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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.2

10th Maths Exercise 6.2 Samacheer Kalvi Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of a tower of height 10\(\sqrt{3}\) m.
Solution:

Ex 6.2 Class 10 Samacheer Question 2.
A road is flanked on either side by continuous rows of houses of height 4\(\sqrt{3}\) m with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is 30°. Find the width of the road.
Solution:

Exercise 6.2 Class 10 Samacheer Kalvi Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are 60° and 45° respectively. If the height of the man is 180 cm and if he is 5 m away from the wall, what is the height of the window? (\(\sqrt{3}\) = 1.732)
Solution:

Let ‘H’ be the fit of the window. Given that elevation of top of the window is 60°.

Given that elevation of bottom of the window is 45°.

∴ Height of the window = 3.66 m

10th Maths Trigonometry Exercise 6.2 Question 4.
A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 40°. Find the height of the pedestal. (tan 40° = 0.8391, \(\sqrt{3}\) = 1.732)
Solution:

Let ‘p’ be the fit of the pedestal and d be the distance of statue from point of cabs, on the ground.
Given the elevation of top of the statue from p^f on ground is 60°.

10th Maths Exercise 6.2 Question 5.
A flag pole ‘h’ metres is on the top of the hemispherical dome of radius V metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°. Find
(i) the height of the pole
(ii) radius of the dome.
Solution:

10th Maths Ex 6.2 Question 6.
The top of a 15 m high tower makes an angle of elevation of 60° with the bottom of an electronic pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Solution:

Let BD be tower of height = 15 m
AE be pole of height = ‘p’

10th Maths 6.2 Question 7.
A vertical pole fixed to the ground is divided in the ratio 1 : 9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 25 m away from the base of the pole, what is the height of the pole?
Solution:

10th Samacheer Kalvi Maths Trigonometry Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are 4° and 8°. What is the height of the peak if the distance between consecutive milestones is 1 mile, (tan 4° = 0.0699, tan 8° = 0.1405).
Solution:

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Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 1 Electrostatics

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Samacheer Kalvi 12th Physics Electrostatics Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electrostatics Multiple Choice Questions

Physics Class 12 Samacheer Kalvi Question 1.
Two identical point charges of magnitude -q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge +q is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement?

(a) A₁ and A₂
(b) B₁ and B₂
(c) both directions
(d) No stable
Answer:
(b) B₁ and B₂

Samacheer Kalvi 12th Physics Chapter 1 Question 2.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Samacheer Kalvi Guru 12th Physics Question 3.
What is the ratio of the charges \(\left|\frac{q_{1}}{q_{2}}\right|\) for the following electric field line pattern?

(a) \(\frac { 1 }{ 5 }\)
(b) \(\frac { 25 }{ 11 }\)
(c) 5
(d) \(\frac { 12 }{ 25}\)
Answer:
(d) \(\frac { 12 }{ 25}\)

Samacheer Kalvi 12th Physics Question 4.
An electric dipole is placed at an alignment angle of 30° with an electric field of 2 x 10⁵ N C^-1. It experiences a torque equal to 8 N m. The charge on the dipole if the dipole length is 1 cm is-
(a) 4 mC
(b) 8 mC
(c) 5 mC
(d) 1 mC
Answer:
(b) 8 mC

12 Physics Samacheer Kalvi Question 5.
Four Gaussian surfaces are given below with charges inside each Gaussian surface. Rank the electric flux through each Gaussian surface in increasing order-

(a) D < C < B < A
(b) A < B = C < D
(c) C < A = B < D
(d)D > C > B > A
Answer:
(a) D < C < B < A

Class 12 Physics Samacheer Kalvi Question 6.
The total electric flux for the following closed surface which is kept inside water-

(a) \(\frac { 80q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 40ε }_{0}}\)
(c) \(\frac { q }{{ 80ε }_{0}}\)
(d) \(\frac { q }{{ 40ε }_{0}}\)
Answer:
(b) \(\frac { q }{{ 40ε }_{0}}\)

12th Physics 1st Chapter Question 7.
Two identical conducting balls having positive charges q₁ and q₂ are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be- (NSEP 04-05)
(a) less than before
(b) same as before
(c) more than before
(d) zero
Answer:
(c) more than before

12th Physics Samacheer Kalvi Question 8.
Rank the electrostatic potential energies for the given system of charges in increasing order

(a) 1 = 4 < 2 < 3
(b) 2 = 4 < 3 < 1
(c) 2 = 3 < 1 < 4
(d) 3 < 1 < 2 < 4
Answer:
(a) 1 = 4 < 2 < 3

Samacheerkalvi.Guru 12th Physics Question 9.
An electric field \(\vec { E } \) = 10x\(\hat{i} \) exists in a certain region of space. Then the potential difference V = V₀ – V_A, Where V₀ is the potential at the origin and V_A is the potential at x = 2 m is-
(a) 10 J
(b) -20 J
(c) + 20 J
(d) – 10 J
Answer:
(a) 10 J

Electrostatics Notes Class 12 State Board Question 10.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. The correct plot for electrostatic potential due to this spherical shell is-

Answer:

12th Physics Lesson 1-Electrostatics Question 11.
Two points A and B are maintained at a potential of 7 V and -4 V respectively. The work done in moving 50 electrons from A to B is-
(a) 8.80 x 10^-17 J
(b) -8.80 x 10^-17 J
(c) 4.40 x 10^-17 J
(d) 5.80 x 10^-17 J
Answer:
(a) 8.80 x 10^-17 J

Electrostatics Class 12 Questions And Answers Pdf Question 12.
If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Electrostatic Problems And Solutions Pdf Question 13.
A parallel plate capacitor stores a charge Q at a voltage V. Suppose the area of the parallel plate capacitor and the distance between the plates are each doubled then which is the quantity that will change?
(a) Capacitance
(b) Charge
(c) Voltage
(d) Energy density
Answer:
(d) Energy density

Samacheer Kalvi Physics Question 14.
Three capacitors are connected in triangle as shown in the figure. The equivalent capacitance between the points A and C is

(a) 1 μF
(b) 2 μF
(c) 3 μF
(d) \(\frac { 1 }{ 4 }\) μF
Answer:
(b) 2 μF

Physics Solution Class 12 Samacheer Kalvi Question 15.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 x 10^-2 C and 5 x 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is (AIIPMT 2012)
(a) 3 x 10^-2 C
(b) 4 x 10^-2 C
(c) 1 x 10^-2 C
(d) 2 x 10^-2 C
Answer:
(a) 3 x 10^-2 C

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Samacheer Kalvi Class 12 Physics Solutions Question 1.
What is meant by quantisation of charges?
Answer:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ±4). This is called quantisation of electric charge.

Samacheer Kalvi Guru Physics Question 2.
Write down Coulomb’s law in vector form and mention what each term represents.
Answer:
The force on a charge q₁ exerted by a point charge q₁ is given by
\(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }{0}}\) \(\frac {{ q }_{1}{ q }_{2}}{{ r }^{2}}\) \(\hat{r} \)₂₁
Here \(\hat{r} \)₂₁ is the unit vector from charge q₁ to q₁.
But \(\hat{r} \)₂₁ = –\(\hat{r} \)₁₂,

Therefore, the electrostatic force obeys Newton’s third law.

Samacheer Kalvi 12th Physics Solutions Question 3.
What are the differences between Coulomb force and gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 x 10^-11 N m² kg^-2. The value of the constant k in Coulomb law is k = 9 x 10⁹ N m² C².
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to coulomb force.

Samacheer Kalvi.Guru 12th Physics Question 4.
Write a short note on superposition principle.
Answer:
According to this superposition principle, the total force acting on a given charge is equal to the vector sum of forces exerted on it by all the other charges.
\({ \vec { F } }_{ 1 }^{ tot }\) = \(\vec { F } \)₁₂ + \(\vec { F } \)₁₃ + \(\vec { F } \)₁₄ + \(\vec { F } \)_1n

Samacheer Kalvi 12 Physics Question 5.
Define ‘Electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced
by a unit charge and is given by
\(\vec { E } \) = \(\frac { \vec { F } }{ { q }_{ 0 } } \)
The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Question 6.
What is mean by ‘Electric field lines’?
Answer:
Electric field vectors are visualized by the concept of electric field lines. They form a set of continuous lines which are the visual representation of the electric field in some region of space.

Question 7.
The electric field lines never intersect. Justify.
Answer:
As a consequence, if some charge is placed in the intersection point, then it has to move in two different directions at the same time, which is physically impossible. Hence, electric field lines do not intersect.

Question 8.
Define ‘Electric dipole’
Answer:
Two equal and opposite charges separated by a small distance constitute an electric dipole.

Question 9.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from -q to + q. The SI unit of dipole moment is coulomb meter (Cm).
\(\vec { P } \) = qa\(\hat{i} \) -qa(\(\hat{-i} \)) = 2 qa\(\hat{i} \)

Question 10.
Define “electrostatic potential”.
Answer:
The electric potential at a point P is equal to the work done by an external force to bring a unit positive charge with constant velocity from infinity to the point P in the region of the external
electric field \(\vec { E } \).

Question 11.
What is an equipotential surface?
Answer:
An equipotential surface is a surface on which all the points are at the same potential.

Question 12.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (V_B – V_A). If the points A and B lie on the same equipotential surface, work done is zero because V_A = V_B.

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to equipotential surface.

Question 13.
Give the relation between electric field and electric potential.
Answer:
Consider a positive charge q kept fixed at the origin. To move a unit positive charge by a small distance dx in the electric field E, the work done is given by dW = -E dx. The minus sign implies that work is done against the electric field. This work done is equal to electric potential difference. Therefore,
dW = dV.
(or) dV = -Edx
Hence E = \(\frac { dV }{ dx }\)
The electric field is the negative gradient of the electric potential.

Question 14.
Define electrostatic potential energy?
Answer:
The potential energy of a system of point charges may be defined as the amount of work done in assembling the charges at their locations by bringing them in from infinity.

Question 15.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 16.
What is meant by electrostatic energy density?
Answer:
The energy stored per unit volume of space is defined as energy density u_E = \(\frac { U }{ Volume }\)
From equation u_E = \(\frac { 1 }{ 2 }\) \(\frac{\left(\varepsilon_{0} A\right)}{d}\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀ (Ad) E² or u_E = \(\frac { 1 }{ 2 }\) ε₀E²

Question 17.
Write a short note on ‘electrostatic shielding’.
Answer:
Consider a cavity inside the conductor. Whatever the charges at the surfaces and whatever the electrical disturbances outside, the electric field inside the cavity is zero. A sensitive electrical instrument which is to be protected from external electrical disturbance is kept inside this cavity. This is called electrostatic shielding.

Question 18.
What is Polarisation?
Answer:
Polarisation \(\vec { P } \) is defined as the total dipole moment per unit volume of the dielectric.
\(\vec { P } \) = X_e \(\vec { P } \)_ext

Question 19.
What is dielectric strength?
Answer:
The maximum electric field the dielectric can withstand before it breakdowns is called dielectric strength.

Question 20.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
C = \(\frac { q }{ V }\) or Q ∝ V.
The SI unit of capacitance is coulomb per volt or farad (F).

Question 21.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge. This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Discuss the basic properties of electric charges.
Answer:
Basic properties of charges
(i) Electric charge:
Most objects in the universe are made up of atoms, which in turn are made up of protons, neutrons and electrons. These particles have mass, an inherent property of particles. Similarly, the electric charge is another intrinsic and fundamental property of particles. The SI unit of charge is coulomb.

(ii) Conservation of charges:
Benjamin Franklin argued that when one object is rubbed with another object, charges get transferred from one to the other. Before rubbing, both objects are electrically neutral and rubbing simply transfers the charges from one object to the other. (For example, when a glass rod is rubbed against silk cloth, some negative charge are transferred from glass to silk. As a result, the glass rod is positively charged and silk cloth becomes negatively charged).

From these observations, he concluded that charges are neither created or nor destroyed but can only be transferred from one object to other. This is called conservation of total charges and is one of the fundamental conservation laws in physics. It is stated more generally in the following way. The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

(iii) Quantisation of charges:
The charge q on any object is equal to an integral multiple of this fundamental unit of charge e.
q = ne
Here n is any integer (0, ±1, ±2, ±3, ± ….). This is called quantisation of electric charge. Robert Millikan in his famous experiment found that the value of e = 1.6 x 10^-19C. The charge of an electron is -1.6 x 10^-19 C and the charge of the proton is +1.6 x 10^-19C. When a glass rod is rubbed with silk cloth, the number of charges transferred is usually very large, typically of the order of 10¹⁰. So the charge quantisation is not appreciable at the macroscopic level. Hence the charges are treated to be continuous (not discrete). But at the microscopic level, quantisation of charge plays a vital role.

Question 2.
Explain in detail Coulomb’s law and its various aspects.
Answer:
Consider two point charges q₁ and q₂ at rest in vacuum, and separated by a distance of r. According to Coulomb, the force on the point charge q₂ exerted by another point charge q₁ is
\(\vec { F } \) ₂₁ = K\(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂,
where [/latex] \(\hat{r} \)₁₂ is the unit vector directed from charge q₁ to charge q₂ and k is the proportionality constant.

Important aspects of Coulomb’s law:
(i) Coulomb’s law states that the electrostatic force is directly proportional to the product of the magnitude of the two point charges and is inversely proportional to the square of the distance between the two point charges.

(ii) The force on the charge q₂exerted by the charge q₁ always lies along the line joining the two charges. \(\hat{r} \)₂₁is the unit vector pointing from charge q₁ to q₂ Likewise, the force on the charge q₁ exerted by q₂ is along – (i.e., in the direction opposite to \(\hat{r} \)₂₁).

(iii) In SI units, k = \(\frac { 1 }{{ 4πε }_{0}}\) and its value is 9 x 10⁹ Nm²C^-2. Here e0 is the permittivity of free space or vacuum and the value of ε₀ = \(\frac { 1 }{{ 4πε }_{0}}\) = 8.85 x 10^-12 C² N^-1 m^-2

(iv) The magnitude of the electrostatic force between two charges each of one coulomb and separated by a distance of 1 m is calculated as follows:
[F] = \(\frac{9 \times 10^{9} \times 1 \times 1}{1^{2}}\) = 9 x 10⁹N. This is a huge quantity, almost equivalent to the weight of one million ton. We never come across 1 coulomb of charge in practice. Most of the electrical phenomena in day-to-day life involve electrical charges of the order of pC (micro coulomb) or nC (nano coulomb).

(v) In SI units, Coulomb’s law in vacuum takes the form \(\vec { F } \) ₂₁ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂. sin Since ε > ε₀, the force between two point charges in a medium other than vacuum is always less than that in vacuum. We define the relative permittivity for a given medium as ε = \(\frac { ε }{{ ε }_{0}}\) .For vacuum or air, ε_r = 1 and for all other media ε_r > 1

(vi) Coulomb’s law has same structure as Newton’s law of gravitation. Both are inversely proportional to the square of the distance between the particles. The electrostatic force is directly proportional to the product of the magnitude of two point charges and gravitational force is directly proportional to the product of two masses.

(vii) The force on a charge q₁ exerted by a point charge q₂ is given by \(\vec { F } \)₁₂ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r_{2}}\) \(\hat{r} \)₁₂ Here \(\hat{r} \)₂₁ is sthe unit vector from charge q₂ to q₁.

Therefore, the electrostatic force obeys Newton’s third law.

(viii) The expression for Coulomb force is true only for point charges. But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the charged spheres in the torsion balance as point charges. The distance between the two charged spheres is much greater than the radii of the spheres.

Question 3.
Define ‘Electric field’ and discuss its various aspects.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by

Here \(\hat{r} \) is the unit vector pointing from q to the point of interest P. The electric field is a vector quantity and its SI unit is Newton per Coulomb (NC^-1).

Important aspects of Electric field:
(i) If the charge q is positive then the electric field points away from the source charge and if q is negative, the electric field points towards the source charge q.

(ii) If the electric field at a point P is \(\vec { E } \), then the force experienced by the test charge qo placed at the point P is \(\vec { F } \) = q₀ \(\vec { E } \). This is Coulomb’s law in terms of electric field. This is shown in the below Figure.

(iii) The equation implies that the electric field is independent of the test charge q0 and it depends only on the source charge q.

(iv) Since the electric field is a vector quantity, at every point in space, this field has unique direction and magnitude as shown in Figures (a) and (b). From equation, we can infer that as distance increases, the electric field decreases in magnitude. Note that in Figures (a) and (b) the length of the electric field vector is shown for three different points. The strength or magnitude of the electric field at point P is stronger than at the point Q and R because the point P is closer to the source charge.

(v) In the definition of electric field, it is assumed that the test charge q₀ is taken sufficiently small, so that bringing this test charge will not move the source charge. In other words, the test charge is made sufficiently small such that it will not modify the electric field of the source charge.

(vi) The expression is valid only for point charges. For continuous and finite size charge distributions, integration techniques must be used. However, this expression can be used as an approximation for a finite-sized charge if the test point is very far away from the finite sized source charge.

(vii) There are two kinds of the electric field: uniform (constant) electric field and non-uniform electric field. Uniform electric field will have the same direction and constant magnitude at all points in space. Non-uniform electric field will have different directions or different magnitudes or both at different points in space. The electric field created by a point charge is basically a non-uniform electric field. This non-uniformity arises, both in direction and magnitude, with the direction being radially outward (or inward) and the magnitude changes as distance increases.

Question 4.
How do we determine the electric field due to a continuous charge distribution? Explain. Electric field due to continuous charge distribution.
Answer:
The electric charge is quantized microscopically. The expressions of Coulomb’s Law, superposition principle force and electric field are applicable to only point charges. While dealing with the electric field due to a charged sphere or a charged wire etc., it is very difficult to look at individual charges in these charged bodies. Therefore, it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus.

Consider the following charged object of irregular shape. The entire charged object is divided into a large number of charge elements ∆q₁, ∆q₂, ∆q₃ ……..∆q_n,…… and each charge element Δq is taken as a point charge.
The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such charge elements.

Here ∆ qi is the i^th charge element, r_ip is the distance of the point P frome the i^th charge element, r_ip is the unit vector from ith charge element to the pont P.
However the equation is only an approximation. To incorporate the continuous distribution of charge, we take the limit ∆q → 0(= dq). In this limit, the summation in the equation becomes an integration and takes the following form

Here r is the distance of the point P from the infinitesimal charge dq and \(\hat{r} \) is the unit vector from dq to point P. Even though the electric field for a continuous charge distribution is difficult to evaluate, the force experienced by some test charge q in this electric field is still given by \(\vec { F } \) = q\(\vec { E } \).

(a) Line charge distribution: If the charge Q is uniformly distributed along the wire of length L, then linear charge density (charge per unit length) is λ = \(\frac { Q }{ L }\). Its unit is colomb per meter (Cm^-1). The charge present in the infinitestimal length dl is dq = λdl.

The electric field due to the line of total charge Q is given by

(b) Surface charge distribution: If the charge Q is uniformly distributed on a surface of area A, then surface charge density (charge per unit area) is σ = \(\frac { Q }{ A }\). Its unit is coulomb per square meter (C m^-2). The charge present in the infinitesimal area dA is dq = σdA. The electric field due to a of total charge Q is given by

(c) Volume charge distribution: If the charge Q is uniformly distributed in a volume V, then volume charge density (charge per unit volume) is given by ρ = \(\frac { Q }{ V }\). Its unit is coulomb per cubic meter (Cm^-3) The charge present in the infinitesimal volume element dV is dq = ρdV. The electric field due to a volume of total charge Q is given by

Question 5.
Calculate the electric field due to a dipole on its axial line and equatorial plane.
Case (I) :
Electric field due to an electric dipole at points on the axial line. Consider an electric dipole placed on the x-ax is as shown in figure. A point C is located at a distance of r from the midpoint O of the dipole along the axial line. Axial line

The electric field at a point C due to +q is

Since the electric dipole moment vector \(\vec { P } \) is from -q to +q and is directed along BC, the above equation is rewritten as

where \(\hat{p} \) is the electric dipole moment unit vector from -q to +q. The electric field at a point C due to -q is

Since +q is located closer to the point C than -q, \(\vec { E } \) _. \(\vec { E } \) + us stronger than \(\vec { E } \) _. Therefore, the length of the E + vector is drawn large than that of \(\vec { E } \) _vector.
The total electric field at point C is calculated using the superposition principle of the electric field.

