answer:
When phosphorous is treated with cone.nitric acid it is oxidised to orthophosphoric acid. This reaction is catalysed by iodine crystals.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-19

Question 24.
Mention the uses of phosphorous?
Answer:

  1. The red phosphorus is used in the match boxes.
  2. It is also used for the production of certain alloys such as phosphor bronze.

Question 25.
Show that phosphine is weakly basic?
Answer:
Phosphine is weakly basic and forms phosphonium salts with halogen acids.
PH3 +HI → PH4I
PH4I + H2O \(\underrightarrow { \triangle }\) PH3+ H3O + I

Question 26.
Draw the structure of PCl3
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-20

Question 27.
How will you prepare PCl3 from white phosphorous?

  1. When a slow stream of chlorine is passed over white phosphorous, PCl3 is formed
  2. It can also be obtained by treating white phosphorous with thionyl chloride.

Question 28.
What happen when PCl3 is treated with cold water?
Answer:
When PCl3 is hydrolysed with cold water it gives phosphorous acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-21

Question 29.
How will you prepare PCl3 ?
Answer:
When PCl3 is treated with excess chlorine, phosphorous pentachloride is obtained.
PCl3+ Cl2 → PCl5

Question 30.
Mention the uses of PCl3 and PCl5?
answer:

  1. Phosphorus trichloride is used as a chlorinating agent and for the preparation of H2PO3.
  2. Phosphorous pentachloride is a chlorinating agent and is useful for replacing hydroxyl groups by chlorine atom.

Question 31.
Draw the structure of H3PO2 and H3PO3?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-22

Question 32.
Draw the structure of H4P2O6 and H4P2O3?
Answwer:
(i) H4P2O6
(ii) H4P2O3

Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-23

Question 33.
Give the method to prepare hypophosphorous acid and pyrophosphoric acid?
Answer:
1. Hypophosphorous acid – Phosphorous reacts with water to give hypophosphorous acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-24

2. Pyrophosphoric acid – Phosphorous acid is heated they produce pyrophosphoric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-25

Question 34.
Complete the reactions

  1. HgO \(\underrightarrow { \triangle }\) ?
  2. BaO2 \(\underrightarrow { \triangle }\) ?

Answer:

  1. HgO \(\underrightarrow { \triangle }\) 2Hg + O2
  2. 2BaO2 \(\underrightarrow { \triangle }\) 2BaO + O2

Question 35.
Give and explain the reaction used to estimation of ozone.
Answer:
Ozone oxidises potassium iodide to iodine.
O3 + 2KI + H2O → 2KOH + O2 + I2
This reaction is quantitative and can be used for estimation of ozone.

Question 36.
Write a short notes on Rhombic sulphur?
Answer:
Rhombic sulphur also known as a sulphur, is the only thermodynamically stable allotropic form at ordinary temperature and pressure. The crystals have a characteristic yellow colour and composed of S8 molecules. When heated slowly above 96°C, it converts into monoclinic sulphur. Upon cooling below 96°C the β form converts back to a form.

Question 37.
Write a notes on monoclinic sulphur?
Answer:
Monoclinic sulphur also contains S8 molecules in addition to small amount of S6 molecules. It exists as a long needle like prism and is also called as prismatic sulphur. It is stable between 96° – 119°C and slowly changes into rhombic sulphur.

Question 38.
What are λ – sulphur?
Answer:
Sulphur also exists in liquid and gaseous states. At around 140° C the monoclinic sulphur melts to form mobile pale yellow liquid called λ sulphur.

Question 39.
How will you prepare sulphurdioxide by laboratory method?
Answer:
Sulphur dioxide is prepared in the laboratory treating a metal or metal sulphite with sulphuric acid.
Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
\({ SO }_{ 3 }^{ – }\) + 2H+ → H2O + SO2

Question 40.
Complete the reactions,
Answer:

  1. Zns + O2 \(\underrightarrow { \triangle }\) ?
  2. FeS2 + O2 \(\underrightarrow { \triangle }\) ?

Answer:

  1. 2Zns + 3O2 \(\underrightarrow { \triangle }\) 2ZnO + 2SO2
  2. 4FeS2 + 11O2 \(\underrightarrow { \triangle }\) 2Fe2O3 + 8SO2

Question 41.
What happen when sulphurdioxdie reacts with sodium hydroxide and sodium carbonate?
Answer:
Sulphur dioxide reacts with sodium hydroxide and sodium carbonate to form sodium bisulphite and sodium sulphite respectively.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-26

Question 42.
Mention the uses of sulphurdioxide?
Answer:

  1. Sulphur dioxide is used in bleaching hair, silk, wool etc…
  2. It can be used for disinfecting crops and plants in agriculture.

Question 43.
Why sulphuric acid is high boiling and viscous liquid ?
Answer:
Sulphuric acid is high boiling and viscous liquid this is due to the association of molecules together through hydrogen bonding.

Question 44.
Explain the reaction between benzene and sulphuric acid.
Answer:
Sulphuric acid reacts with benzene to give benzene sulphuric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-27

Question 45.
Give the uses of sulphuric acid?
Answer:

  1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc…
  2. It is used as a drying agent and also used in the preparation of pigments, explosives etc..

Question 46.
Draw the structure of H2SO3 and H2SO4 ?
Answer:
1. H2SO3
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-28

2. H2SO4
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-29

Question 48.
Draw the structure of

  1. Pyrosulphuric acid
  2. Peroromonosulphuric acid?
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-30
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

Answer:
1. Pyrosulphuric acid H2S2O7

2. Peroromonosulphuric acid H2SO5
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-31

Question 49.
What happen when chlorine reacts with trupentine?
Answer:
When chlorine burnt with turpentine it forms carbon and hydrochloric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-32

Question 50.
How will you prepare hydrochloric acid by laboratory method?
Answer:
It is prepared by the action of sodium chloride and concentrated sulphuric acid

  • NaCl + H2SO4 → NaHSO4 + HCl
  • NaHSO4 + NaCl → Na2SO4 + HCl

Dry hydrochloric acid is obtained by passing the gas through conc. sulphuric acid

Question 51.
Mention the uses of hydrochloric acid?
Answer:

  1. Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride, glucose from com starch etc.,
  2. It is used in the extraction of glue from bone and also for purification of bone black.

Question 52.
Complet the following reaction?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-33

Question 53.
Why noble gases have the largest ionisation energy?
Answer:
Noble gases have the largest ionisation energy compared to any other elements in a given row as they have completely filled orbital (ns2np6) in their outer most shell. They are extremely stable and have a small tendency to gain or lose electrons. Hence noble gases have the largest ionisation energy.

Question 54.
Mention the uses of Neon?
Answer:
Neon is used in advertisement as neon sign and the brilliant red glow is caused by passing electric current through neon gas under low
pressure.

Question 55.
Give the uses of Krypton?
Answer:

  1. Krypton is used in fluorescent bulbs, flash bulbs etc…
  2. Lamps filed with krypton are used in airports as approaching lights as they can penetrate through dense fog.

Question 56.
Mention the application of Xenon?
Answer:

  1. Xenon is used in fluorescent bulbs, flash bulbs and lasers.
  2. Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed electronic flash bulbs used by photographers.

Question 57.
Give the uses of Radon?
Answer:

  1. Radon is radioactive and used as a source of gamma rays
  2. Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e. cancer growth.

Question 58.
Why are pentahalides more covalent than trihalides?
Answer:
Since elements in the +5 oxidation state have less tendency to lose electrons than in the +3 oxidation state, therefore elements in the +5 oxidation state are more covalent than in the +3 oxidation state. In other words, pentahalides are more covalent than trihalides.

59. Why is N2 less reactive at room temperature?
Answer:
Due to the presence of triple bond between the two nitrogen atoms, the bond dissociation energy of N2 (941.4 kJ mol) is very high. Therefore, N2 is less reactive at room temperature.

Question 60.
Why is ICl more reactive than I2 ?
Answer:
ICl bond is weaker than I – I bond. Therefore ICl can break easily to form halogen atoms which readily brings about the reaction.

Question 61.
Why is helium used in diving apparatus?
Answer:
It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

Question 62.
Give the disproportionation reaction of H3PO3?
Answer:
H3PO3 on heating undergoes self-oxidation reduction as:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-34

Question 63.
Why is red Phosphorus less reactive than white Phosphorus?
Answer:
White Phosphorus is more reactive than red Phosphorus under normal conditions because of angular strain in the P4 molecule where the angles are only 60°.

Question 64.
Nitrogen does not form any pentahalide like Phosphorus. Why?
Answer:
Nitrogen does not form pentahalide due to non-availability of the d-orbitals in its valence shell.