Note that the total electric field is along \(\vec { E } \)₊, since +q is closer to C than -q.

The direction of \(\vec { E } \)_tot is shown in Figure
If the point C is very far away from the dipole then (r >> a). Under this limit the term(r² – a²)² ≈ r⁴ Substituting this into equation, we get

If the point C is chosen on the left side of the dipole, the total electric field is still in the

Case (II) :
Electric field due to an electric dipole at a point on the equatorial plane
Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure. Since the point C is equi-distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of E+ is along BC and the direction of E is along CA. E+ and E_ are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it.

The perpendicular components \(\left|\vec{E}_{+}\right|\) sin θ and \(\left|\vec{E}_{-}\right|\) sin θ are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the paralle component of \(\vec { E } \)₊ and \(\vec { E } \)_– and its direction is along \(\hat{-p} \).


The magnitudes \(\vec { E } \)₊ and \(\vec { E } \)_– are the same and are given by

By substituting equation (1) into equation (2), we get

At very large distances (r >> a), the equation becomes
\(\vec { E } \)_tot \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{ r }^{3}}\) (r >>) …… (4)

Question 6.
Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field:
Consider an electric dipole of dipole moment \(\vec { p } \) placed in a uniform electric field E whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec { E } \) in the direction of the field and charge -q will experience a force -q\(\vec { E } \) in a direction opposite to the field.

Since the external field \(\vec { E } \) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O
\(\vec { τ } \) = \(\overrightarrow{\mathrm{OA}}\) × (-q\(\vec { E } \)) + \(\overrightarrow{\mathrm{OB}}\) × q\(\vec { E } \)
Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.

The magnitude of the total torque
\(\vec { τ } \) = \(|\overrightarrow{\mathrm{OA}}|\)(-q\(\vec { E } \)) sin θ + \(|\overrightarrow{\mathrm{OB}}|\) \(|q \overrightarrow{\mathrm{E}}|\) sin θ
where θ is the angle made by \(\vec { P } \) with \(\vec { E } \). Since p = 2aq, the torque is written in terms of the vector product as
\(\vec { τ } \) = \(\vec { p } \) x \(\vec { E } \)
The magnitude of this torque is τ = pE sin θ and is maximum Torque on dipole
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field \(\vec { E } \). Once \(\vec { E } \) is aligned with \(\vec { E } \), the total torque on the dipole becomes zero.

Question 7.
Derive an expression for electrostatic potential due to a point charge.
Answer:
Electric potential due to a point charge:
Consider a positive charge q kept fixed at the origin. Let P be a point at distance r from the charge q.
The electric potential at the point P is


Electric field due to positive point charge q is

The infinitesimal displacement vector, d\(\vec { r } \) = dr\(\hat{r} \) and using \(\hat{r} \) . \(\hat{r} \) = 1, we have

After the integration,

Hence the electric potential due to a point charge q at a distance r is
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\) …… (2)
Important points (If asked in exam)
(i) If the source charge q is positive, V > 0. If q is negative, then V is negative and equal to
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{ r }\)
(ii) The description of motion of objects using the concept of potential or potential energy is simpler than that using the concept of field.
(i) From expression (2), it is clear that the potential due to positive charge decreases as the distance increases, but for a negative charge the potential increases as the distance is increased. At infinity (r = ∞) electrostatic potential is zero (V = 0).
(iv) The electric potential at a point P due to a collection of charges q₁,q₂,q₃… q_n is equal to sum of the electric potentials due to individual charges.

Where r₁, r₂,r₃,…..r_n are the distances of q₁,q₂,q₃… q_n
respectively from P

Question 8.
Derive an expression for electrostatic potential due to an electric dipole.
Answer:
Electrostatic potential at a point due to an electric dipole:
Consider two equal and opposite charges separated by a small distance 2a. The point P is located at a distance r from the midpoint of the dipole. Let 0 be the angle between the line OP and dipole axis AB.
Let r₁ be the distance of point P from +q and r₁ be the distance of point P from -q.
Potential at P due to charge +q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{1}}\)
Potential at P due to charge -q = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }_{2}}\)
Total Potential at the point P,
V = \(\frac { 1 }{{ 4πε }_{0}}\)q \(\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\) ….. (1)
Suppose if the point P is far away from the dipole, such that r >> a, then equation can be expressed in terms of r. By the cosine law for triangle BOP,

\(r_{1}^{2}\) = r² + a² – 2ra cos θ = r² \(\left(1+\frac{a^{2}}{r^{2}}-\frac{2 a}{r} \cos \theta\right)\)
Since the point P is very far from dipole, then r >> a. As a result the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) is very small and can be neglected. Therefore

since \(\frac { a }{ r }\) << 1, we can use binominal theorem and retain the terms up to first order
\(\frac { 1 }{{ r_{1}} }\) = \(\left(1+\frac{a}{r} \cos \theta\right)\) ……. (2)
Similarly applying the cosine law for triangle AOP,
\(r_{2}^{2}\) = r² + a² – 2ra cos (180 – θ)
Since cos (180 – θ) = cos θ we get
\(r_{2}^{2}\) = r² + a² + 2ra cos θ
Neglecting the term \(\frac {{ a }^{ 2 }}{{ r }^{ 2 }}\) (because r >> a)
\(r_{2}^{2}\) = r² \(\left(1+\frac{2 a \cos \theta}{r}\right)\) (or) r₂ = r \(\left(1+\frac{2 a \cos \theta}{r}\right)^{\frac{1}{2}}\)
Using Binomial theorem, we get
\(\frac { 1 }{{ r_{2}} }\) = \(\frac { 1 }{ r }\) \(\left(1-a \frac{\cos \theta}{r}\right)\)
Substituting equations (3) and (2) in equation (1)

But the electric dipole moment p = 2qa and we get,
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\left(\frac{p \cos \theta}{r^{2}}\right)\)
Now we can write p cos θ = \(\vec { P } \), \(\hat{r} \) where \(\hat{r} \) is the unit vector from the point O to point P. Hence the electric potential at a point P due to an electric dipole is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\vec{p} \cdot \hat{r}}{r^{2}}\) (r >> a) ….. (4)
Equation (4) is valid for distances very large compared to the size of the dipole. But for a point dipole, the equation (4) is valid for any distance.
Special cases:

Case (I):
If the point P lies on the axial line of the dipole on the side of +q, then θ = 0. Then the electric potential becomes
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (II):
If the point P lies on the axial line of the dipole on the side of -q, then θ = 180°, then
V = – \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { p }{{r}^{ 2 }}\)

Case (III):
If the point P lies on the equatorial line of the dipole, then θ = 90°. Hence, V = 0.

Question 9.
Obtain an expression for potential energy due to a collection of three point charges which are separated by finite distances.
Answer:
Electrostatic potential energy for collection of point charges:
The electric potential at a point at a distance r from point charge ql is given by
V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 1 }}{r}\) …… (1)
This potential V is the work done to bring a unit positive charge from infinity to the point. Now if the charge q₂ is brought from infinity to that point at a distance r from qp the work done is the product of q₂ and the electric potential at that point. Thus we have W = q₂V …… (2)
This work done is stored as the electrostatic potential energy U of a system of charges q₁ and q₂ separated by a distance r. Thus we have
U = q₂ V = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r}\) …… (3)
The electrostatic potential energy depends only on the distance between the two point charges. In fact, the expression (3) is derived by assuming that q₁ is fixed and q₂ is brought from infinity. The equation (3) holds true when q₂ is fixed and q₁ is brought from infinity or both q₂and q₂ are simultaneously brought from infinity to a distance r between them.
Three charges are arranged in the following configuration as shown in Figure.

To calculate the total electrostatic potential energy, we use the following procedure. We bring all the charges one by one and arrange them according to the configuration.
(i) Bringing a charge q₁ from infinity to the point A requires no work, because there are no other charges already present in the vicinity of charge q₁

(ii) To bring the second charge q₂ to the point B, work must be done against the electric field created by the charge q₁ So the work done on the charge q₁ is W = q₂V_1B. Here V_1B is the electrostatic potential due to the charge q₁ at point B.
U = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{{r}_{12}}\) ….. (4)
Note that the expression is same when q₂ is brought first and then q₁ later.

(iii) Similarly to bring the charge q3 to the point C, work has to be done against the total electric field due to both charges q₁ and q₂. So the work done to bring the charge q₃ is = q₃ (V_1C + V_2C). Here V_1C is the electrostatic potential due to charge q₁ at point C and V_2C is the electrostatic potential due to charge q₂ at point C. The electrostatic potential is

(iv) Adding equations (4) and (5), the total electrostatic potential energy for the system of three charges q₁,q₂ and q₃ is

Note that this stored potential energy U is equal to the total external work done to assemble the three charges at the given locations. The expression (6) is same if the charges are brought to their positions in any other order. Since the Coulomb force is a conservative force, the electrostatic potential energy is independent of the manner in which the configuration of charges is arrived at.

Question 10.
Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Answer:
Electrostatic potential energy of a dipole in a uniform electric field:
Consider a dipole placed a torque when kept in an uniform electric field \(\vec { E } \). A dipole experiences a torque when kept in an uniform electric field \(\vec { E } \). This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole.

The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

Since τ_ext is equal and opposite to τ_E = \(\vec { P } \) x \(\vec { E } \), we have
\(\left|\overrightarrow{\mathrm{r}}_{\mathrm{ext}}\right|\) = \(\left|\overrightarrow{\mathrm{r}}_{\mathrm{E}}\right|\)= \(|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{E}}|\) …. (2)
Substituting equation (2) in equation (1) We get,

This work done is equal to the potential energy difference between the angular positions θ and θ’.
U(θ) – (Uθ’) = AU = -pE cos θ +PE cos θ’.
If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos θ’ = θ.
The potential energy stored in the system of dipole kept in the uniform electric field is given by El = -pE cos θ = –\(\vec { P } \) . \(\vec { E } \) ….. (3)
In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.
The potential energy is maximum when the dipole is aligned anti-parallel (θ = π) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

Question 11.
Obtain Gauss law from Coulomb’s law.
Answer:
Gauss law: Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux Φ_E through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\frac{\mathrm{Q}_{\mathrm{end}}}{\varepsilon_{0}}\)
A positive point charge Q is surrounded by an imaginary sphere of radius r as shown in figure. We can calculate the total electric flux through the closed surface of the sphere using the equation.

Φ_E = \(\oint { \vec { E } } \) .d \(\vec { A } \) = \(\oint { EdA } \) cos θ …… (1)
The electric field of the point charge is directed radially outward at all points on the surface of the sphere. Therefore, the direction of the area element d \(\vec { A } \) is along the electric field \(\vec { E } \) and θ = 0°.
Φ_E = \(\oint { EdA } \) since cos 0° = 1 ….. (2)
E is uniform on the surface of the sphere,
Φ_E = \(\oint { EdA } \) ….. (3)
Substituting for
\(\oint { dA } \) = 4π² and E = \(\frac { 1 }{{ 4πε }_{0}}\) Q in equation 3, we get
Φ_E = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { q }{{ r }^{2}}\) × 4π² = 4π \(\frac { 1 }{{ 4πε }_{0}}\) = \(\frac { q }{{ ε }_{0}}\) ……. (4)
The equation (4) is called as Gauss’s law. The remarkable point about this result is that the equation (4) is equally true for any arbitrary shaped surface which encloses the charge Q.

Question 12.
Obtain the expression for electric field due to an infinitely long charged wire.
Answer:
Electric field due to an infinitely long charged wire:
Consider an infinitely long straight wire having uniform linear charge density λ. Let P be a point located at a perpendicular distance r from the wire. The electric field at the point P can be found using Gauss law. We choose two small charge elements A₁ and A₁ on the wire which are at equal distances from the point P.

The resultant electric field due to these two charge elements points radially away from the charged wire and the magnitude of electric field is same at all points on the circle of radius r. From this property, we can infer that the charged wire possesses a cylindrical symmetry.

Let us choose a cylindrical Gaussian surface of radius r and length L. The total electric flux in this closed surface is

It is seen that for the curved surface, \(\vec { E } \) is parallel to \(\vec { A } \) and \(\vec { E } \).d \(\vec { A } \) = EdA. For the top and bottom surface, \(\vec { E } \) is perpendicular to \(\vec { A } \) and \(\vec { E } \).d\(\vec { A } \) = 0
Substituting these values in the equation (2) and applying Gauss law


Since the magnitude of the electric field for the entire curved surface is constant, E is taken out of the integration and Q_encl is given by Q_encl = λL.

Here,

dA = total area of the curved surface = 2πrL. Substituting this in
equation (4), We get

The electric field due to the infinite charged wire depends on \(\frac { 1 }{ r }\) rather than \(\frac { 1 }{{r}^{ 2 }}\) for a point charge.
Equation (6) indicates that the electric field is always along the perpendicular direction (\(\hat{r} \) ) to wire. In fact, if λ > 0 then E points perpendicular outward (\(\hat{r} \) ) from the wire and if λ < 0, then E points perpendicular inward (- \(\hat{r} \) ).

Question 13.
Obtain the expression for electric field due to an charged infinite plane sheet.
Answer:
Electric field due to charged infinite plane sheet: Consider an infinite plane sheet of charges with uniform surface charge density o. Let P be a point at a distance of r from the sheet. Since the plane is infinitely large, the electric field should be same at all points equidistant from the plane and radially directed at all points. A cylindrical shaped Gaussian surface of length 2r and area A of the flat surfaces is chosen such that the infinite plane sheet passes perpendicularly through the middle part of the Gaussian surface.

Applying Gauss law for this cylindrical surface,

The electric field is perpendicular to the are a element at all points on the curved surface and is parallel to the surface areas at P and P’. Then,

Since the magnitude of the electric field at these two equal surfaces is uniform, E is taken out of the integration and Q_encl is given by Q_encl = σA, we get

The total area of surface either at P or P’

Hence 2EA = \(\frac { σA }{{ ε }_{0}}\) or E = \(\frac { σ }{{ 2ε }_{0}}\) …… (3)
In vector from, E = \(\frac { σ }{{ 2ε }_{0}}\) \(\hat{n} \) ….. (4)
Hence \(\hat{n} \) is the outward unit vector normal to the plane. Note that the electric field due to an infinite plane sheet of charge depends on the surface charge density and is independent of the distance r.

The electric field will be the same at any point farther away from the charged plane. Equation (4) implies that if o > 0 the electric field at any point P is outward perpendicular n to the plane and if σ < 0 the electric field points inward perpendicularly (\(\hat{n} \) ) to the plane. For a finite charged plane sheet, equation (4) is approximately true only in the middle region of the plane and at points far away from both ends.

Question 14.
Obtain the expression for electric field due to an uniformly charged spherical shell.
Answer:
Electric field due to a uniformly charged spherical shell:
Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a):
At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
\(\oint { \vec { E } } .d\vec { A } \) = \(\frac { Q }{{ ε }_{0}}\) …….(1)
The electric field \(\vec { E } \) and d\(\vec { A } \) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\vec { E } \) is also the same at all points due to the spherical symmetry of the charge distribution.

But

dA = total area of Gaussian surface = 4πr². Substituting this value in equation (2).

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b):
At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\vec { E } \) = \(\frac{\mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\) \(\hat{r} \) …… (4)

Case (c):
At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law

Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes E = 0 (r < R) …(6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

Question 15.
Discuss the various properties of conductors in electrostatic equilibrium.
Answer:
Properties of conductors in electrostatic equilibrium:
(i) The electric field is zero everywhere inside the conductor. This is true regardless of whether the conductor is solid or hollow. This is an experimental fact. Suppose the electric field is not zero inside the metal, then there will be a force on the mobile charge carriers due to this electric field.

As a result, there will be a net motion of the mobile charges, which contradicts the conductors being in electrostatic equilibrium. Thus the electric field is zero everywhere inside – the conductor. We can also understand this fact by applying an external uniform electric field on the conductor.

Before applying the external electric field, the free electrons in the conductor are uniformly distributed in the conductor. When an electric field is applied, the free electrons accelerate to the left causing the left plate to be negatively charged and the right plate to be positively charged.

Due to this realignment of free electrons, there will be an internal electric field created inside the conductor which increases until it nullifies the external electric field. Once the external electric field is nullified the conductor is said to be in electrostatic equilibrium. The time taken by a conductor to reach electrostatic equilibrium is in the order of 10^-6s, which can be taken as almost instantaneous.

(ii) There is no net charge inside the conductors. The charges must reside only on the surface of the conductors. We can prove this property using Gauss law. Consider an arbitarily shaped conductor. A Gaussian surface is drawn inside the conductor such that it is very close to the surface of the conductor.

Since the electric field is zero everywhere inside the conductor, the net electric flux is also zero over this Gaussian surface. From Gauss’s law, this implies that there is no net charge inside the conductor. Even if some charge is introduced inside the conductor, it immediately reaches the surface of the conductor.

(iii) The electric field outside the conductor is perpendicular to the surface of the conductor and has a magnitude of \(\frac { σ }{{ ε }{0}}\) where a is the surface charge density at that point. If the electric field has components parallel to the surface of the conductor, then free electrons on the surface of the conductor would experience acceleration. This means that the conductor is not in equilibrium. Therefore at electrostatic equilibrium, the electric field must be perpendicular to the surface of the conductor.

We now prove that the electric field has magnitude \(\frac { σ }{{ ε }{0}}\) just outside the conductor’s surface. Consider a small cylindrical Gaussian surface. One half of this cylinder is embedded inside the conductor. Since electric field is normal to the surface of the conductor, the curved part of the cylinder has zero electric flux. Also inside the conductor, the electric field is zero. Hence the bottom flat part of the Gaussian surface has no electric flux. Therefore the top flat surface alone contributes to the electric flux. The electric field is parallel to the area vector and the total charge inside the surface is σA. By applying Gauss’s law,
EA = \(\frac { σA }{{ ε }{0}}\)
In vector from, \(\vec { E } \) = \(\frac { σ }{{ ε }{0}}\) \(\hat{n} \)
Here n represents the unit vector outward normal to the surface of the conductor. Suppose σ < 0, then electric field points inward perpendicular to the surface.

(iv) The electrostatic potential has the same value on the surface and inside of the conductor. We know that the conductor has no parallel electric component on the surface which means that charges can be moved on the surface without doing any work. This is possible only if the electrostatic potential is constant at all points on the surface and there is no potential difference between any two points on the surface. Since the electric field is zero inside the conductor, the potential is the same as the surface of the conductor. Thus at electrostatic equilibrium, the conductor is always at equipotential.

Question 16.
Explain the process of electrostatic induction.
Answer:
Whenever a charged rod is touched by another conductor, charges start to flow from charged rod to the conductor. This type of charging without
actual contact is called electrostatic induction:
(i) Consider an uncharged (neutral) conducting sphere at rest on an insulating stand. Suppose a negatively charged rod is brought near the conductor without touching it, as shown in figure (a). The negative charge of the rod repels the electrons in the conductor to the opposite side.

Various steps in electrostatic induction
As a result, positive charges are induced near the region of the charged rod while negative charges on the farther side. Before introducing the charged rod, the free electrons were distributed uniformly on the surface
of the conductor and the net charge is zero. Once the charged rod is brought near the conductor, the distribution is no longer uniform with more electrons located on the farther side of the rod and positive charges are located closer to the rod. But the total charge is zero.

(ii) Now the conducting sphere is connected to the ground through a conducting wire. This is called grounding. Since the ground can always receive any amount of electrons, grounding removes the electron from the conducting sphere. Note that positive charges will not flow to the ground because they are attracted by the negative charges of the rod (figure (b)).

(iii) When the grounding wire is removed from the conductor, the positive charges remain near the charged rod (figure (c)).

(iv) Now the charged rod is taken away from the conductor. As soon as the charged rod is removed, the positive charge gets distributed uniformly on the surface of the conductor (figure (d)). By this process, the neutral conducting sphere becomes positively charged.

Question 17.
Explain dielectrics in detail and how an electric field is induced inside a dielectric.
Answer:
Induced Electric field inside the dielectric:
When an external electric field is applied on a conductor, the charges are aligned in such a way that an internal electric field is created which cancels the external electric field. But in the case of a dielectric, which has no free electrons, the external electric field only realigns the charges so that an internal electric field is produced.

The magnitude of the internal electric field is smaller than that of external electric field. Therefore the net electric field inside the dielectric is not zero but is parallel to an external electric field with magnitude less than that of the external electric field. For example, let us consider a rectangular dielectric slab placed between two oppositely charged plates (capacitor) as shown in the figure.

The uniform electric field between the plates Induced electric field lines inside the dielectric acts as an external electric field \(\vec { E } \)_ext which polarizes the dielectric placed between plates. The positive charges are induced on one side surface and negative charges are induced on the other side of surface But inside the dielectric, the net charge is zero even in a small volume. So the dielectric in the external field is equivalent to two oppositely charged sheets with the surface charge densities +σb and -σb. These charges are called bound charges. They are not free to move like free electrons in conductors. This is shown in the figure.

(a) Balloon sticks to the wall
(b) Polarisation of wall due to the electric field created by the balloon
For example, the charged balloon after rubbing sticks onto a wall. The reason is that the negatively charged balloon is brought near the wall, it polarizes opposite charges on the surface of the wall, which attracts the balloon.

Question 18.
Obtain the expression for capacitance for a parallel plate capacitor.
Answer:
Capacitance of a parallel plate capacitor:
Consider a capacitor with two parallel plates each of cross-sectional area A and separated by a distance d. The electric field between two infinite parallel plates is uniform and is given by E = \(\frac { σ }{{ ε }{0}}\) where σ is the surface charge density on the plates σ = \(\frac { Q }{ A }\) .If the separation distance d is very much smaller than the size of the plate (d² << A), then the above result is used even for finite-sized
parallel plate capacitor.

Capacitance of a parallel plate capacitor
The electric field between the plates is
E = \(\frac { Q }{{ Aε }{0}}\) ….. (1)
Since the electric field is unifonn, the electric potential between the plates having separation d is given by
V = Ed = \(\frac { Qd }{{ Aε }{0}}\) ….. (2)
Therefore the capacitance of the capacitor is given by
C = \(\frac { Q }{ V }\) = \(\frac{\mathrm{Q}}{\left(\frac{\mathrm{Q} d}{\mathrm{A} \varepsilon_{0}}\right)}\) = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) ….. (3)
From equation (3), it is evident that capacitance is directly
proportional to the area of cross section and is inversely proportional to the distance between the plates. This can be understood from the following.
(i) If the area of cross-section of the capacitor plates is increased, more charges can be distributed for the same potential difference. As a result, the capacitance is increased.

(ii) If the distance d between the two plates is reduced, the potential difference between the plates (V = Ed) decreases with E constant.

Question 19.
Obtain the expression for energy stored in the parallel plate capacitor.
Answer:
Energy stored in the capacitor:
Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor. To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = VdQ ….. (1)
Where V = \(\frac { Q }{ C }\)
The total work done to charge a capacitor is

This work done is stored as electrostatic potential energy (U_E) in the capacitor.
U_E = \(\frac {{ Q }^{2}}{ 2C }\) = \(\frac { 1 }{ 2 }\) CV² (∴ Q = CV) ….. (3)
where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor? To understand this question, the equation (3) is rewritten as follows using the results
C = \(\frac{\varepsilon_{0} \mathrm{A}}{d}\) and V = Ed
U_E = \(\frac { 1 }{ 2 }\) \(\left(\frac{\varepsilon_{0} \mathrm{A}}{d}\right)\) (Ed)² = \(\frac { 1 }{ 2 }\) ε₀(Ad)² …… (4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit volume of space is defined as energy density \(\overline { Volume } \). Frome equation (4) we get
u_E = \(\frac { 1 }{ 2 }\) ε₀E²
From equation (5), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

Question 20.
Explain in detail the effect of a dielectric placed in a parallel plate capacitor.
Answer:
(i) When the capacitor is disconnected from the battery:
Consider a capacitor with two parallel plates each of cross-sectional area A and are separated by a distance d. The capacitor is charged by a battery of voltage V₀ and the charge stored is Q₀. The capacitance of the capacitor without the dielectric is
C₀ = \(\frac {{ Q }_{0}}{{ V }_{0}}\) ….. (1)
The battery is then disconnected from the capacitor and the dielectric is inserted between the plates. The introduction of dielectric between the plates will decrease the electric field. Experimentally it is found that the modified electric field is given by

(a) Capacitor is charged with a battery
(b) Dielectric is inserted after the battery is disconnected
E = \(\frac {{ E }_{0}}{{ ε }_{r}}\) …… (2)
Here E₀ is the electric field inside the capacitors when there is no dielectric and ε_r is the relative permeability of the dielectric or simply known as the dielectric constant. Since ε_r > 1, the electric field E < E₀. As a result, the electrostatic potential difference between the plates (V = Ed) is also reduced. But at the same time, the charge Q₀ will remain constant once the battery is disconnected. Hence the new potential difference is
V = Ed = \(\frac {{ E }_{0}}{{ ε }_{r}}\)d = \(\frac {{ V }_{0}}{{ ε }_{r}}\) ….. (3)
We know that capacitance is inversely proportional to the potential difference. Therefore as V decreases, C increases. Thus new capacitance in the presence of a dielectric is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (4)
Since ε_r > 1, we have C > C₀. Thus insertion of the dielectric constant ε_r increases the capacitance. Using equation,
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
C = \(\frac{\varepsilon_{r} \varepsilon_{o} A}{d}\) = \(\frac { εA }{ d }\) …… (5)
where ε = ε_rε₀ is the permittivity of the dielectric medium. The energy stored in the capacitor before the insertion of a dielectric is given by U₀ = \(\frac { 1 }{ 2 }\) \(\frac{\mathrm{Q}_{0}^{2}}{\mathrm{C}_{0}}\) ….. (6)
After the dielectric is inserted, the charge Q₀ remains constant but the capacitance is increased. As a result, the stored energy is decreased.

Since ε_r> 1 we get U < U₀. There is a decrease in energy because, when the dielectric is inserted, the capacitor spends some energy in pulling the dielectric inside.

(ii) When the battery remains connected to the capacitor: Let us now consider what happens when the battery of voltage V₀ remains connected to the capacitor when the dielectric is inserted into the capacitor.
The potential difference V₀ across the plates remains constant. But it is found experimentally (first shown by Faraday) that when dielectric is inserted, the charge stored in the capacitor is increased by a factor ε_r.

(a) Capacitor is charged through a battery
(b) Dielectric is inserted when the battery is connected.
Q = ε_rQ₀ ….. (1)
Due to this increased charge, the capacitance is also increased. The new capacitance is
C = \(\frac {{ Q }_{0}}{ V }\) = ε_r \(\frac {{ Q }_{0}}{{ V }_{0}}\) = ε_r C₀ …… (2)
However the reason for the increase in capacitance in this case when the battery remains connected is different from the case when the battery is disconnected before introducing the dielectric.
Now, C₀ = \(\frac {{ ε }{_0}A}{ d }\) and, C = \(\frac { εA }{ d }\) …… (3)
U₀ = \(\frac { 1 }{ 2 }\) C₀ \({ V }_{ 0 }^{ 2 }\) ….. (4)
Note that here we have not used the expression
U₀ = \(\frac { 1 }{ 2 }\)\({{ V }_{ 0 }^{ 2 }}{{C}_{0}}\)
because here, both charge and capacitance are changed, whereas in equation 4, V₀ remains constant. After the dielectric is inserted, the capacitance is increased; hence the stored energy is also increased.
U = \(\frac { 1 }{ 2 }\) \({ CV }_{ 0 }^{ 2 }\) = \(\frac { 1 }{ 2 }\) ε_r \({ CV }_{ 0 }^{ 2 }\) = ε_r U₀
Since e_r > 1 we have U > U₀
It may be noted here that since voltage between the capacitor V₀ is constant, the electric field between the plates also remains constant.

Question 21.
Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.
Answer:
Capacitor in series and parallel:
(i) Capacitor in series:
Consider three capacitors of capacitance C₁, C₂ and C₃ connected in series with a battery of voltage V as shown in the figure (a).
As soon as the battery is connected to the capacitors in series, the electrons of charge -Q are transferred from negative terminal to the right plate of C₃which pushes the electrons of same amount -Q from left plate of C₃ to the right plate of C₂ due to electrostatic induction. Similarly, the left plate of C₂ pushes the charges of Q to the right plate of which induces the positive charge +Q on the left plate of C₁ At the same time, electrons of charge -Q are transferred from left plate of C₁ to positive terminal of the battery.

By these processes, each capacitor stores the same amount of charge Q. The capacitances of the capacitors are in general different, so that the voltage across each capacitor is also different and are denoted as V₁, V₂ and V₃ respectively.
The total voltage across each capacitor must be equal to the voltage of the a battery.
V = V₁ + V₂ + V₃ ….. (1)
Since Q = CV, we have V = \(\frac { Q }{{ C }_{1}}\) + \(\frac { Q }{{ C }_{2}}\) + \(\frac { Q }{{ C }_{3}}\)
Q = \(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \) ….. (2)
If three capacitors in series are considered to form an equivalent single capacitor Cs shown in figure (b), then we have V = \(\frac { Q }{{ C }_{s}}\)
Substituting this expression into equation (2) we get
V = \(\frac { Q }{{ C }_{s}}\) = Q\(\left( \frac { 1 }{ { C }_{ 1 } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } \right) \)
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) + \(\frac { 1 }{{ C }_{3}}\) ….. (3)
Thus, the inverse of the equivalent capacitance C_s of three capacitors connected in series is equal to the sum of the inverses of each capacitance. This equivalent capacitance C_s is always less than the smallest individual capacitance in the series.

(ii) Capacitance in parallel:
Consider three capacitors of capacitance C₁,C₂ and C₃ connected in parallel with a battery of voltage V as shown in figure (a).

Since corresponding sides of the capacitors are connected to the same positive and negative terminals of the battery, the voltage across each capacitor is equal to the battery’s voltage. Since capacitance of the capacitors is different, the charge stored in each capacitor is not the same. Let the charge stored in the three capacitors be Q₁,Q₂, and Q₂ respectively. According to the law of conservation of total charge, the sum of these three charges is equal to the charge Q transferred by the battery,
Q = Q₁ + Q₂ + Q₃ ….. (1)
Now, since Q = CV, we have
Q = C₁V + C₂ V + C₃ V ….. (2)
If these three capacitors are considered to form a single capacitance C_P which stores the total charge Q as shown in the figure (b), then we can write Q = CPV. Substituting this in equation (2), we get
C_p V = C₁ V + C₂ V + C₃ V
C_p = C₁ + C₂ + C₃
Thus, the equivalent capacitance of capacitors connected in parallel is equal to the sum of the individual capacitance. The equivalent capacitance C_p in a parallel connection is always greater than the largest individual capacitance. In a parallel connection, it is equivalent as area of each capacitance adds to give more effective area such that total capacitance increases.

Question 22.
Explain in detail how charges are distributed in a conductor, and the principle behind the lightning conductor.
Answer:
Distribution of charges in a conductor: Consider two conducting spheres A and B of radii r₁ and r₂ respectively connected to each other by a thin conducting wire as shown in the figure. The distance between the spheres is much greater than the radii of either spheres.

If a charge Q is introduced into any one of the spheres, this charge Q is redistributed into both the spheres such that the electrostatic potential is same in both the spheres. They are now uniformly charged and attain electrostatic equilibrium. Let q₁ be the charge residing on the surface of sphere A and q₂ is the charge residing on the surface of sphere B such that Q = q₁ + q₂ The charges are distributed only on the surface and there is no net charge inside the conductor. The electrostatic potential at the surface of the sphere A is given by
V_A = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) …. (1)
The electrostatic potential at the surface of the sphere B is given by
V_B = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. 2)
The surface of the conductor is an equipotential. Since the spheres are connected by the conducting wire, the surfaces of both the spheres together form an equipotential surface. This implies that
V_A = V_B or \(\frac {{ q }_{ 1 }}{{ r }_{ 1 }}\) = \(\frac {{ q }_{ 2 }}{{ r }_{ 2 }}\) ….. (3)
Let us take the charge density on the surface of sphere A is σ₁ and charge density on the surface of sphere B is σ₁. This implies that q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₁ and q₁ = \({ 4\pi r }_{ 1 }^{ 2 }\)σ₂. Substituting these values into equation (3), we get
σ₁ r₁ = σ₂r₂ ….. (4)
from which we conclude that
σr = constant …. (5)
Thus the surface charge density o is inversely proportional to the radius of the sphere. For a smaller radius, the charge density will be larger and vice versa.

Lightning arrester or lightning conductor:
This is a device used to protect tall buildings from lightning strikes. It works on the principle of action at points or corona discharge. The device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle.

The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike. Since the induced charge density on thin sharp spke is large, it results in a corona discharge.

This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud. The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the Earth. The lightning arrester does not stop the •lightning; rather it divers the lightning to the ground safety.

Question 23.
Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction:
A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor.

Two comb shaped metallic conductors E and D are fixed near the pulleys. The comb D is maintained at a positive potential of 104 V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working:
Due to the high electric field near comb D, air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 10⁷ which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

Samacheer Kalvi 12th Physics Electrostatics Numarical Problems

Question 1.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Solution:
Charge produced in each object q = 50 nC
q = 50 x 10^-9 C
Charge of electron (e) = 1.6 x 10^-9 C
Number of electron transferred, n = \(\frac { q }{ e }\) = \(\frac {{ 50 × 10 }^{-9}}{{ 1.6 × 10 }^{-19}}\)
=31. 25 × 10^-9 × 10¹⁹
n = 31.25 x 10¹⁰ electrons
Ans. n = 31.25 x 10¹⁰ electrons

Question 2.
The total number of electrons in the human body is typically in the order of 10²⁸. Suppose, due to some reason, you and your friend lost 1% of this number of electrons. Calculate the electrostatic force between you and your friend separated at a distance of 1 m. Compare this with your weight. Assume mass of each person is 60 kg and use point charge approximation.
Solution:
Number of electrons in the human body = 10²⁸
Number of electrons in me and my friend after lost of 1% = 10²⁸ x 1%
= 10²⁸ x \(\frac { 1 }{ 100 }\)
n = 10²⁶ electrons
Separation distance d = 1 m
Charge of each person q = 10²⁶ x 1.6 x 10^-19
q = 1.6 x 10⁷ C
Electrostatic force, F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{q_{1} q_{2}}{r^{2}}\) = \(\frac{9 \times 10^{9} \times 1.6 \times 10^{7} \times 1.6 \times 10^{7}}{1^{2}}\)
F = 2.304 x 10²⁴N
Mass of the person, M = 60 kg
Acceleration due to gravity, g = 9.8 ms^-2
Weight (W) = mg
= 60 x 9.8
W = 588 N
Comparison: Electrostatic force is equal to 3.92 x 10²¹ times of weight of the person.

Question 3.
Five identical charges Q are placed equidistant on a semicircle as shown in the figure. Another point charge q is kept at the center of the circle of radius R. Calculate the electrostatic force experienced by the charge q.

Solution:
Force acting on q due to Q₁ and Q₅ are opposite direction, so cancel to each other.
Force acting on q due to Q₃ is F₃ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{3}}{{ R }^{2}}\)
Force acting on q due to Q₂ and Q₄

Resolving in two component method:
(i) Vertical Component:
Q₂ Sin θ and Q₄ Sinθ are equal and opposite direction, so they are cancel to each other.

(ii) Horizontal Component:
Q₂ Sin θ and Q₄ cos θ are equal and same direction, so they can get added.
F₂₄ = F_2q + F_4q = F₂ cos 55° + F₄ cos 45°
F₂₄ = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ qQ }_{2}}{{ R }^{2}}\) cos 45° + \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac { qQ 4}{{ R }^{2}}\) cos 45°
Resultant net force F

Question 4.
Suppose a charge +q on Earth’s surface and another +q charge is placed on the surface of the Moon, (a) Calculate the value of q required to balance the gravitational attraction between Earth and Moon (b) Suppose the distance between the Moon and Earth is halved, would the charge q change? (Take m_E = 5.9 x 10²⁴ kg, m_M = 7.348 x 10²² kg)
Solution:
Mass of the Earth, M_E = 5.9 x 10²⁴ kg
Mass of the Moon, M_M = 7.348 x 10²² kg
Charge placed on the surface of Earth and Moon = q
(a) Required charge to balance the F_G between Earth and Moon
F_C = F_G (or) \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac {{ q }^{2}}{{ r }^{2}}\) = \(\frac{\mathrm{G} \mathrm{M}_{\mathrm{E}} \times \mathrm{M}_{\mathrm{M}}}{r^{2}}\)
q² = G × M_E × M_M × 4πε₀ = 320.97 × 10²⁵
q = \(\sqrt { 320.97\times { 10 }^{ 25 } } \) = 5.665 x 10¹³ = 5.67 x 10¹³ C

(b) The distance between Moon and Earth is

so q = 5.67 x 10¹³ C
There is no change.

Question 5.
Draw the free body diagram for the following charges as shown in the figure (a), (b) and (c).

Solution:

Question 6.
Consider an electron travelling with a speed V_Φ and entering into a uniform electric field \(\vec { E } \) which is perpendicular to \(\overrightarrow{\mathrm{V}_{0}}\) as shown in the Figure. Ignoring gravity, obtain the electron’s acceleration, velocity and position as functions of time.
Solution:

Speed of an electron = V₀
Uniform electric field = \(\vec { E } \)
(а) Electron’s acceleration:
Force on electron due to uniform electric field, F = Ee
Downward acceleration of electron due to electric field, a = \(\frac { F }{ m }\) = – \(\frac { eE }{ M }\)
Vector from, \(\vec { a } \) = – \(\frac { eE }{ M }\) \(\hat{j} \)

(b) Electron’s velocity:
Speed of electron in horizontal direction, u = V₀ From the equation of motion, V = u + at
V = V₀ \(\frac { eE }{ M }\) t
Vector from \(\vec { V } \) = V₀ \(\hat{j} \) – \(\frac { eE }{ M }\) t \(\hat{j} \)

(c) Electron’s position:
Position of electron, s = r
From equation of motion, r = V₀ t + \(\frac { 1 }{ 2 }\) \(\left(-\frac{e \mathrm{E}}{\mathrm{M}}\right)\) t²
r = V₀ t + \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)
Vector from,
\(\vec { r } \) = V₀ t \(\hat{j} \) \(\frac { 1 }{ 2 }\) \(\frac { eE }{ M }\) t² \(\hat{j} \)

Question 7.
A closed triangular box is kept in an electric field of magnitude E = 2 × 10³ N C^-1 as shown in the figure.