Question 65.
Give reason for the following. Among the noble gases only Xenon is well known to form chemical compounds?
Answer:
Xe is largest in size and has the highest polarising power.

Question 66.
Why is hydrogen sulphide, with greater molar mass a gas, while water a liquid at room temperature?
Answer:
H2O molecules are associated with intermolecular H-bonding, H2S is not because oxygen is more electronegative and smaller in size than sulphur. That is why H2O is a liquid and H2S is a gas.

Question 67.
Noble gases are chemically inert. Give reasons.
Answer:
Noble gases are chemically inert because they have their octet complete except Helium, i.e. they have a stable electronic configurations.

Question 68.
Why do noble gases exist as monoatomic?
Answer:
Noble gases have stable electronic configuration, that is why they have no tendency to lose or gain electrons.Therefore they do not form covalent bond.

Question 69.
Nitrogen exists as diatomic molecule and Phosphrus as P4. Why?
Answer:
Nitrogen has a triple bond between its two atoms because of its small size and high electro negativity. Phosphorus P4 has single bond, that is why it is tetra-atomic.

Question 70.
Why is H2S more acidic than water?
Answer:
It is because bond dissociation energy of S – H bond is less than O – H bond due to longer bond length.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 3 Marks Questions and Answers

Question 1.
Complete the following reactions.
(i) 6Li + N2 → ?
(ii) 3Ca + N2 → ?
(iii) 2B + N2 → ?
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-35

Qustion 2.
How is ammonia prepared?
Answer:
1. Ammonia is formed by the hydrolysis of urea.
NH2CONH2 + H2O → 2NH3 + CO2

2. Ammonia is prepared in the laboratory by heating an ammonium salt with a base.
\({ 2NH }_{ 4 }^{ + }\) + OH → 2NH3+ H2O
2NH4Cl + CaO → CaCl2 + 2NH3 + H2O

3. It can also be prepared by heating a metal nitrides such as magnesium nitride with water.
Mg3N2+ 6H2O → 3Mg(OH)2 + 2NH3

Question 3.
Explain the structure of ammonia.
Answer:
Ammonia molecule is pyramidal in shape N – H bond distance is 1.016 A and H – H bond distance is 1.645 A with a bond angle 107°. The structure of ammonia may be regarded as a tetrahedral with one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape as shown in the figure.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-36

Question 4.
How will you prepare nitric acid?
Answer:
Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate with concentrated sulphuric acid.
KNO3 + H2SO4 → KHSO4 + HNO3

The temperature is kept as low as possible to avoid decomposition of nitric acid. The acid condenses to a fuming liquids which is coloured brown by the presence of a little nitrogen dioxide which is formed due to the decomposition of nitric acid.
4HNO3 → 4NO2 + 2H2O + O2

Question 5.
Discuss the Commercial method to prepare Nitric acid.
(Or)
How will you prepare nitric acid by Ostwald’s process?
Answer:
Nitric acid prepared in large scales using Ostwald’s process. In this method ammonia from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.
4NH3 + 5O2 → 4NO + 6H3O + 120kJ
2NO + O2 → 2NO3

The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.
6NO2 + 3H2O → 4HNO3 + 2NO + H2O

Question 6.
Draw the structure of the following compounds.
(a) Nitrous acid
(b) Nitric acid
(c) Pernitric acid
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-37

Question 7.
Identify A, B and C from the following reactions.

  1. P4 + Mg → A
  2. P4+ Ca → B
  3. P4 + Na → C

Answer:

  1. P4 + 6Mg → 2Mg3P2 (Magnesium phosphide)
  2. P4 + 6Ca → 2Ca3P2 (Calcium phosphide)
  3. P4+ 12Na → 2Na3P (Sodium phosphide)

Question 8.
How will you prepare phosphine and explain the purification of phosphine?
Answer:
Phosphine is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-38
Phosphine is freed from phosphine dihydride (P2H4) by passing through a freezing mixture. The dihydride condenses while phosphine does not. Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-39
Phosphine is prepared in pure form by heating phosphorous acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-40
A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-41

Question 9.
What happens when PH3 reacts with oxygen or air?
Answer:
When phosphine air or oxygen it bums to give meta phosphoric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-42

Question 10.
Explain the structure of phosphine.
Answer:
In phosphine, phosphorus shows sp3 hybridisation. Three orbitals are occupied by bond pair and fourth comer is occupied by lone pair of electrons. Hence, bond angle is reduced to 94°. Phosphine has a pyramidal shape.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-43

Question 11.
Discuss the uses of phosphine.
Answer:
Phosphine is used for producing smoke screen as it gives large smoke. In a ship, a pierced container with a mixture of calcium carbide and calcium phosphide, liberates phosphine and acetylene when thrown into sea. The liberated phosphine catches fire and ignites acetylene. These burning gases serves as a signal to the approaching ships. This is known as Holmes signal.