Calculate the electric flux through the (a) vertical rectangular surface (b) slanted surface and (c) entire surface.
Answer:
Electric field of magnitude E = 2 × 10³ NC^-1
(a) Vertical rectangular surface:
Rectangular area A= 5 × 10^-2 × 15 × 10^-2
A= 75 × 10^-24 m²
θ =180°
⇒ cos 180° = -1

Electric flux, Φ_v.s = EA cos θ
= 2 × 10³ × 75 × 10^-4 × cos 180°
= -150 × 10^-1
Φ_v.s = -15 Nm² C^-1

(b) Slanted surface:
cos θ = cos 60° = 0.5
sin θ = sin 30° = \(\frac { Opposite }{ hyp }\)

hyp = \(\frac {{ 5 × 10 }^{2}}{ 0.5 }\)
hyp = 0.1m
Area of slanted surface A₂ = (0.1 × 15 × 10^-2)
A₂ = 0.015 M²
Electric flux, Φ_v.s = EA = cos θ
= 2 × 10³ × 0.015 × cos 60°
= 2 × 10³ × 0.015 × 10³
= 0.015 × 10³
Φ_v.s = 15 Nm² C^-1
Horizontal surface
θ = 90° ; cos 90° = 0
Electric flux, Φ_H.S = E. A₃ Cos 90° = 0

(c) Entire surface:
Φ_Total = Φ_V.S + Φ_S.S + Φ_H.S = -15 + 15 + 0
Φ_Total = 0

Question 8.
The electrostatic potential is given as a function of x in figure (a) and (b). Calculate the corresponding electric fields in regions A, B, C and D. Plot the electric field as a function of x for the figure (b).

Answer:
The relation between electric field and potential
E = – \(\frac { dv }{ dx }\)

(a) Region A :
dv = -3V ; dx = 0.2 m
Electric field, E_A = \(\frac { (-3) }{ 0.2 }\) = 15 V m^-1
Region B:
dv = 0V ; dx = 0.2 m
Electric field, E_B = \(\frac { 0 }{ 0.2 }\) = 0
Region C:
dv = 2V ; dx = 0.2 m
Electric field, E_C = \(\frac { -2 }{ 0.2 }\) = 10 V m^-1
Region D:
dv = -6V ; dx = 0.2 m
Electric field, E_D = \(-\left(\frac{-6}{0.2}\right)\) = 10 V m^-1 = 30 V m^-1

Electric field, E_A = 15 V m^-1
Electric field, E_B = 0
Electric field, E_C = \(\frac { (-3) }{ 0.2 }\) = 10 V m^-1
Electric field, E_D = 30 V m^-1

(b)

Question 9.
A spark plug in a bike or a car is used to ignite the air-fuel mixture in the engine. It consists of two electrodes separated by a gap of around 0.6 mm gap as shown in the figure.

To create the spark, an electric field of magnitude 3 x 10⁶Vm^-1 is required, (a) What potential difference must be applied to produce the spark? (b) If the gap is increased, does the potential difference increase, decrease or remains the same? (c) find the potential difference if the gap is 1 mm.
Answer:
Separation gap between two electrodes, d = 0.6 mm
d = 0.6 × 10^-3 m
Magnetude of electric field Electric field = E = 3 × 10⁶ V m^-1
Electric field E = \(\frac { V }{ d }\)
(a) Applied potential difference, V = E . d
= 3 × 10⁶ × 0.6 10^-13 = 1.8 × 10³
V = 1800 V

(b) From equation, V = E . d
If the gap (distance) between the electrodes increased, the potential difference also increases.

(c) Gap between the electrodes, d = 1mm = 1 x 10^-3 m
Potential difference, V = E.d
= 3 × 10⁶ × 1 × 10^-3 = 3 × 103
V = 3000 V

Question 10.
A point charge of +10 μC is placed at a distance of 20 cm from another identical point charge of +10 μC. A point charge of -2 μC is moved from point a to b as shown in the figure. Calculate the
change in potential energy of the system? Interpret your result.

q₁ = 10μC = 10 x 10^-6 C
q₂ = 2μC = -2 x 10^-6 C
distance, r = 5cm = 5 x 10^-2 m
Answer:
Change in potential energy,

= -36 × 1 × 10⁹ × 10^-12 × 10² = -36 × 10^-1
∆ U = -3.6 J

Negative sign implies that to move the charge -2pC no external work is required. System spends its stored energy to move the charge from point a to point b.
Ans:
∆ U = -3.6 J, negative sign implies that to move the charge -2μC no external work is required. System spends its stored energy to move the charge from point a to point b.

Question 11.
Calculate the resultant capacitances for each of the following combinations of capacitors.

Answer:

Parallel combination of capacitor 1 and 2
C_p = C₀ + C₀ = 2C₀
Series combination of capacitor C_p and 3
\(\frac { 1 }{{ C }_{S}}\) = \(\frac { 1 }{{ C }_{p}}\) + \(\frac { 1 }{{ C }_{3}}\) = \(\frac { 1 }{{ 2C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = (or) \(\frac { 1 }{{ C }_{S}}\) = \(\frac { 3 }{ 2 }\) C₀  (or)C_S = \(\frac { 2 }{ 3 }\) C₀ 

\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 1 }{{ C }_{1}}\) + \(\frac { 1 }{{ C }_{2}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 1 }{{ C }_{0}}\) (or)
\(\frac { 1 }{ { C }_{ { S }_{ 1 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 1 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 1 }{{ C }_{3}}\) + \(\frac { 1 }{{ C }_{4}}\) = \(\frac { 1 }{{ C }_{0}}\) + \(\frac { 1 }{{ C }_{0}}\) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }{0}}{ 2 }\)

\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination
C_p = \({ C }_{ { S }_{ 1 } }\) + \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\) + \(\frac {{ C }_{0}}{ 2 }\) (or) C_p = \(\frac {{ 2C }_{0}}{ 2 }\) C_p = C₀

(c) Capacitor 1, 2 and 3 are in parallel combination
C_p = C₀ + C₀ + C₀ = 3C₀
C_p = 3C₀

(d) Capacitar C₁ and C₂ are in combination

Similarly C₃ and C₄ are in series combination


\({ C }_{ { S }_{ 1 } }\) and \({ C }_{ { S }_{ 2 } }\) are in parallel combination across RS:

(e) Capacitor 1 and 2 are series combination

Similarly 3 and 4 are series combination
\(\frac { 1 }{ { C }_{ { S }_{ 2 } } } \) = \(\frac { 2 }{{ C }_{0}}\) (or) \({ C }_{ { S }_{ 2 } }\) = \(\frac {{ C }_{0}}{ 2 }\)
Three capacitors are in parallel combination

Question 12.
An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure.

(a) Calculate the time of flight for both electron and proton
(b) Suppose if a neutron is allowed to fall, what is the time of flight?
(c) Among the three, which one will reach the bottom first?
(Take m_p = 1.6 x 10^-27 kg, m_e= 9.1 x 10^-31 kg and g = 10 m s^-2)
Answer:
Potential difference between the parallel plates V = 5 V
Separation distance, h = 1 mm =1 x 10^-3 m
Mass of proton, mp = 1.6 x 10^-27 kg
Mass of proton, m =9.1 x 10^-31 kg
Charge of an a proton (or) electron, e— 1.6 x 10^-19 C
[u = 0; s = h]
From equation of motion, S = ut + \(\frac { 1 }{ 2 }\) at²
From equation of motion, h = \(\frac { 1 }{ 2 }\) at²
t = \(\sqrt { \frac { 2h }{ a } } \)
Acceleration of an electron due to electric field, a = \(\frac { F }{ m }\) = \(\frac { eE }{ m }\)
[E = \(\frac { V }{ d }\)]

(a) Time of flight for both electron and proton,

t_p = 63 ns……. (2)

(b) time of flight of neutron t_n = \(\sqrt { \frac { 2h }{ g } } \) = \(\sqrt{\frac{2 \times 1 \times 10^{3}}{10}}\) = \(\sqrt{0.2 \times 10^{-3}}\)
t_n = 0.0141 s = 14.1 x 10^-3 s
t_n = 14.1 x 10^-3 ms ……. (3)
(c) Compairision of values 1,2 and 3. The electron will reach the bottom first.

Question 13.
During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 x 10⁶ Vm^-1 ), lightning will occur.

(a) If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
(b) In a typical lightning phenomenon, around 25C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?
Answer:
(a) Electric field between the cloud and ground,
V = E.d
V= 3 x 10⁶ x 1000 = 3 x 10⁹V
(a) Electrons transfered from cloud to ground,
q = 25 C
Electron static potential energy,
U = \(\frac { 1 }{ 2 }\) CV²
[C = \(\frac { q }{ V }\)]
= \(\frac { 1 }{ 2 }\) qV = \(\frac { 1 }{ 2 }\) x 25 x 3 x 10⁹
U = 37.5 x 10⁹ J

Question 14.
For the given capacitor configuration
(a) Find the charges on each capacitor
(b) potential difference across them
(c) energy stored in each capacitor.

Answer:
Capacitor b and c in parallel combination
C_p = C_b + C_c = (6 + 2) μF = 8 μF
Capacitor a, c_p and d are in series combination, so the resulatant copacitance
\(\frac { 1 }{{ C }_{s}}\) = \(\frac { 1 }{{ C }_{a}}\) + \(\frac { 1 }{{ C }_{cp}}\) + \(\frac { 1 }{{ C }_{d}}\) = \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) + \(\frac { 1 }{ 8 }\) = \(\frac { 3 }{ 8 }\)
C_s  = \(\frac { 8 }{ 3 }\) μF

(a) Charge on each capacitor,
Charge on capacitor a, Q_a = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_a = 24 μC
Charge on capacitor, d, Q_d = C_s V = \(\frac { 8 }{ 3 }\) x 9
Q_d = 24 μC
Capacitor b and c in parallel
Charge on capacitor, b, Q_b = \(\frac { 6 }{ 3 }\) x 9 = 18
Q_b = 18 μC
Charge on capacitor, c, Q_c = \(\frac { 2 }{ 3 }\) x 9 = 6
Q_c = 6 μC

(b) Potential difference across each capacitor, V = \(\frac { q }{ C }\)
Capacitor C_a, V_a = \(\frac{ { q }_{a}}{{ C }_{a}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V
Capacitor C_b, V_b = \(\frac{ { q }_{b}}{{ C }_{b}}\) = \(\frac {{ 18 × 10 }^{6}}{{ 6 × 10 }^{6}}\) = 3 V
Capacitor C_c, V_c = \(\frac{ { q }_{c}}{{ C }_{c}}\) = \(\frac {{ 6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) = 3 V
Capacitor C_d, V_d = \(\frac{ { q }_{d}}{{ C }_{d}}\) = \(\frac {{ 24 × 10 }^{6}}{{ 8 × 10 }^{6}}\) = 3 V

(c) Energy stores in a capacitor, U = \(\frac { 1 }{ 2 }\) CV²
Energy in capacitor C_a, U_a = \(\frac { 1 }{ 2 }\) C_a \({ V }_{ a }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ
Capacitor C_b, U_b = \(\frac { 1 }{ 2 }\) C_b \({ V }_{ b }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 6 x 10^-6 x (3)²
U_a = 27 μJ
C_c, U_c = \(\frac { 1 }{ 2 }\) C_c \({ V }_{ c }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 2 x 10^-6 x (3)²
U_a = 9 μJ
C_d, U_d = \(\frac { 1 }{ 2 }\) C_d \({ V }_{ d }^{ 2 }\) = \(\frac { 1 }{ 2 }\) x 8 x 10^-6 x (3)²
U_a = 36 μJ

Question 15.
Capacitors P and Q have identical cross sectional areas A and separation d. The space between the capacitors is filled with a dielectric of dielectric constant er as shown in the figure. Calculate the capacitance of capacitors P and Q.

Answer:
Cross-sectional area of parallel plate capacitor = A
Each area of different medium between parallel plate capacitor = \(\frac { A }{ 2 }\)
Separation distance = d
Capacitance of parallel plate capacitor, C = \(\frac { εA }{ d }\)
Air medium of dielectric constant, ε_r = 1
dielectric medium of dielectric constant = ε_r

Case 1:
Capacitance of air filled capacitor

Capacitance of dielectric filled capacitor

Capacitance of parallel plate capacitor

Case 2:
Each distance of different medium between the parallel plate capacitor = \(\frac { d }{ 2 }\)
Capacitance of dielectric filled capacitor

Capacitance of air filled capacitor,

Capacitance of parallel plate capacitor,

Samacheer Kalvi 12th Physics Electrostatics Additional Questions Solved

I. Multiple Choice Questions

Question 1.
When a solid body is negatively charged by friction, it means that the body has
(a) acquired excess of electrons
(b) lost some, problems
(c) acquired some electrons and lost a lesser number of protons
(d) lost some positive ions
Answer:
(a) acquired excess of electrons

Question 2.
A force of 0.01 N is exerted on a charge of 1.2 x 10^-5 G at a certain point. The electric field at that point is
(a) 5.3 x 10⁴ NC^-1
(b) 8.3 x 10^-4 NC^-1
(c) 5.3 x 10² NC^-1
(d) 8.3 x 10⁴ NC^-1
Answer:
(d) 8.3 x 10⁴ NC^-1
Hint:
E = \(\frac { F }{ q }\) = \(\frac { 0.01 }{{ 1.2 × 10 }^{-5}}\) = 8.3 x 10² NC^-1

Question 3.
The electric field intensity at a point 20 cm away from a charge of 2 x 10 5 C is
(a) 4.5 x 10⁶ NC^-1
(b) 3.5 x 10⁵ NC^-1
(c) 3.5 x 10⁶ NC^-1
(d) 4.5 x 10⁵ NC^-1
Answer:
(a) 4.5 x 10⁶ NC^-1
Hint:
E = \(\frac{q}{4 \pi \varepsilon_{0} r^{2}}\) = \(\frac{9 \times 10^{9} \times 2 \times 10^{-5}}{(0.2)^{2}}\) = 4.5 x 10⁶ NC^-1

Question 4.
How many electrons will have a charge of one coulomb?
(a) 6.25 x 10¹⁸
(b) 6.25 x 10¹⁹
(c) 1.6 x 10¹⁸
(d) 1.6 x 10¹⁹
Answer:
(a) 6.25 x 10¹⁸
Hint:
Number of electron, n = \(\frac { q }{ e }\) = \(\frac { 1 }{{ 1.6 × 10 }^{-19}}\) = 6.25 × 10¹⁸

Question 5.
The ratio of the force between two charges in air and that in a medium of dielectric constant K is
(a) K : 1
(b) 1 : K
(c) K² : 1
(d) 1 : K²
Answer:
(a) K : 1

Question 6.
The work done in moving a positive charge on an equipotential surface is
(a) finite and positive
(b) infinite
(c) finite and negative
(d) zero
Answer:
(d) zero

Question 7.
If a charge is moved against the coulomb force of an electric field.
(a) work is done by the electric field
(b) energy is used from some outside source
(c) the strength of the field is decreased
(d) the energy of the system is decreased
Answer:
(b) energy is used from some outside source

Question 8.
No current flows between two charged bodies when connected
(a) if they have the same capacitance
(b) if they have same quantity of charge
(c) if they have the same potential
(d) if they have the same charge density
Answer:
(c) if they have the same potential

Question 9.
Electric field lines about a negative point charge are
(a) circular, anticlockwise
(b) circular, clockwise
(c) radial, inwards
(d) radial, outwards
Answer:
(c) radial, inwards

Question 10.
Two plates are 1 cm apart and the potential difference between them is 10 V. The electric field between the plates is
(a) 10 NC^-1
(b) 250 NC^-1
(c) 500 N^-1
(d) 1000 NC^-1
Answer:
(d) 1000 NC^-1
Hint:
E = \(\frac { V }{ d }\) = \(\frac { 10 }{{ 1 × 10 }^{-2}}\) = 8.3 x 10² NC^-1

Question 11.
At a large distance (r), the electric field due to a dipole varies as
(a) \(\frac { 1 }{ r }\)
(b) \(\frac { 1 }{{ r }^{2}}\)
(c) \(\frac { 1 }{{ r }^{3}}\)
(d) \(\frac { 1 }{{ r }^{4}}\)
Answer:
(c) \(\frac { 1 }{{ r }^{3}}\)

Question 12.
Two thin infinite parallel plates have uniform charge densities +c and -σ. The electric field in the space between then is
(a) \(\frac { σ }{{ 2ε }_{0}}\)
(b) \(\frac { σ }{{ ε }_{0}}\)
(c) \(\frac { 2σ }{{ 2ε }_{0}}\)
(d) Zero
Answer:
(b) \(\frac { σ }{{ ε }_{0}}\)

Question 13.
Two isolated, charged conducting spheres of radii R₁, and R₂ produce the same electric field near their surfaces. The ratio of electric potentials on their surfaces is-
(a) \(\frac {{ R }_{1}}{{ R }_{2}}\)
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)
(c) \(\frac { { R }_{ 1 }^{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
(d) \(\frac { { R }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 2 } } \)
Answer:
(b) \(\frac {{ R }_{2}}{{ R }_{1}}\)

Question 14.
A 100 μF capacitor is to have an energy content of 50 J in order to operator a flash lamp. The voltage required to charge the capacitor is
(a) 500 V
(b) 1000 V
(c) 1500 V
(d) 2000 V
Answer:
(b) 1000 V
Hint:

Question 15.
A 1 μF capacitor is placed in parallel with a 2 μF capacitor across a 100 V supply. The total charge on the system is
(a) \(\frac { 100 }{ 3 }\) μC
(b) 100 μC
(c) 150 μC
(d) 300 μC
Answer:
(d) 300 μC
Hint:
Equivalent capacitor = 1 + 2 = 3 μF
Total charge, q = CV = 3 x 100 = 300 μF

Question 16.
A parallel plate capacitor of capacitance 100 μF is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes
(a) 250 V
(b) 500 V
(c) 1000V
(d) 2000 V
Answer:
(a) 250 V
Hint:
Here, C’ = 2C, since the charge remains the same.
q = C’V’ = CV ⇒ V = \(\frac { CV }{ 2C }\) = \(\frac { 500 }{ 2 }\) = 250 V

Question 17.
A point charge q is placed at the midpoint of a cube of side L. The electric flux emerging from the cube is ‘
(a) \(\frac { q }{{ ε }_{0}}\)
(b) \(\frac { q }{{ 6Lε }_{0}}\)
(c) \(\frac { 6Lq }{{ ε }_{0}}\)
(d) zero
Answer:
(a) \(\frac { q }{{ ε }_{0}}\)

Question 18.
The capacitor C of a spherical conductor of radius R is proportional to
(a) R²
(b) R
(c) R^-1
(d) R⁰
Answer:
(b) R

Question 19.
Energy of a capacitor of capacitance C, when subjected to a potential V, is given by
(a) \(\frac { 1 }{ 2 }\) CV²
(b) \(\frac { 1 }{ 2 }\) C²V
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { 1 }{ 2 }\) \(\frac { C }{ V }\)
Answer:
(a) \(\frac { 1 }{ 2 }\) CV²

Question 20.
The electric field due to a dipole at a distance r from its centre is proportional to
(a) \(\frac { 1 }{{ r }^{3/2}}\)
(b) \(\frac { 1 }{{ r }^{3}}\)
(c) \(\frac { 1 }{ r }\)
(d) \(\frac { 1 }{{ r }^{3}}\)
Answer:
(b) \(\frac { 1 }{{ r }^{3}}\)

Question 21.
A point charge q is rotating around a charge Q in a circle of radius r. The workdone on it by the coulomb force is
(a) 2πrq
(b) 2πQq
(c) \(\frac { Q }{{ 2ε }^{0}r}\)
(d) zero
Answer:
(d) zero

Question 22.
The workdone in rotating an electric dipole of moment P in an electric field E through an angle 0 from the direction of the field is
(a) pE (1 – cos θ)
(b) 2pE
(c) zero
(d) -pE cos θ
Answer:
(a) pE (1 – cos θ)
Hint:
W = pE(cos θ₀ – cos θ)
[θ₀ = cos 0, cos 0 = 1]
W = pE(1 – cos θ)

Question 23.
Capacitance of a parallel plate capacitor can be increased by
(a) increasing the distance between the plates
(b) increasing the thickness of the plates
(c) decreasing the thickness of the plates
(d) decreasing the distance between the plates
Answer:
(d) decreasing the distance between the plates

Question 24.
Two charges are placed in vacuum at a distance d apart. The force between them is F. If a medium of dielectric constant 2 is introduced between them, the force will now be
(a) 4F
(b) 2F
(c) F/2
(d) F/4
Answer:
(d) F/4

Question 25.
An electric charge is placed at the centre of a cube of side a. The electric flux through one of its faces will be
(a) \(\frac { q }{{ 6ε }^{0}}\)
(b) \(\frac { q }{ { ε }_{ 0 }{ a }^{ 2 } } \)
(c) \(\frac { q }{ { 4πε }_{ 0 }{ a }^{ 2 } } \)
(a) \(\frac { q }{{ ε }^{0}}\)
Answer:
(a) \(\frac { q }{{ 6ε }^{0}}\)
Hint:
According to Gauss’s law, the electric flux through the cube is \(\frac { q }{{ ε }^{0}}\). Since there are six faces, the flux through one face is \(\frac { q }{{ 6ε }^{0}}\).

Question 26.
The electric field in the region between two concentric charged spherical shells-
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 27.
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 V. The potential at the centre of the sphere is-
(a) 800 V
(b) zero
(c) 8 V
(d) 80 V
Answer:
(d) 80 V

Question 28.
A 4 μF capacitor is charged to 400 V and then its plates are joined through a resistance of 1 K Ω. The heat produced in the resistance is-
(a) 0.16 J
(b) 0.32 J
(c) 0.64 J
(d) 1.28 J
Answer:
(b) 0.32 J
Hint:
The energy stored in capacitor is converted into heat
U = H = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 4 x 10^-6 x (400)² = 0.32 J

Question 29.
The workdone in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is-
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) zero
(c) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}}\)
(d) infinite
Answer:
(b) zero
Hint:
The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 30.
Two plates are 2 cm apart. If a potential difference of 10 V is applied between them. The electric field between the plates will be
(a) 20 NC^-1
(b) 500 NC^-1
(c) 5 NC^-1
(d) 250 NC^-1
Answer:
(b) 500 NC^-1
Hint:
\(\frac { V }{ d }\) = \(\frac { 10 }{{ 2 ×10 }^{-2}}\) 500 NC^-1

Question 31.
The capacitance of a parallel plate capacitor does not depend on
(a) area of the plates
(b) metal of the plates
(c) medium between the plates
(d) distance between the plates
Answer:
(b) metal of the plates

Question 32.
A capacitor of 50 μF is charged to 10 volts. Its energy in joules is
(a) 2.5 x 10^-3
(b) 5 x 10^-3
(c) 10 x 10^-4
(d) 2.5 x 10^-4
Answer:
(a) 2.5 x 10^-3
Hint:
U = \(\frac { 1 }{ 2 }\) CV² = \(\frac { 1 }{ 2 }\) x 50 x 10^-6 x (10)² = 2.5 x 10^-3 J

Question 33.
A cube of side b has a charge q at each of its vertices. The electric field due to this charge distribution at the centre of the cube is
(a) \(\frac { q }{{b}^{ 2 }}\)
(b) \(\frac { q }{{2b}^{ 2 }}\)
(c) \(\frac { 32q }{{b}^{ 2 }}\)
(d) zero
Answer:(d) zero
Hint:
There
is an equal charge at diagonally opposite comer. The fields due the these at the centre cancel out. Therefore, the net field at the centre is zero.

Question 34.
Total electric fulx coming out of a unit positive charge put in air is
(a) ε₀
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)
(c) (4πε₀)^-1
(d) 4πε₀
Answer:
(b) \({ \varepsilon }_{ 0 }^{ -1 }\)

Question 35.
Electron volt (eV) is a unit of
(a) energy
(b) potential
(c) current
(d) charge
Answer:
(a) energy

Question 36.
A point Q lies on the perpendicular bisector of an electric dipole of dipole moment P. If the distance of Q from the dipole is r, then the electric field at Q is proportional to-
(a) p^-1 and r^-2
(b) p and r^-2
(c) p and r^-3
(d) p² and r^-3
Answer:
(c) p and r^-3

Question 37.
A hollow insulated conducting sphere is given a positive charge of 10 μC. What will be the electric field at the centre of the sphere is its radius is 2 metres?
(a) zero
(b) 8 μCm^-2
(c) 20 μCm^-2
(d) 5 μCm^-2
Answer:
(d) zero

Question 38.
A particle of charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is-
(a) qE²y
(b) q²Ey
(c) qEy²
(d) qEy
Answer:
(d) qEy
Hint:
Force on the particle = qE
KE = Workdone by the force = F.y = qEy

Question 39.
Dielectric constant of metals is-
(a) 1
(b) greater then 1
(c) zero
(d) infinite
Answer:
(d) infinite

Question 40.
When a positively charged conductor is earth connected
(a) protons flow from the conductor to the earth
(b) electrons flow from the earth to the conductor
(c) electrons flow from the conductor to the earth
(d) no charge flow occurs
Answer:
(b) electrons flow from the earth to the conductor

Question 41.
The SI unit of electric flux is
(a) volt metre²
(b) newton per coulomb
(c) volt metre
(d) joule per coulomb
Answer:
(c) volt metre

Question 42.
Twenty seven water drops of the same size are charged to the same potential. If they are combined to form a big drop, the ratio of the potential of the big drop to that of a small drop is-
(a) 3
(b) 6
(c) 9
(d) 27
Answer:
(c) 9
Hint:
V’ = n^2/3 V
⇒ \(\frac { V’ }{ V }\) = (27)^2/3 = 9

Question 43.
A point charge +q is placed at the midpoint of a cube of side l. The electric flux emerging ’ from the cube is-
(a) \(\frac { q }{{ ε }^{0}}\)
(b) \(\frac {{ 6ql }^{2}}{{ ε }^{0}}\)
(c) \(\frac { q }{ { 6l }^{ 2 }{ { ε }^{ 0 } } } \)
(d) \(\frac { { C }^{ 2 }{ V }^{ 2 } }{ 2 } \)
Answer:
(a) \(\frac { q }{{ ε }^{0}}\)

Question 44.
The energy stored in a capacitor of capacitance C, having a potential difference V between the plates, is-
Answer:
(c)

Question 45.
The electric potential at the centre of a charged conductor is-
(a) zero
(b) twice that on the surface
(c) half that on the surface
(d) same as that on the surface
Answer:
(d) same as that on the surface

Question 46.
The energy stored in a capacitor is given by
(a) qV
(b) \(\frac { 1 }{ 2 }\)qV
(c) \(\frac { 1 }{ 2 }\) CV
(d) \(\frac { q }{ 2C }\)
Answer:
(b) \(\frac { 1 }{ 2 }\)qV

Question 47.
The unit of permitivity of free space so is
(a) coulomb/newton-metre
(b) newton-metre²/coulomb²
(c) coulomb²/newton-metre²
(d) coulomb/(newton-metre)²
Answer:
(c) coulomb²/newton-metre²

Question 48.
An electric dipole has the magnitude of its charge as q and its dipole moment is p. It is placed in a uniform electric field E. It its dipole moment is along the direction of the field, the force on it and its potential energy are, respectively.
(a) 2qE and minimum
(b) qE and pE
(c) zero and minimum
(d) qE and maximum
Answer:
(c) zero and minimum
Hint:
Potential energy, U = -pE cos θ
For q = 0°; U = -pE, which is minimum.

Question 49.
An electric dipole of moment \(\vec { P } \) is lying along a uniform electric field \(\vec { E } \) . The workdone in rotating the dipole by 90° is
(a) \(\frac { pE }{ 2 }\)
(b) 2pE
(c) pE
(d) √2pE
Answer:
(c) pE

Question 50.
Aparallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates
(a) does not charge
(b) becomes zero
(c) increases
(d) decreases
Answer:
(c) increases

Question 51.
When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance
(a) increases K times
(b) increases K^-1 times
(c) decreases K times
(d) remains constant
Answer:
(c) decreases K times

Question 52.
A comb run through one’s dry hair attracts small bits of paper. This is due to the fact that
(a) comb is a good conductor
(b) paper is a good conductor
(c) the atoms in the paper gets polarised by the charged comb
(d) the comb posseses magnetic properties
Answer:
(c) the atoms in the paper gets polarised by the charged comb

Question 53.
Which of the following is not a property of equipotential surfaces?
(a) they do not cross each other
(b) they are concentric spheres for uniform electric field
(c) the rate of change of potential with distance on them is zero
(d) they can be imaginary spheres.
Answer:
(b) they are concentric spheres for uniform electric field

Question 54.
A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will be
(a) reduced to half
(b) doubled
(c) becomes 4 times
(d) remains the same
Answer:
(d) remains the same

Question 55.
If the electric field in a region is given by \(\vec { E } \) = 5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \) , then the electric flux through a surface of area 20 units lying in the y-z plane will be-
(a) 20 units
(b) 80 units
(c) 100 units
(d) 180 units
Answer:
(c) 100 units
Hints:
The area vector \(\vec { A } \) = 20\(\hat{j} \); \(\vec { E } \) = (5\(\hat{j} \) + 4\(\hat{j} \) + 9\(\hat{k} \))
Flux (Φ) = \(\vec { E } \) – \(\vec { A } \) = 5 x 20 =100 units

Question 56.
A, B and C are three points in a uniform electric field. The electric potential is-

(a) maximum at A
(b) maximum at B
(c) maximum at B
(d) same at all the three points A, B and C
Answer:
(b) maximum at B
Hint:
The potential decreases in the direction of the field. Therefore V_B > V_C>C_A.

Question 57.
A conducting sphere of radius R is give a charge Q. The electric potential and the electric field at the centre of the sphere are, respectively-
(a) zero, \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R }^{ 2 } } \)
(b) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \)
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero
(d) zero,zero
Answer:
(c) \(\frac { Q }{ { 4\pi ε }_{ 0 }{ R } } \), zero.

II. Fill in the blanks

Question 1.
A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences …………………
Answer:
neither a net force nor a torque

Question 2.
The unit of permittivity is…………………
Answer:
C²N^-1m^-2

Question 3.
The branch of physics which deals with static electric charges or charges at rest is …………………
Answer:
electrostatics

Question 4.
The charges in a electrostatics field are analogous to ………………… in a gravitational field.
Answer:
mass

Question 5.
The substances which acquire charges on rubbing are said to be …………………
Answer:
electrified

Question 6.
Electron means …………………
Answer:
amber

Question 7.
A glass rod rubbed with a silk cloth. Glass rod and silk cloth acquires …………………
Answer:
positive and negative charge respectvely .

Question 8.
When ebonite rod is rubbed with fur, ebonite rod and fur acquires …………………
Answer:
negative and positive charge respectively

Question 9.
………………… termed the classification of positive and negative charges.
Answer:
Franklin

Question 10.
Applications such as electrostatic point spraying and powder coating, are based on the property of ………………… between charged bodies.
Answer:
attraction and repulsion

Question 11.
Bodies which allow the charge to pass through them are called …………………
Answer:
conductor

Question 12.
Bodies which do not allow the charge to pass through them are called …………………
Answer:
insulators

Question 13.
The unit of electric charge is …………………
Answer:
coulomb

Question 14.
Total charge in an isolated sysem …………………
Answer:
remains a constant

Question 15.
The force between two charged bodies was studied by …………………
Answer:
coulomb

Question 16.
The unit of permittivity in free space (s0) is …………………
Answer:
C²N^-1m^-2

Question 17.
The value of s, for air or vacuum is …………………
Answer:1

Question 18.
Charges can neither be created nor be destroyed is the statement of law of conservation of …………………
Answer:
charge

Question 19.
The space around the test charge, in which it experiences a force is known as field …………………
Answer:
electric

Question 20.
Electric field at a point is measued in terms of …………………
Answer:
electric field intensity

Question 21.
The unit of electric field in tensity is …………………
Answer:
NC^-1.

Question 22.
The lines of force are far apart, when electric field E is …………………
Answer:
small

Question 23.
The lines of force are close together, when electric field E is …………………
Answer:
large

Question 24.
Electric dipole moment …………………
Answer:
P = 2qd

Question 25.
Torque experienced by electric dipole is …………………
Answer:
x = PE sin θ

Question 26.
An electric dipole placed in a non-uniform electric field at an angle 0 experiences …………………
Answer:
both torque and force

Question 27.
When thee dipole is aligned parallel to the field, its electric potential energy is …………………
Answer:
u = -PE

Question 28.
Change of potential with distance is known as …………………
Answer:
potential distance

Question 29.
The number of electric lines of force crossing through the given area is …………………
Answer:
electric flux

Question 30.
The process of isolating a certain region of space from external field is called …………………
Answer:
electrostatic shielding

Question 31.
Capacitor is a device to store …………………
Answer:
charge

Question 32.
The charge density in maximum at …………………
Answer:
pointed

Question 33.
The principle made use of in lightning arrestor is …………………
Answer:
action of points

Question 34.
Van de Graaff generator producers large electrostatic potential difference of the order of …………………
Answer:
10⁷ V

III. Match the following

Question 1.

Answer:
(i) → (d)
(ii) → (a)
(iii) → (b)
(iv) → (c)

Question 2.

Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 3.

Answer:
(i) → (b)
(ii) → (d)
(iii) → (a)
(iv) → (c)

Question 4.

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 is 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: Electric lines of force cross each other.
Reason: Electric field at a point supermpose to give one resultant electric field.
Answer:
(e) Both assertion and reason are true but the reason is not correct explanation of the assertion.
Explanation: If electric lines of forces cross each other, then the electric field at the point of intersection will have two direction simultaneously which is not possible physically.

Question 2.
Assertion: Charge is quantized.
Reason: Charge, which is less than 1 C is not possible.
Answer:
(c) If assertion is true but reason is false.
Explanation: Q = ±ne and charge lesser than 1 C is possible.

Question 3.
Assertion:
A point charge is brought in an electric field. The field at a nearby point will increase, whatever be the nature of the charge.
Reason: The electric field is independent of the nature of charge.
(d) If the assertion and reason both are false.
Explanation: Electric field at the nearby-point will be resultant of existing field and field due to the charge brought. It may increase or decrease if the charge is positive or negative depending on the position of the point with respect to the charge brought.

Question 4.
Assertion: The tyre’s of aircraft’s are slightly conducting.
Reason: If a conductor is connected to ground, the extra charge induced on conductor will flow to ground.
Answer:
(b) Both assertion and reason are true but the reason is not correct explanation of the assertion.
Explanation: During take off and landing, the friction between treys and the run way may cause electrification of treys. Due to conducting to a ground and election sparking is avoided.

Question 5.
Assertion: The lightening conductor at the top of a high building has sharp ends.
Reason: The surface density of charge at sharp points is very high, resulting in setting up of electric wind.
Answer:
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion.

Samacheer Kalvi 12th Physics Electrostatics Short Answer Questions

Question 1.
What is meant by triboelectric charging?
Answer:
Charging the objects through rubbing is called triboelectric charging.

Question 2.
What is meant by conservation of total charges?
Answer:
The total electric charge in the universe is constant and charge can neither be created nor be destroyed. In any physical process, the net change in charge will always be zero.

Question 3.
State Gauss’s Law?
Answer:
Definition:
Gauss’s law states that if a charge Q is enclosed by an arbitrary closed surface, then the total electric flux OE through the closed surface is
Φ_E = \(\oint { \vec { E } } \) .d\(\vec { A } \) = \(\frac {{ q }_{encl}}{{ ε }_{0}}\)

Question 4.
What is meant by electrostatic shielding?
During lightning accompanied by a thunderstorm, it is always safer to sit inside a bus than in open ground or under a tree. The metal body of the bus provides electrostatic shielding, since the electric field inside is zero. During lightning, the charges flow through the body of the conductor to the ground with no effect on the person inside that bus.

Question 5.
What is meant by dielectric?
Answer:
A dielectric is a non-conducting material and has no free electrons. The electrons in a dielectric are bound within the atoms. Ebonite, glass and mica are some examples of dielectrics.

Question 6.
What are non-polar molecules? Give examples.
A non-polar molecule is one in which centers of positive and negative charges coincide. As a result, it has no permanent dipole moment. Examples of non-polar molecules are hydrogen (H₂), oxygen (O₂), and carbon dioxide (CO₂) etc.

Question 7.
What are polar molecules? Give examples.
Answer:
In polar molecules, the centers of the positive and negative charges are separated even in the absence of an external electric field. They have a permanent dipole moment.
The net dipole moment is zero in the absence of an external electric field. Examples of polar molecules are H₂O, N₂O, HCl, NH₃.

Question 8.
What is a capacitors?
Answer:
Capacitor is a device used to store electric charge and electrical energy. Capacitors are widely used in many electronic circuits and have applications in many areas of science and technology.

Samacheer Kalvi 12th Physics Electrostatics Long Answer Questions

Question 1.
Derive an expression for electric field due to the system of point charges?
Answer:
Electric field due to the system of point charges:
Suppose a number of point charges are distributed in space. To find the electric field at some point P due to this collection of point charges, superposition principle is used. The electric field at an arbitrary point due to a collection of point charges is simply equal to the vector sum of the electric fields created by the individual point charges. This is called superposition of electric fields.
Consider a collection of point charges q₁, q₂, q₃,…., q_n located at various points in space. The ‘ total electric field at some point P due to all these n charges is given by

Here r_1p, r_2p, r_3p,…., r_np, are the distance of the charges ₁, q₂, q₃,…., q_n from the point respectively. Also \(\hat{r} \)_1p + \(\hat{r} \)_2p + \(\hat{r} \)_3p,…., \(\hat{r} \)_np are the corresponding unit vectors directed from q₁, q₂, q₃,…., q_n tpo P.

Equation (2) can be re-written as,

For example in figure, the resultant electric field due to three point charges q₁, q₂, q₃ at point P is shown. Note that the relative lengths of the electric field vectors for the charges depend on relative distantes of the charges to the point P.

Question 2.
Derive an expression for electric flux of rectangular area placed in uniform electric field.
Answer:
(i) Electric flux for uniform Electric field:
Consider a uniform electric field in a region of space. Let uschoose an area A normal to the electric field lines as shown in figure (a). The electric flux for this case is
Φ_E = EA ….. (1)
Suppose the same area A is kept parallel to the uniform electric field, then no electric field lines pierce through the area A, as shown in figure (b). The electric flux for this case is zero.
Φ_E = 0 ….. (2)
If the area is inclined at an angle θ with the field, then the component of the electric field perpendicular to the area alone contributes to the electric flux. The electric field component parallel to the surface area will not contribute to the electric flux. This is shown in figure (c). For this case, the electric flux
ΦE = (E cosθ) A …(3)
Further, θ is also the angle between the electric field and the direction normal to the area. Hence in general, for uniform electric field, the electric flux is defined as
Φ_E= \(\vec { E } \).\(\vec { A } \) = EA cos θ …(4)

Here, note that \(\vec { A } \) is the area vector \(\vec { A } \) = A\(\hat{n} \). Its magnitude is simply the area A and the direction is along the unit vector h perpendicular to the area. Using this definition for flux, Φ_E= \(\vec { E } \).\(\vec { A } \), equations (2) and (3) can be obtained as special cases.
In figure (a), θ = 0° so Φ_E= \(\vec { E } \).\(\vec { A } \) = EA
In figure (b), θ = 90° so Φ_E= \(\vec { E } \).\(\vec { A } \) = 0

(ii) Electric flux in a non uniform electric field and an arbitrarily shaped area: Suppose the electric field’is not uniform and the area A is not flat, then the entire area is divided
into n small area segments ∆\(\vec { A } \)₁ ∆\(\vec { A } \)₂, ∆\(\vec { A } \)₃,…..∆\(\vec { A } \)_n, such that each area element is almost flat and the electric field over each area element is considered to be uniform.
The electric flux for the entire area A is approximately written as

By taking the limit ∆\(\vec { A } \)₁ → 0 (for all i) the summation in equation (5) becomes integration. The total electric flux for the entire area is given by
Φ_E = ∫\(\vec { E } \).d\(\vec { A } \) ….. (6)
From Equation (6), it is clear that the electric flux for a given surface depends on both the electric field pattern on the surface area and orientation of the surface with respect to the electric field.

(iii) Electric flux for closed surfaces: In the previous section, the electric flux for any arbitrary curved surface is discussed. Suppose a closed surface is present in the region of the non-uniform electric field as shown in figure (a).