Question 12.
How does PCl3 reacts with the following reagents?

  1. C2H5OH
  2. C2H5COOH

Answer:

  1. C2H5OH + PCl3 → C2H5Cl + H3PO3
  2. C2H5COOH + PCl3 → C2H5COCl + H3PO3

Question 13.
Explain the reaction between PCl5 and water.
Answer:
Phosphorous pentachloride reacts with water to give phosphoryl chloride and orthophosphoric acid.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-44

Question 14.
Explain the structure of phosphorous trioxide (P2O3).
Answer:
In phosphorous trioxide four phosphorous atoms lie at the comers of a tetrahedron and six oxygen atoms along the edges. The P – O bond distance is 165.6 pm which is shorter than the single bond distance of P – O (184 pm) due to pπ – dπ bonding and results in considerable double bond character.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-45

Question 15.
Discuss the structure of phosphorous pentaoxide (P2O3).
Answer:
In P4O10 each P atoms form three bonds to oxygen atom and also an additional coordinate bond with an oxygen atom. Terminal coordinate P – O bond length is 143 pm, which is less than the expected single bond distance. This may be due to lateral overlap of filled p orbitals of an oxygen atom with empty d-orbital on phosphorous.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-46

Question 16.
Mention the uses of oxygen.
Answer:

  1. Oxygen is one of the essential component for the survival of living organisms.
  2. It is used in welding (oxyacetylene welding)
  3. Liquid oxygen is used as fuel in rockets etc.

Question 17.
Give and explain reducing behaviour of sulphur dioxide.
Answer:
As Sulphur dioxide can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
SO2 + 2H2O + Cl2 → H2SO4 + 2HCl
It also reduces potassium permanganate and dichromate to Mn2+ and Cr3+ respectively.

  • 2KMnO4 + 5SO2 + 2H2O → K2SO4 + 2MnSO4 + 2H2SO4
  • K2Cr2O7 + 3SO2 + H2SO4 → K2SO4 + Cr2(SO4)3 + H2O

Question 18.
Why bleaching action of sulphur dioxide is temporary?
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property.
SO2 + 2H2O → 2H2SO4 + 2(H)
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-47

However, the bleached product (colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour. Hence bleaching action of sulphur dioxide is temporary.

Question 19.
Explain the structure of sulphur dioxide.
Answer:
In sulphur dioxide, sulphur atom undergoes sp2 hybridisation. A double bond arises between S and O is due to pπ- dπ overlapping.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-48

Question 20.
How will you manufacture sulphuric acid by contact process?
Answer:
Manufacture of sulphuric acid by contact process involves the following steps:
1. Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen/air.
S + O2 → SO2
4FeS2 + 11O2 → 2Fe2O3 + 8SO2

2. Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as V2O5 or platinised asbestos.

3. The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum (H2S2O7). The oleum is converted into sulphuric acid by diluting it with water.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-49
To maximise the yield the plant is operated at 2 bar pressure and 720 K. The sulphuric acid obtained in this process is over 96 % pure.

Question 21.
Prove that H2SO4 is a strong dibasic acid.
Answer:
Sulphuric acid forms two types of salts namely sulphates and bisulphates.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-50

Question 22.
Draw the structure of
(a) Marshall’s acid
(b) polythionic acid
(c) Dithionic acid
Answer:
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-51

Question 23.
What happens when chlorine reacts with ammonia?
Answer:
1. If chlorine reacts with excess ammonia, it gives nitrogen and ammonium chloride.
8NH3 + 3Cl2 → N2 + 6NH4Cl

2. If ammonia reacts with excess chlorine, it gives nitrogen trichloride and ammonium chloride.
4NH2 + 3Cl2 → NCl2 + 3NH4Cl

Question 24.
Bleaching action of chlorine is permanent. Justify this statement.
Answer:
Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.
Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements - II img-52
Colouring matter + Nascent oxygen → Colourless oxidation product
Therefore, bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and hydrogen sulphide to sulphur.