The total electric flux over this closed surface is written as
Φ_E = \(\oint { \vec { E } } \).d\(\vec { A } \) …… (7)
Note the difference between equations (6) and (7). The integration in equation (7) is a closed surface integration and for each areal element, the outward normal is the direction of d\(\vec { A } \) as shown in the figure (b).
The total electric flux over a closed surface can be negative,
positive or zero. In the figure (b), it is shown that in one area element, the angle between d\(\vec { A } \) and \(\vec { E } \) is less than 90°, then the electric flux is positive and in another areal element, the angle between dA and E is greater than 90°, then the electric flux is negative. In general, the electric flux is negative if the electric field lines enter the closed surface and positive if the electric field lines leave the closed surface.

Samacheer Kalvi 12th Physics Electrostatics Numerical Problems

Question 1.
Electrons are caused to fall through a potential difference of 1500 volts. If they were initially at rest. Then calculate their final speed.
Solution:
The electrical potential energy is converted into kinetic energy. If v is the final speed then

Question 2.
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential.
Solution:
Let r be the radius of a small drop and R that of the large drop. Then, since the volume remains conserved,
\(\frac { 1 }{ 2 }\) πR² = \(\frac { 4 }{ 3 }\) πR³n
⇒ R³ = r³n
R = r³(n)^1/3
Further, since the total charge remains conserved, we have, using Q = CV
C_large V = n C_small v
Where V is the potential of the large drop.
4πε₀ RV = n (4πε₀r)v
V = \(\frac { nrv }{ R }\) = \(\frac { nrv }{{ r(n) }^{1/3}}\)
V = vn^2/3

Question 3.
Two particles having charges Q₁ and Q₂ when kept at a certain distance, exert a force F on each other. If the distance between the two particles is reduced to half and the charge on each particle is doubled. Find the force between the particles.
Solution:
F = \(\frac { 1 }{{ 4πε }_{0}}\) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{r^{2}}\)
If the distance is educed by half and two particles of charges are doubled.

Question 4.
Two charged spheres, separated by a distance d, exert a force F on each other. If they are immersed in a liquid of dielectric constant 2, then what is the force.
Solution:
Force between the charges (vacuum)

Force between the charges (medium)

Question 5.
Find the force of attraction between the plates of a parallel plate capacitor.
Solution:
Let d be the distance between the plates. Then the capacitor is
C = \(\frac { { \varepsilon }_{ 0 }A }{ d } \)
Energy stored in a capacitor,

Energy magnitude of the force is,

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You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.17

10th Maths Exercise 3.17 Answers Question 1.
If A = \(\left[\begin{array}{cc}{1} & {9} \\ {3} & {4} \\ {8} & {-3}\end{array}\right]=\left[\begin{array}{ll}{5} & {7} \\ {3} & {3} \\ {1} & {0}\end{array}\right]\) then verify that
(i) A + B = B + A
(ii) A + (-A) = (-A) + A = 0
Solution:

Exercise 3.17 Class 10 Question 2.

Solution:

Ex 3.17 Class 10 Question 3.
Find X and Y if X + Y = \(\left[\begin{array}{ll}{7} & {0} \\ {3} & {5}\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}{3} & {0} \\ {0} & {4}\end{array}\right]\)
Solution:

10th Maths Exercise 3.17 Question 4.
If A = \(\left[\begin{array}{lll}{0} & {4} & {9} \\ {8} & {3} & {7}\end{array}\right]\), B = \(\left[\begin{array}{lll}{7} & {3} & {8} \\ {1} & {4} & {9}\end{array}\right]\) find the value of
(i) B – 5A
(ii) 3A – 9B
Solution:

10th New Syllabus Maths Exercise 3.17 Question 5.
Find the values of x, y, z if

Solution:
(i) \(\left(\begin{array}{cc}{x-3} & {3 x-z} \\ {x+y+7} & {x+y+z}\end{array}\right)=\left(\begin{array}{ll}{1} & {0} \\ {1} & {6}\end{array}\right)\)
x – 3 = 1 ⇒ x = 4
3x – z = 0
3(4) – z = 0
-z = -12 ⇒ z = 12
x + y + 7 = 1
x + y = -6
4 + y = -6
y = -10
x = 4, y = -10, z = 12

(ii) \(\left[\begin{array}{ccc}{x} & {y-z} & {z+3}\end{array}\right]+\left[\begin{array}{lll}{y} & {4} & {3}\end{array}\right]=\left[\begin{array}{lll}{4} & {8} & {16}\end{array}\right]\)
x + y = 4 ……………. (1)
y – z + 4 = 8 ………….. (2)
z + 3 + 3 = 16 ………….. (3)
From (3), we get z = 10
From (2), we get y – 10 + 4 = 8
From (2), we get y = 14
From (1) we get x + 14 = 4
x = -10
x = -10, y = 14, z = 10

10th Maths Exercise 3.17 Samacheer Kalvi Question 6.

Solution:

10th Samacheer Maths Question 7.
Find the non-zero values of x satisfying the matrix equation

Solution:

Samacheer Kalvi 10th Guide Maths Question 8.

Solution:
x² – 4x = 5
y² – 2y = 8
y² – 2y – 8 = 0
(y – 4)(y + 2) = 0
y = 4, -2
x² – 4x – 5 = 0
(x – 5)(x + 1) = 0
x = 5, -1
x = -1, 5, y = 4, -2

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science Civics Solutions Chapter 2 Election, Political Parties and Pressure Groups

Election, Political Parties and Pressure Groups Textual Exercise

I. Choose the correct answer.

Election Political Parties And Pressure Groups Question 1.
India has adapted the electoral system followed in the
(a) USA
(b) United Kingdom
(c) Canada
(d) Russia
Answer:
(b) United Kingdom

Election Political Parties And Pressure Groups Questions And Answers Question 2.
The Election Commission of India is a/an
(a) Independent body
(b) Statutory body
(c) Private body
(d) Public corporation
Answer:
(a) Independent body

Class 9 Civics Chapter 2 Question 3.
Which Article of the Constitution provides for an Election Commission?
(a) Article 280
(b) Article 315
(c) Article 324
(d) Article 325
Answer:
(c) Article 324

Political Parties And Pressure Groups Question 4.
Which part of the constitution of India says about the election commission?
(a) Part III
(b) Part XV
(c) Part XX
(d) Part XXII
Answer:
(b) Part XV

Chapter 2 Civics Class 9 Question 5.
Who accords recognition to various political parties as national or regional parties?
(a) The President
(b) The Election Commission
(c) The Parliament
(d) The President in consultation with the Election Commission
Answer:
(b) The Election Commission

Question 6.
Assertion (A): Indian Constitution provides for an independent Election Commission
Reason (R): To ensure free and fair elections in the country.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 7.
NOT A was introduced in the year ………..
(a) 2012
(b) 2013
(c) 2014
(d) 2015
Answer:
(c) 2014

Question 8.
The term pressure groups originated in …….
(a) USA
(b) UK
(c) USSR
(d) India
Answer:
(a) USA

Question 9.
Assertion (A): A large number of pressure groups exist in India.
Reason (R): Pressure Groups are not developed in India to the same extent as in the USA.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A)
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

II. Fill in the blanks.

1. The Election Commission of India is a body of ……… members.
2. National Voters day has been celebrated on ………
3. In India ……… party system is followed.
4. In 2017, there were ……. recognised national parties.
5. Narmada Bachao Andolan is a …………
Answers:
1. 3
2. 25^th January
3. Multi
4. Seven
5. Pressure group
III. Match the following.

Answers:
1. (d)
2. (c)
3. (b)
4. (a)

IV. Give short answers.

Question 1.
Explain the electoral system in India.
Answer:

1. The electoral system in India has been adapted from the system followed in the United Kingdom.
2. India is a socialist, secular, democratic, republic and the largest democracy in the world.
3. The modem Indian nation state came into existence on 15th August 1947.

Question 2.
Give the meaning of a political party.
Answer:
A political party is an organisation formed by a group of people with a certain ideology and agenda to contest elections and hold power in the government.

A political party has three components: a leader, active members and the followers.

Question 3.
Distinguish between two-party system and the multi-party system.
Answer:

Two party system	Multi-party system
Two party system in which only two major parties exist, for example, USA, UK.	Multi-party system in which there are more than two political parties, for example India, Srilanka, France and Italy.

Question 4.
What is a pressure group?
Answer:

1. A pressure group is a group of people who are organised actively for promoting and defending their common interest. It is so called as it attempts to bring a change in the public policy by exerting pressure on the government.
2. The pressure groups are also called ‘interest groups’ or vested groups.
3. They are different from the political parties in that they neither contest elections nor try to capture political power.

V. Answer in detail.

Question 1.
Discuss merits and demerits of direct elections?
Answer:
Merits:

1. As the voters elect their representatives directly, direct elections are considered to be a more democratic method of election.
2. It educates people regarding the government activities and helps in choosing the appropriate candidates. Also, it encourages people to play an active role in politics.
3. It empowers people and makes the rulers accountable for their actions.

Demerits:

1. Direct elections are very expensive.
2. Illiterate voters sometimes get misguided by false propaganda and sometimes campaigning based on caste, religious and various other sectarian consideration spose serious challenges.
3. Since conducting direct elections is a massive exercise, ensuring free and fair elections at every polling station is a major challenge to the Election Commission.
4. There are instances of some political candidates influencing the voters through payments in the form of cash, goods or services.
5. Election campaigns sometimes results in violence, tension, law and order problems and
   affects the day-to-day life of people.

Question 2.
What are the functions of political parties?
Answer:

1. Parties contest elections. In most democracies, elections are fought mainly among the candidates put up by political parties.
2. Parties put forward their policies and programmes before the electorate to consider and choose.
3. Parties play a decisive role in making laws for a country. Formally, laws are debated and passed in the legislature.
4. Parties form and run the governments.
5. Those parties that lose in the elections play the role of the Opposition to the party or a group of coalition parties in power, by voicing different views and criticising the government for its failures or wrong policies.
6. Parties shape public opinion. They raise and highlight issues of importance.
7. Parties function as the useful link between people and the government machinery.

Question 3.
What are the functions of Pressure groups in India?
Answer:
Pressure groups are the interest groups that work to secure certain interest by influencing the public policy. They are non-aligned with any political party and work as an indirect yet powerful group to influence the policy decisions. Pressure groups carry out a range of functions including representation, political participation, education, policy formulation and policy implementation.

Political Participation: Pressure groups can be called the informal face of politics. They exert influence precisely by mobilising popular support through activities such as petitions, marches, demonstrations and other forms of political protest. Such forms of political participation have been particularly attractive to young people.

Education: Many pressure groups devote significant resources by carrying out research, maintaining websites, commenting on government policy and using high-profile academics, scientists and even celebrities to get their views across, with an emphasis to cultivate expert authority.

Policy Formulation: Though the pressure groups themselves are not policy-makers, yet it does not prevent many of them from participating in the policy-making process. Many pressure groups are vital sources of information and render advice to the government and therefore they are regularly consulted in the process of policy formulation.

VI. Project and Activity

Question 1.
Compare the policies, programmes and achievements of a national party and a state party.
Answer:

1. Refer the National policies, programmes and achievements from the Internet and library books.
2. The students are instructed to compare the policies programmes and achievements.
3. This is a group Activity.

VII. HOTS

Question 1.
“Elections are considered essential for any representative democracy”. Why?
Answer:
“A democracy requires a mechanism by which people can choose their representatives at regular intervals and change them if they wish to do so. Therefore, elections are considered essential for any representative democracy. In an election the voters make many choices.

1. This helps the public in choosing the development course which they want and want to adopt.
2. This directly provide the public the opportunity to select their representatives and these representatives decision can be legitimacy.
3. This also ensures the transparency as well as the accountability as the public representatives are chosen directly.

Question 2.
What is the principle of universal adult franchise? What is its importance?
Answer:
Principle: Universal Adult Franchise means that the right to vote should be given to all adult citizens without the discrimination of caste, class, colour, religion (or) gender. It is based on equality, which is a basic principle of democracy.

Importance: Under this system a government is elected that is accountable to the people it governs. Because every vote counts, issues in a society receive their appropriate weight in terms of importance and urgency.

Question 3.
Discuss merits and demerits of democracy.
Answer:

	Merits		Demerits
1.	Safeguards the interests of the people.	1.	More emphasis on quantity than on quality.
2.	Based on the principle of equality.	2.	Rule of the incompetent.
3.	Stability and responsibility in administration.	3.	Based on unnatural equality.                •
4.	Political education to the people.	4.	Voters do not take interest in election.
5.	Little chance of revolution.	5.	Lowers the moral standard.
6.	Stable government.	6.	Democracy is a government of the rich.
7.	Helps in making people good citizens.	7.	Misuse of public funds and time.
8.	Based on public opinion.	8.	No stable government.
		9.	Dictatorship of majority.
		10.	Bad influence of political parties.

Question 4.
Discuss the multi-party system.
Answer:

1. A multi-party system is a system where multiple political parties that have ideas participate in the national elections.
2. A lot of countries that use this system have a coalition government, meaning many parties are in control, and they all work together to make laws.
3. Countries with a multi-party political system tend to have greater voter participation.
4. No democracy can survive without multi-party system.

VIII. Life Skill

Question 1.
Conduct a mock poll in your classroom.
Answer:

1. Help the students to understand the process of electing officials and the power of vote by holding a mock election.
2. These are great activities to enjoy during the presidential election.
3. Students explain the steps taken from party formation to National election.
4. The students will act out the campaigning and voting process by stimulating a real election in their own class room.

Steps set for a mock-poll in the class room:

1. Setting up the political parties.
2. Preparing a manifesto.
3. Running a campaign.
4. Holding the class room election.

Election, Political Parties and Pressure Groups Additional Questions

I. Choose the correct answer.

Question 1.
India is the democracy in the world.
(a) largest
(b) smallest
(c) strongest
(d) None of the above
Answer:
(a) largest

Question 2.
Kudavolai was the system of voting followed during the ……. period in Tamil Nadu.
(a) Chera
(b) Chola
(c) Pandya
(d) Pallava
Answer:
(b) Chola

Question 3.
Which country has single party system?
(a) USA
(b) UK
(c) Cuba
(d) France
Answer:
(c) Cuba

Question 4.
Assertion (A): Parties shape public opinion.
Reason (R): They raise and highlight issues of importance.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true and (R) does not explain (A).
(c) (A) is correct and (R) is false
(d) (A) is false and (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 5.
India is the ………… th country in the world to introduce NOTA.
(a) 10
(b) 12
(c) 14
(d) 16
Answer:
(c) 14

Question 6.
The ……. is elected by members of the Lok Sabha.
(a) Prime Minister
(b) President
(c) Governor
(d) Cabinet Minister
Answer:
(a) Prime Minister

II. Fill in the blanks.
1. ……. in elections are the best way to make your ‘voice’ heard.
2. Indirect elections are less ……..
3. ……… parties are an essential part of Democracy.
4. …… treats all the parties equally.
5. The pressure groups are also called ……. groups.
6. A political party has three components: a ………. and the ……….
Answers:
1. Voting
2. expensive
3. Political
4. Election Commission
5. Interest
6. a leader, acting members, followers

III. Match the following.

Answers:
1. (d)
2. (a)
3. (b)
4. (c)

IV. Give short answers.

Question 1.
What do you know about Voters Verified Paper Audit Trail?
Answer:

1. Voters Verified Paper Audit Trail (WPAT) is the way forward to enhance credibility and transparency of the election process.
2. This system was first introduced in 2014 General election.

Question 2.
Mention the merits and demerits of Indirect elections.
Answer:
Merits:

1. Indirect elections are less expensive.
2. It is more suited to elections in large countries.

Demerits:

1. If the number of voters is very small, there exists the possibility of corruption, bribery, horse trading and other unfair activities.
2. It is less democratic because people do not have a direct opportunity to elect, but they instead do it through their representatives. So, this may not reflect the true will of the people.

Question 3.
Write a short note on “State Parties”.
Answer:
Other than the seven national parties, most of the major parties of the country are classified by the Election Commission as ‘state parties’. These are commonly referred to as regional parties. A party is recognised as a state party by the Election Commission of India based on certain percentage of votes secured or a certain number of seats won in the Assembly or Lok Sabha elections.

Question 4.
Classify pressure groups in India.
Answer:
The pressure groups in India can be broadly classified into the following categories:

1. Business groups
2. Trade unions
3. Agrarian groups
4. Professional associations
5. Student organisations
6. Religious organisations
7. Tribal organisations
8. Linguistic groups
9. Ideology-based groups
10. Environmental protection groups.

V. Answer in detail.

Question 1.
Give an account of Mobilisation and Democratic Participation.
Answer:
Mobilising people towards socially productive activities that lead to the overall betterment of people’s lives is essential. Sometimes earthquakes, tsunamis, floods and other such natural disasters on a massive scale occur and people’s immediate mobilisation for evacuation and emergency relief becomes most essential.

Democratic Participation: Democracy can-succeed only when smaller local groups and, in fact, every citizen can take action that supports the tax and revenue collection systems, observance of national norms in environmental protection, cleanliness, health and hygiene, sanitary drives and immunisation programmes like pulse polio.

However, we must keep in mind that there is no better form of government than Democratic government. To create a better society and nation, the people of India along with the union and state governments should come together to fight against the miseries of human life.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1

11th Maths Exercise 1.1 Answers Question 1.
Write the following in roster form.
(i) {x ∈ N : x² < 121 and x is a prime}.
(ii) the set of all positive roots of the equation (x – 1)(x + 1)(x² – 1) = 0.
(iii) {x ∈ N : 4x + 9 < 52}.
(iv) {x : \(\frac{x-4}{x+2}\) = 3, x ∈ R – {-2}}
Solution:
(i) A = {2, 3, 5, 7}
(ii) B = {1}
(iii) 4x + 9 < 52
4x + 9 – 9 < 52 – 9
4x < 43
x < \(\frac{43}{4}\) (i.e.) x < 10.75 4
But x ∈ N
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(i.e.) x – 4 = 3(x + 2)
x – 4 = 3x + 6
– 4 – 6 = 3x – x
2x = -10 ⇒ x = -5
A = {-5}

11th Std Maths Exercise 1.1 Answers Question 2.
Write the set {-1,1} in set builder form.
Solution:
A = {x: x² = 1}

11th Maths Book Exercise 1.1 Answers Question 3.
State whether the following sets are finite or infinite.
(i) {x ∈ N : x is an even prime number}
(ii) {x ∈ N : x is an odd prime number}
(iii) {x ∈ Z : x is even and less than 10}
(iv) {x ∈ R : x is a rational number}
(v) {x ∈ N : x is a rational number}
Solution:
(i) Finite set
(ii) Infinite set
(iii) Infinite
(iv) and
(v) infinite

Exercise 1.1 Class 11 Maths State Board Question 4.
By taking suitable sets A, B, C, verify the following results:
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(if) A × (B ∪ C) = (A × B) ∪ (A × C).
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B).
(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A).
Solution:
To prove the following results let us take U = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
B = {2, 7, 8, 9}
C = {1, 5, 8, 7}
(i) To prove: A × (B ∩ C) = (A × B) ∩ (A × C)
B ∩ C = {8}; A = {1, 2, 5, 7}
So A × (B ∩ C) = {1, 2, 5, 7} × {8}
= {(1, 8), (2. 8), (5, 8), (7, 8)}
Now A x B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)} …. ( 1)
A × C = {(1, 1), (1, 5),(1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∩ (A × C) = {(1, 8), (2, 8), (5, 8), (7, 8)} ……… (2)
(1) = (2)
⇒ A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To prove A × (B ∪ C) = (A × B) (A × C)
B = {2, 7, 8, 9}, C = {1, 5, 8, 10)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
A × (B ∪ C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)) …. (1)
A × B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9),
(7, 2), (7, 7), (7, 8), (7, 9)}
A × C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∪ (A × C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)} …… (2)
(1) = (2) ⇒ A × (B ∪ C) = (A × B) ∪ (A × C)

(iii) A × B = {(1, 2), (1, 7), (1, 8), (1, 9) (2, 2), (2, 7), (2, 8), (2, 9) (5, 2), (5, 7), (5, 8), (5, 9) (7, 2), (7, 7), (7, 8), (7, 9)}
B × A = {(2, 1), (2, 2), (2, 5), (2, 7) (7, 1), (7, 2), (7, 5), (7, 7) (8, 1), (8, 2), (8, 5), (8, 7) (9,1), (9, 2), (9, 5), (9, 7)}
L.H.S. (A × B) ∩ (B × A) = {(2, 2), (2, 7), (7, 2), (7, 7)} …. (1)
R.H.S. A ∩ B = {2, 7}
B ∩ A = {2, 7}
(A ∩ B) × (B ∩ A) = {2, 7} × {2, 7}
= {(2, 2), (2, 7), (7, 2), (7, 7)} ……… (2)
(1) = (2) ⇒ LHS = RHS

(iv) To prove C – (B – A) = (C ∩ A) ∪ (C ∩ B)
B – A = {8, 9}
C = {1, 5, 8, 10}
∴ LHS = C – (B – A) = {1, 5, 10} …… (1)
C ∩ A = {1}
U = {1, 2, 5, 7, 8, 9, 10}
B = {2, 7, 8, 9} ∴ B’ = {1, 5, 10}
C ∩ B = {1, 5, 10}
R.H.S. (C ∩ A) ∪ (C ∩ B) = {1} ∪ {1, 5, 10}
= {1, 5, 10} ……. (2)
(1) = (2) ⇒ LHS = RHS

(v) To prove (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A)
A= {1, 2, 5, 7}, B = {2, 7, 8, 9}, C = {1, 5, 8, 10}
Now B – A = {8, 9}
(B – A) ∩ C = {8} ……. (1)
B ∩ C = {8}
A = {1, 2, 5, 7}
So (B ∩ C) – A = {8} …… (2)
C – A = {8, 10}
B = {2, 7, 8, 9}
B ∩ (C – A) = {8} …. (3)
(1) = (2) = (3)

(vi) To prove (B – A) ∪ C ={1, 5, 8, 9, 10}
B – A = {8, 9},
C = {1, 5, 8, 10}
(B – A) ∪ C = {1, 5, 8, 9, 10} ……. (1)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A – C = {2, 7}
(B ∪ C) – (A – C) = {1, 5, 8, 9, 10} ……… (2)
(1) = (2)
⇒ (B – A) ∪ C = (B ∪ C) – (A – C)

11th Maths Exercise 1.1 Question 5.
Justify the trueness of the statement.
“An element of a set can never be a subset of itself.”
Solution:
A set itself can be a subset of itself (i.e.) A ⊆ A. But it cannot be a proper subset.

11th Maths Chapter 1 Exercise 1.1 Question 6.
If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).
Solution:
n(P( A)) = 1024 = 2¹⁰ ⇒ n( A) = 10
n(A ∪ B) = 15
n(P(B)) = 32 = 2⁵ ⇒ n(B) = 5
We know n(A ∪ B) = n{A) + n(B) – n(A ∩ B)
(i.e.) 15 = 10 + 5 – n(A ∩ B)
⇒ n(A ∩ B) = 15 – 15 = 0

11 Maths Exercise 1.1 Question 7.
If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A(A ∆ B)).
Solution:
n(A ∪ B) = 10; n(A ∩ B) = 3
n(A ∆ B) = 10 – 3 = 7
and n(P(A ∆ B)) = 27 = 128

11th Maths 1.1 Exercise Question 8.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.
Solution:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

11th Maths Exercise 1.1 Answers In Tamil Question 9.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.
Solution:
n(A) = 3 ⇒ set A contains 3 elements
n(B) = 2 ⇒ set B contains 2 elements –
we are given (x, 1), (y, 2), (z, 1) are elements in A × B ⇒ A = {x, y, z} and B = {1, 2}

11th Maths Exercise 1.1 Answers State Board Question 10.
If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (-1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.
Solution:
n(A × A) = 16 ⇒ n( A) = 4
S ={(-1, 0), (-1, 1), (0, 2), (1, 2)}

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 Additional Questions

Samacheer Kalvi 11th Maths Example Sums Question 1.
Write the following sets in roster form
(a) {x ∈ N; x³ < 1000}
(b) {The set of positive roots of the equation (x² – 4) (x³ – 27) = 0}
Solution:
(a) A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) B = {2, 3}

11th Maths Exercise 1.1 4th Sum Question 2.
By taking suitable sets A, B, C verify the following results
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) (B – A) ∪ C = (B ∪ C) – (A – C)
Solution:
Prove by yourself

11th Maths 1st Chapter Exercise 1.1 Question 3.
Given n(A) = 7; n(B) = 8 and n(A ∪ B) = 10 find n[P(A ∩ B)].
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i.e.,) 10 = 7 + 8 – n(A ∩ B)
⇒ n(A ∩ B) = 7 + 8 – 10 = 5
So n[P(A ∩ B)] = 25 = 32

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.2

9th Maths Exercise 1.2 Samacheer Kalvi Question 1.
Find the cardinal number of the following sets.
(i) M = {p, q, r, s, t, u}
(ii) P = {x : x = 3n + 2, n ∈ W and x < 15}
(iii) Q = {v : v = \(\frac { 4 }{ 3n }\) ,n ∈ N and 2 < n ≤ 5}
(iv) R = {x : x is an integers, x ∈ Z and -5 ≤ x < 5}
(v) S = The set of all leap years between 1882 and 1906.
Solution:
(i) n(M) = 6
(ii) W = {0, 1, 2, 3, ……. }
if n = 0, x = 3(0) + 2 = 2
if n = 1, x = 3(1) + 2 = 5
if n = 2, x = 3(2) + 2 = 8
if n = 3, x = 3(3)+ 2 =11
if n = 4, x = 3(4) + 2=14
∴ P= {2, 5, 8, 11, 14}
n(P) = 5

(iii) N = {1,2, 3, 4, …..}
n ∈ {3, 4, 5}

n(Q) = 3

(iv) x ∈ z
R = {-5, – 4, -3, -2, -1, 0, 1, 2, 3, 4}
n(R)= 10.

(v) S = {1884, 1888, 1892, 1896, 1904}
n (S) = 5.

9th Maths Set Language Exercise 1.2 Question 2.
Identify the following sets as finite or infinite.
(i) X = The set of all districts in Tamilnadu.
(ii) Y = The set of all straight lines passing through a point.
(iii) A = {x : x ∈ Z and x < 5}
(iv) B = {x : x² – 5x + 6 = 0, x ∈ N}
Solution:
(i) Finite set
(ii) Infinite set
(iii) A = { ……. , -2, -1, 0, 1, 2, 3, 4}
∴ Infinite set

(iv) x² – 5x + 6 = 0
(x – 3) (x – 2) = 0
B = {3, 2}
∴ Finite set.

9th Maths Exercise 1.2 Question 3.
Which of the following sets are equivalent or unequal or equal sets?
(i) A = The set of vowels in the English alphabets.
B = The set of all letters in the word “VOWEL”
(ii) C = {2, 3, 4, 5}
D = {x : x ∈ W, 1 < x < 5}
(iii) X = A = { x : x is a letter in the word “LIFE”}
Y = {F, I, L, E}
(iv) G = {x : x is a prime number and 3 < x < 23}
H = {x : x is a divisor of 18}
Solution:
(i) A = {a, e, i, o, u}
B = {V, O,W, E, L}
The sets A and B contain the same number of elements.
∴ Equivalent sets

(ii) C ={2, 3, 4, 5}
D = {2, 3, 4}
∴ Unequal sets

(iii) X = {L, I, F, E}
Y = {F, I, L, E}
The sets X and Y contain the exactly the same elements.
∴ Equal sets.

(iv) G = {5, 7, 11, 13, 17, 19}
H = {1, 2, 3, 6, 9, 18}
∴ Equivalent sets.

9th Maths Set Language Exercise 1.2 Solutions Question 4.
Identify the following sets as null set or singleton set.
(i) A = (x : x ∈ N, 1 < x < 2}
(ii) B = The set of all even natural numbers which are not divisible by 2.
(iii) C = {0}
(iv) D = The set of all triangles having four sides.
Solution:
(i) A = { } ∵ There is no element in between 1 and 2 in Natural numbers.
∴ Null set

(ii) B = { } ∵ All even natural numbers are divisible by 2.
∴ B is Null set

(iii) C = {0}
∴ Singleton set

(iv) D = { }
∵ No triangle has four sides.
∴ D is a Null set.

9th Maths Exercise 1.2 In Tamil Question 5.
State which pairs of sets are disjoint or overlapping?
(i) A = {f, i, a, s} and B = {a, n, f, h, s)
(ii) C = {x : x is a prime number, x > 2} and D = {x : x is an even prime number}
(iii) E = {x: x is a factor of 24} and F = {x : x is a multiple of 3, x < 30}
Solution:
(i) A = {f, i, a, s}
B = {a, n, f, h, s}
A ∩ B ={f, i, a, s} ∩ {a, n,f h, s} = {f, a, s}
Since A ∩ B ≠ ϕ , A and B are overlapping sets.

(ii) C = {3, 5, 7, 11, ……}
D = {2}
C ∩ D = {3, 5, 7, 11, …… } ∩ {2} = { }
Since C ∩ D = Ø, C and D are disjoint sets.

(iii) E = {1, 2, 3, 4, 6, 8, 12, 24}
F = {3, 6, 9, 12, 15, 18, 21, 24, 27}
E ∩ F = {1, 2, 3, 4, 6, 8, 12, 24} ∩ {3, 6, 9, 12, 15, 18, 21, 24, 27}
= {3, 6, 12, 24}
Since E ∩ F ≠ ϕ, E and F are overlapping sets.

9th Standard Maths Exercise 1.2 In Tamil Question 6.
If S = {square,rectangle,circle,rhombus,triangle}, list the elements of the following subset of S.
(i) The set of shapes which have 4 equal sides.
(ii) The set of shapes which have radius.
(iii) The set of shapes in which the sum of all interior angles is 180°
(iv) The set of shapes which have 5 sides.
Solution:
(i) {Square, Rhombus}
(ii) {Circle}
(iii) {Triangle}
(iv) Null set.

9th Standard Maths Exercise 1.2 Question 7.
If A = {a, {a, b}}, write all the subsets of A.
Solution:
A= {a, {a, b}} subsets of A are { } {a}, {a, b}, {a, {a, b}}.

9th Std Maths Exercise 1.2 Question 8.
Write down the power set of the following sets.
(i) A = {a, b}
(ii) B = {1, 2, 3}
(iii) D = {p, q, r, s}
(iv) E = Ø
Solution:
(i) The subsets of A are Ø, {a}, {b}, {a, b}
The power set of A
P(A ) = {Ø, {a}, {b}, {a,b}}

(ii) The subsets of B are ϕ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
The power set of B
P(B) = {Ø, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}

(iii) The subset of D are Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s},{p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}}
The power set of D
P(D) = {Ø, {p}, {q}, {r}, {s}, {p, q}, {p, r}, {p, s}, {q, r}, {q, s}, {r, s}, {p, q, r}, {q, r, s}, {p, r, s}, {p, q, s}, {p, q, r, s}

(iv) The power set of E
P(E) = { }.

9th Maths 1.2 In Tamil Question 9.
Find the number of subsets and the number of proper subsets of the following sets.
(i) W = {red,blue, yellow}
(ii) X = { x² : x ∈ N, x² ≤ 100}.
Solution:
(i) Given W = {red, blue, yellow}
Then n(W) = 3
The number of subsets = n[P(W)] = 2³ = 8
The number of proper subsets = n[P(W)] – 1 = 2³ – 1 = 8 – 1 = 7

(ii) Given X ={1,2,3, }
X² = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}
n(X) = 10
The Number of subsets = n[P(X)] = 2¹⁰ = 1024
The Number of proper subsets = n[P(X)] – 1 = 2¹⁰ – 1 = 1024 – 1 = 1023.

9th Maths 1.2 Exercise Question 10.
(i) If n(A) = 4, find n[P(A)].
(ii) If n(A) = 0, find n[P(A)].
(iii) If n[P(A)] = 256, find n(A).
Solution:
(i) n( A) = 4
n[ P(A)] = 2n = 2⁴ = 16
(ii) n(A) = 0
n[P(A)] = 2⁰ = 1
(iii) n[P(A)] = 256

n[P(A)] = 28
∴ n(A) = 8.

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 6 Gaseous State

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Samacheer Kalvi 11th Chemistry Gaseous State Textual Evaluation Solved

I. Choose the correct answer from the following:

11th Chemistry Lesson 6 Book Back Answers Question 1.
Gases deviate from ideal behavior at high pressure. Which of the following statement (s) is correct for non – ideality?
(a) at high pressure the collision between the gas molecule become enormous
(b) at high pressure the gas molecules move only in one direction
(c) at high pressure, the volume of gas become insignificant
(d) at high pressure the inter molecular interactions become significant
Answer:
(d) at high pressure the inter molecular interactions become significant

11th Chemistry Chapter 6 Book Back Answers Question 2.
Rate of diffusion of a gas is …………
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to its square root of its molecular weight
(d) inversely proportional to the square root of its molecular weight
Answer:
(d) inversely proportional to the square root of its molecular weight

Gaseous State 11th Chemistry Question 3.
Which of the following is the correct expression for the equation of state of van der Waals gas?

Answer:

11th Chemistry Unit 6 Book Back Answers Question 4.
When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules ………….
(a) are above inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to the loss in kinetic energy
(d) colide without loss of energy
Answer:
(b) exert no attractive forces on each other

11th Chemistry Gaseous State Question 5.
Equal weights of methane and oxygen is mixed in an empty container at 298 K. The fraction of total pressure exerted by oxygen ………..
(a) 1/3
(b) 1/2
(c) 2/3
(d) 1/3 × 273 × 298
Answer:
(a) 1/3
Hint:
mass of methane = mass of oxygen = a
number of moles of methane = \(\frac {a}{16}\)
number of moles of Oxygen = \(\frac {a}{32}\)
mole fraction of Oxygen =
Partial pressure of oxygen = mole fraction x Total Pressure = \(\frac {1}{3}\)P

Gaseous State Class 11 Question 6.
The temperatures at which real gases obey the ideal gas laws over a wide range of pressure is called …………
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
(d) Reduced temperature
Answer:
(b) Boyle temperature
Hint:
The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called Boyle temperature

Gaseous State Class 11 Notes Pdf Question 7.
In a closed room of 1000 m³ a perfume bottle is opened up. The room develops a smell. This is due to which property of gases?
(a) Viscosity
(b) Density
(c) Diffusion
(d) None
Answer:
(c) Diffusion

Samacheer Kalvi Guru 11th Chemistry Question 8.
A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be ………….
(a) At the center of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Answer:
(b) Near the hydrogen chloride bottle
Hint:
Rate of diffusion α 1/√m
m_NH3  = 17
m_HCl = 36.5
γ_NH3  > γ_HCl
Hence white fumes first formed near hydrogen chloride.

Samacheer Kalvi Class 11 Chemistry Solutions Question 9.
The value of universal gas constant depends upon ………..
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Samacheer Kalvi.Guru 11th Chemistry Question 10.
The value of the gas constant R is …………
(a) 0.082 dm³ atm.
(b) 0.987 cal mol^-1 K^-1
(c) 8.3 J mol^-1K^-1
(d) 8 erg mol^-1K^-1
Answer:
(c) 8.3 J mol^-1K^-1

Samacheerkalvi.Guru 11th Chemistry Question 11.
Use of hot air balloon in sports at meteorological observation is an application of
(a) Boyle’s law
(b) Newton’s law
(c) Kelvin’s law
(d) Brown’s law
Answer:
(a) Boyle’s law

Class 11 Chemistry Solutions Samacheer Kalvi Question 12.
The table indicates the value of van der Waals constant ‘a’ in (dm³)² atm. mol^-2

The gas which can be most easily liquefied is ………….
(a) O₂
(b) N₂
(c) NH₃
(d) CH₄
Answer:
(c) NH₃
Hint:
Higher the value of ‘a’, greater the intermolecular force of attraction, easier the liquefaction. Option (c) is correct

Gaseous State Questions And Answers Pdf Question 13.
Consider the following statements.
(i) Atmospheric pressure is less at the top of a mountain than at sea level
(ii) Gases are much more compressible than solids or liquids
(iii) When the atmospheric pressure increases the height of the mercury column rises Select the correct statement.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (i), (ii) and (iii)
Answer:
(d) (i), (ii) and (iii)

Class 11 Gaseous State Question 14.
Compressibility factor for CO₂ at 400 K and 71.0 bar is 0.8697. The molar volume of CO₂ under these conditions is ………..
(a) 22.04 dm³
(b) 2.24 dm³
(c) 0.41 dm³
(d) 19.5 dm³
Answer:
(c) 0.41 dm³
Compressibility factor (z) = \(\frac {Pv}{nRT}\)
V = \(\frac {z x nRT }{p}\)

V = 0.41 dm³

Samacheer Kalvi 11th Chemistry Solution Question 15.
If temperature and volume of an ideal gas is increased to twice its values, the initial pressure P becomes ………….
(a) 4P
(b) 2P
(c) P
(d) 3P
Answer:
(c) P
Hint:

P₂ = P₂ Option (c)

Question 16.
At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3\(\sqrt{3}\) times that of a hydrocarbon having molecular formula What is the value of n?
(a) 8
(b) 4
(c) 3
(d) 1
Answer:
(b) 4.
Hint:

Squaring on both sides and rearranging
27 x 2 = m_CnH_2n-2
54 = n(12) + (2n-2)(l)
54 = 12n+2n – 2
54 = 14n – 2
n = (54 + 2)/14 = 56/14 = 4

Question 17.
Equal moles of hydrogen and oxygen gases are placed in a container, with a pin-hole through which both can escape what fraction of oxygen escapes in the time required for one-half of the hydrogen to escape. (NEET phase 1)
(a) 3/8
(b) 1/2
(c) 1/8
(d) 1/4
Answer:
(c) 1/8
Hint:

The fraction of oxygen that escapes in the time required for one half of the hydrogen to escape is 1/8

Question 18
The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion ie α = \(\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\)For an ideal gas α is equal to ………..
(a) T
(b) 1/T
(c) P
(d) none of these
Answer:
(b) 1/T
Hint:

Question 19.
Four gases P, Q, R and S have almost same values of ‘b’ but their ‘a’ values (a, b are Van der Waals Constants) are in the order Q < R < S < P. At a particular temperature, among the four gases the most easily liquefiable one is ………….
(a) P
(b) Q
(c) R
(d) S
Answer:
(a) P
Hint:
Greater the ‘a’ value, casier the liquefaction

Question 20.
Maximum deviation from ideal gas is expected from (NEET)
(a) CH_4(g)
(b) NH_3(g)
(c) H_2(g)
(d) N_2(g)
Answer:
(b) NH_3(g)

Question 21.
The units of Van der Waals constants ‘b’ and ‘a’ respectively
(a) mol L^-1 and L atm² mol^-1
(b) mol L and L atm mol²
(c) mol^-1 L and L² atm mol^-1
(d) none of these
Answer:
(c) mol^-1 L and L² atm mol^-1
Hint:
an²/V² atm
a = atm L²/mol² = L² mol^-2 atm
nb = L
b = L /mol = L mol^-1

Question 22.
Assertion : Critical temperature of CO₂ is 304 K, it can be liquefied above 304 K.
Reason : For a given mass of gas, volume is to directly proportional to pressure at constant temperature.
(a) both assertion and reason arc 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
Hint:
Correct Statement: Critical temperature of CO₂ is 304 K. It means that CO₂ cannot be liquefied above 304 K, whatever the pressure may applied. Pressure is inversely proportional to volume.

Question 23.
What is the density of N, gas at 227°C and 5.00 atm pressure? (R = 0.082 L atm K^-1 mol^-1)
(a) 1.40 g/L
(b) 2.81 g/L
(c) 3.41 g/L
(d) 0.29 g/L
Answer:
(c) 3.41 g/L
Hint:
Density = \(\frac {Mass}{Volume}\)

Question 24.
Which of the following diagrams correctly describes the behaviour of a fixed mass of an ideal gas ? (T is measured in K)

Answer:

For a fixed mass of an ideal gas V α T
P α 1/V
and PV = Constant

Question 25.
25 g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?
(a) HBr
(b) HCl
(c) HF
(d) HI
Answer:
(d) HI
Hint:
At a given temperature and pressure
Volume α number of moles
Volume α Mass / Molar mass
Volume α 28 / Molar mass
i.e. if molar mass is more , volume is less. Hence Hl has the least volume.

II. Answer these questions briefly.

Question 26.
State Boyle’s law.
Answer:
Boyle’s law states that at a given temperature. the volume occupied by a fixed mass of a gas is inversely proportional to its pressure.
V α \(\frac {1}{P}\);
where T and n are fixed or PV = Constant = k

Question 27.
A balloon filled with air at room temperature and cooled to a much lower temperalure can be used as a model for Charles’ law.
Answer:
Charles’ law:

• V T at constant P and n (or) \(\frac {V}{T}\) = Constant
• A balloon filled with air at room temperature and cooled to a much lower temperature. the size of the balloon is reduced. Because if the temperature of the gas decreases, the volume also decreases in a direct proportion.
• When temperature is reduced, the gas molecules inside in over slower due to decreased temperature and hence the volume decreases.

Question 28.
Name two items that can serve as a model for Gay Lussac’s law and explain.
Answer:
Gay Lussac’s law:
1. P α T at constant volume (or) = \(\frac {V}{T}\)
2. Example – 1:
You fill the car type completely full of air on the hottest day of summer. The type cannot change it shape and volume. But when winter comes, the pressure inside the lyre is reduced and the shape is also reduced. This confirms that pressure and temperature are direct related to each other.
3. Example – 2:
The egg in the bottle experiment.

A glass bottle is taken, inside the bottle put some pieces of cotton with fire. Then place a boiled egg (shell removed) at the top of the bottle. The temperature inside the bottle increases from the fire, rising (he pressure. By scaling the bottle with egg, the fire goes on, dropping the temperature and pressure. This causes the egg to be sucked into the bottle.
P α T is proved (or) = \(\frac{P_{1}}{V_{1}}=\frac{P_{2}}{V_{2}}\)

Question 29.
Give the mathematical expression that relates gas volume and moles. Describe in words what the mathematical expression means.
Answer:

1. The mathematical relationship betwêen the volume of a gas and the number of moles is V α n
2. \(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant
   Where V₁ and n₁ are the volume and number of moles of a gas and V₂ and n₂ are the values of volume and number of moles of same gas at a different set of conditions.
3. If the volume of the gas increase then the number of moles of the gas also increases.
4. At a certain temperature and pressure, the volume of a gas is dirctly proportional to the number of the moles of the gas.

Question 30.
What are ideal gases? In what way real gases differ from ideal gases.
Answer:

1. Ideal gases are the gases that obey gas laws or gas equation PV = nRT.
2. Real gases do not obey gas equation. PV = nRT.
3. The deviation of real gases from ideal behaviour is measure in terms of a ratio of PV to nRT. This is termed as compression factor (Z). Z = \(\frac {PV}{nRT}\)
4. For ideal gases Z = 1.
5. For real gases Z > 1 or Z < 1. For example, at high pressure real gases have Z >1 and at intermediate pressure Z < 1.
6.  Above the Boyle point Z> 1 for real gases and below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increases with the increase in pressure.
7. So, it is clear that at low pressure and high temperature, the real gases behave as ideal gases.

Question 31.
Can a Van der Waals gas with a = 0 be liquefied? Explain.
Answer:

• a = 0 for a Van der Waals gas i.e. for a real gas. Van der Waals constant a = 0. It cannot be liquefied.
• If a = 0, there is a very less interaction between the molecules of gas.
• ‘a’ is the measure of strength of Van der Waals force of attraction between the molecules of the gas.
• If a is equal to zero, the Van der Waals force of attraction is very less and the gas cannot be liquefied.

Question 32.
Suppose there ¡s a tiny sticky area on the wan of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?
Answer:

• Molecules hitting the tiny sticky area on the wall of the container of gas moves faster as they get closer to adhesive surface, but this effect is not permanent.
• The pressure on the sticky wall is greater than on the ordinary area of walls.

Question 33.
Explain the following observations
(a) Aerated water bottles are kept under water during summer
(b) Liquid ammonia bottle is cooled before opening the seal
(c) The type of an automobile is inflated to slightly lesser pressure in summer than in winter
(d) The size of a weather balloon becomes larger and larger as it ascends up into larger altitude
Answer:
(a) In aerated water bottles, CO₂ gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more of gas will be present above the liquid surface in the glass bottle.

In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept under water. As a result, the temperature is likely to decrease and the solubility of CO₂ is likely to increase in aqueous solution resulting in decreased pressure.

(b) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes accident.

However, if the bottle is cooled under tap water for sometime, there will be a decrease in the volume of a gas to a large extent. if the seal is opened now, the gas will corne out of the bottle at a slower rate, reduces the chances of accident.

(c) The pressure of air is directly proportional to the temperature. Since the temperature is higher in summer than in higher, the pressure of the air in the tube of the lyre is likely to be quite high as compared to winter. It is quite likely that the tube may burst under high pressure in summer, Therefore, it is advisable to inflate the types to lesser pressure in summer than in winter.

(d) The volume of the gas is inversely proportional to pressure at a given temperature according to Boyle’s law. As the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon or the size of the balloon is likely to increase.

Question 34.
Give suitable explanation for the following facts about gases.
(a) Gases don’t settle at the bottom of a container
(b) Gases diffuse through all the space available to them and
(c) Explain with an increase in temperature
Answer:
(a) Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.

(b) When a sample of a gas introduced to one part of a closed container, its molecules very quickly disperse throughout the container, this process by which molecules disperse in space in response to differences in concentration is called diffusion. For e.g., you can smell perfume in a room, because it difluses into the air totally inside the room.

(c) Diffusion is faster at higher temperature because the gas molecules have greater kinetic energy. Since heat increase the motion, then diffusion happens faster.

Question 35.
Suggest why there ¡s no hydrogen (H₂) in our atmosphere. Why does the moon have no atmosphere?
Answer:
1. Hydrogen is the lightest element thus when produced in free state, it rises above all the other gases to the top of the atmosphere, where it is open to cosmic storms and solar flares. There it literally leaks from the atmosphere to the empty space. Hydrogen easily gains velocity required to escape Earth’s magnetic field. Hydrogen is very reactive in nature. So it would have reacted with O₂ , in its way to produce H₂O. So majority portion of H₂ reacts and very less amount of it present in the upper level of atmosphere and gains velocity to escape the atmosphere.

2. Moon has no atmosphere because the value of acceleration due to gravity ‘g’ on the surface of the moon is small. Therefore, the value of escape velocity on the surface of the moon is very small. The molecule of the atmospheric gases on the surface of the moon have thermal velocities greater than the escape velocity. That’s why all the molecules of gases have escaped and there is no atmosphere in the moon. The moon has insufficient gravity to retain an atmosphere. So we conclude that moon has no atmosphere.

Question 36.
Explain whether a gas approaches ideal behaviour or deviates from ideal behaviour if –
(a) it is compressed to a smaller volume at constant temperature
(b) the temperature is raised while keeping the volume constant
(c) more gas is introduced into the same volume and at the same temperature
Answer:
(a) it a gas is compressed to a smaller volume at constant temperature, pressure is increased. At high pressure with a smaller volume, the gas deviates from ideal behaviour.

(b) If a gas temperature is raised keeping the volume constant, the pressure of the gas will increase. At high pressure, the gas deviates from ideal behaviour.

(c) if more gas is introduced into the same volume and at the same temperature, the number of moles are increasing. if the volume remains same, the increased number of moles collide with each other and kinetic energy increases and pressure decreases. At increased pressure, the gas deviates from ideal behaviour.

Question 37.
Which of the following gases would you expect to deviate from ¡deal behaviour under conditions of low temperature F Cl₂, or Br₂? Explain.
Answer:
1. Bromine deviates (Br₂) from the ideal gas maximum than Cl₂ and F₂. Because Br₂ has biggest size (atomic weight 79.9) provides maximum attraction between bromine molecules which is directly proportional to the size of the molecule and the boiling point of the liquid made from those molecules.

2. Br₂ deviates from ideal behaviour because it has largest atomic radii compared to Cl₂ and F₂. So it contains more electrons than other two, and the Vander Waals forces are stronger in Br₂ than in Cl₂ and F₂. So Br₂ deviates from ideal behaviour.

Question 38.
Distinguish between diffusion and effusion.
Answer:
Diffusion:

• Diffusion is the spreading of molecules of a substance throughout a space or a second substance.
• Diffusion refers to the ability of the gases to mix with each other.
• E.g.. Spreading of something such as brown tea liquid spreading through the water in a tea cup.

Effusion:

• Effusion is the escape of gas molecules through a very small hole in a membrane into an evacuated area.
• Effusion is a ability of a gas to travel through a small pin-hole.
• E.g., pouring out something like the soap studs bubbling out from a bucket of water.

Question 39.
Aerosol cans carry clear warning of heating of the can. Why?
Answer:
Aerosol cans carry clear warning of heating of the can. As the temperature rises, pressure in the can will increase and ambient temperatures about 120°F may lead to explosions. So aerosol cans should always be stored in dry areas where they will not be exposed to excessive temperatures. You should never throw an aerosol can onto a fire or leave it in the direct sunlight. even it is empty. This is because the pressure will build up so much that the can will burst. It is due to 2 reasons.

• The gas pressure increases.
• More of the liquefied propellant turns into a gas.

Question 40.
When the driver of an automobile applies brake, the passengers are pushed toward the front of the car but a helium balloon is pushed toward back of the car. Upon forward acceleration the passengers are pushed toward the front of the car. Why?
Answer:
1. When the driver of an automobile applies brake, the passengers are pushed toward the front of the car due to inertia of the body, but a helium balloon pushed toward back of the car. Helium balloon responds to the air around it. Helium molecules are lighter than air of our atmosphere, and so they move toward back by gravity as a result of the accelerating frame.

2. Upon forward acceleration, the passengers arc pushed toward the front of the car, because the body in motion tends to stay in motion until acted upon by an outside force. Helium balloon is going to move opposite to this pseudo gravitational force.

Question 41.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It is difficult to drink water with a straw on the top of Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water into the straw at the top of the mountain because gravity falls off gradually with height. The air pressure falls off, there isn’t enough atmospheric pressure to push the water up in the straw all the way to the mouth.

Question 42.
Write the Van der Waals equation for a real gas. Explain the correction term for pressure and volume.
Answer:
Van der Waals equation of state for real gases is –
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT

Correction term for pressure:
\(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\) is the pressure correction. It represents the intermolecular interaction that causes the non ideal behaviour.

Correction term for Volume:
V – nb is the volume correction. it is the effective volume occupied by real gas.

Question 43.
Derive the values of critical constants from the Van der Waals constants.
Answer:
Derivation of critical constants from the Van der Waals constants:
Van der Waals equation is,
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V-n b)\) = nRT for 1 mole
From this equation, the values of critical constant P_CV_C and T_C arc derived in terms of a and b the Van der Waals constants.
\(\left(P+\frac{a n^{2}}{V^{2}}\right)(V – b)\) = RT ………..(1)
On expanding the equestion (1)
P V + \(\frac {a}{V}\) – pb – \(\frac{\mathrm{ab}}{\mathrm{V}^{2}}\) – RT = 0 ………(2)
Multiplying eqestion (2) by \(\frac{V^{2}}{P}\),

equation (3) is rearranged in the powers of V
V³ – \(\left[\frac{\mathrm{RT}}{\mathrm{P}}+\mathrm{b}\right]\) V² + \(\frac {aV}{P}\) –
= 0 ………..(4)
The above equation (4) is an cubic equation of V, which can have three roots. At the critical point. all the three values of V are equal to the critical volume V_C.
i.e. V = V_C.
V – V_C = O ……….(5)
(V – V_C)³ = O ………(6)
(V³ – 3V_CV² + 3V_C³V – V_C³ = 0 ………(7)
As the equation (4) is identical with equation (7), comparing the ‘V’ ternis in (4) and (7),

Divide equation (11) by (10)

When equation (12) is substituted in (10)

substituting the values of Vc and Pc in equation (9)

Critical constant a and b can be calculated using Van der Waals Constant as follows:

Question 44.
Why do astronauts have to wear protective suits when they are on the surface of moon?
Answer:
In space, there is no pressure, if we do wear a pressurised suit, our body will die. In space, we have to wear a pressurised suit, otherwise our body will continue to push out and blow up like a balloon. It would look cool, but we will be dead. So the astronauts in space must wear a pressurised suit (protective suits).

Question 45.
When ammonia combines with HCl, NH₄ Cl is formed as white dense fumes. Why do more funies appear near HCl?
Answer:

1. When ammonia combines with HCl, NH₄ Cl is formed as white dense fumes. The reaction takes place in neutralization between a weak base and a strong acid.
2. The property of the gas is diffusion.
3. Diffusion of gases Ammonia and hydrogen chloride. Concentrated ammonia solution is placed on a pad in one end of a tube and concentrated HCl on the pad at the other. After about a minute, the gases diffuses far enough to meet and a ring of solid ammonium chloride is formed near the HCl end.

Question 46.
A sample of gas at 15°C at 1 atm has a volume of 2.58 dm³. Vhen the temperature is raised to 38°C at I atm does the volume of the gas increase? if so, calculate the final volume.
Answer:

V₂ = 2.78 dm³ i.e. volume increased from 2.58 dm³ to 2.78 dm³.

Question 47.
A sample of gas has a volume of 8.5 dm³ at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 6.37 dm³. What ¡s its initial temperature?
Answer:

T₁ = 364.28 K

Question 48.
Of two samples of nitrogen gas, sample A contains 1.5 moles of nitrogen in a vessel of volume of 37.6 dm³ at 298K, and the sample B is in a vessel of volume 16.5 dm³ at 298 K. Calculate the number of moles in sample B.
Answer:
n_A = 1.5 mol n_B = ?
V_A = 37.6 dm³ V_B = 16.5 dm³
(T = 298 K constant)

Question 49.
Sulphur hexafluoride is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 dm³ at 69.5°C, assuming ¡deal gas behaviour.
Answer:
n = 1.82 mole
V = 5.43 dm³
T = 69.5 + 273 = 342.5
P = ?
PV = nRT
P = \(\frac {nRT}{V}\)

P = 94.25 atm.

Question 50.
Argon is an inert gas used in light bulbs to retard the vapourlzation of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure in atm.
Answer:
P₁ = 1.2 atm
T₁ = 18°C + 273 = 291 K
T₂ = 85°C + 273 = 358 K
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

Question 51.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 6°C and 4 atm. to the water surface, where the temperature is 25°C and pressure is 1 atm. Calculate the final volume in (mL) of the bubble, if its initial volume is 1.5 mL.
Answer:
T₁ = 6°C + 273 = 279 K
P₁ = 4 atm V₁ = 1.5m
T₂ = 25°C + 273 = 298 K
P₂ = 1 atm V ₁ = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V₂ = 6.41 mol.

Question 52.
Hydrochloric acid is treated with a metal to produce hydrogen gas. Suppose a student carries out this reaction and collects a volume of 154.4 x 10^-3 dm³ of a gas at a pressure of 742 mm of Hg at a temperature of 298 K. What mass of hydrogen gas (in mg) did the student collect?
Answer:
V = 154.4 x 10^-3dm³
P = 742 mm of Hg
T = 298 K m = ?
n = \(\frac{PV}{RT}\) =
= 0.006 mol
n = \(\frac{PV}{RT}\)
n = \(\frac{Mass}{Molar Mass}\)
Mass = n x Molar mass
= 0.006 x 2.016
= 0.0121 g = 12.1 mg.

Question 53.
It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?
Answer:

Question 54.
A tank contains a mixture of 52.5 g of oxygen and 65.1 g of CO2 at 300 K the total pressure in the tank ¡s 9.21 atm. Calculate the partial pressure (in atm.) of each gas in the mixture.
Answer:
m_O2 = 52.5 g
P_O2 = ?
m_CO2 = 65.1 g
P_CO2 = ?
T = 300 K P = 9.21 atm
P_O2 = X_O2 x total pressure

P_O2 = X_O2 x Total pressure
= 0.53 x 9.21 atm = 4.88 atm
P_CO2 = X_CO2 x Total pressure
= 0.47 x 9.21 atm = 4.33 atm

Question 55.
A combustible gas Is stored in a metal tank at a pressure of 2.98 atm at 25 °C. The tank can withstand a maximum pressure of 12 atm after which it will explode. The building in which the tank has been stored catches fire. Now predict whether the tank will blow up first or start melting? (Melting point of the metal = 1100 K).
Answer:
T₁ = 298 K;
P₁ = 2.98 atm;
T₂ = 1100K;
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)
P₂ = \(\frac{P_{1}}{T_{1}} \times T_{2}\)
= \(\frac{2.98 \mathrm{atm}}{298 \mathrm{K}} \times 1100 \mathrm{K}\) = 11 atm
At 1100 K, the pressure of the gas inside the tank will become 11 atm. Given that tank can withstand a maximum pressure of 12 atm, the tank will start melting first.

In Text Questions – Evaluate Yourself

Question 1.
Freon-I 2, the compound widely used in the refrigerator system as coolant causes depletion of ozone layer. Now It has been replaced by eco-friendly compounds. Consider 1.5 dm³ sample of gaseous freon at a pressure of 0.3 atm. If the pressure is changed to 1.2 atm. at a constant temperature, what will be the volume of the gas increased or decreased?
Answer:
Volume of freon (V₁) = 1.5 dm³
Pressure (P₁) = 0.3 atm
‘T’ is constant
P₂ = 1.2 atm
V₂ = ?
P₁V₁ = P₂V₂
V₂ = \(\frac{P_{1} V_{1}}{P_{2}}\)

= 0.375 dm³
Volume decreased from 1.5 dm³ to 0.375 dm³

Question 2.
Inside a certain automobile engine, the volume of air in a cylinder is 0.375 dm³, when the pressure is 1.05 atm. When the gas is compressed to a volume of 0.125 dm³ at the same temperature, what is the pressure of the compressed air?
Answer:
V₁ = 0.375dm³
V₂ = 0.125 dm³
P₁ = 1.05 atm
P₂ = ?
T – Constant
P₁V₁ = P₂V₂
P₂ = \(\frac{P_{1} V_{1}}{V_{2}}=\frac{10.5 \times 0.375}{0.125}\)
= 3.15 atm.

Question 3.
A sample of gas has a volume of 3.8 dm³ at an unknown temperature. When the sample is submerged in ice water at 0°C, its volume gets reduced to 2.27 dm³. What is its initial temperature?
Answer:
V₁ = 3.8 dm³
T₂ = 0°C = 273K
T₁ = ? V₂ = 2.27dm³
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)

T₁ = 457 K

Question 4.
An athlete in a kinesiology research study has his lung volume of 7.05 dm³ during a deep inhalation. At this volume the lungs contain 0.312 mole of air. During exhalation the volume of his Jung decreases to 2.35 dm³ How many moles of air does the athlete exhale during exhalation? (assume pressure and temperature remain constant)
Answer:
V₁ = 7.05 dm³
V₂ = 2.35 dm³
n₁ = 0.312 mol
n₁ =?
‘P’ and ‘T’ are constant
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\)

n₂ = 0.104 mol
Number of moles exhaled = 0.312 – 0.104 = 0.208 moles

Question 5.
A small bubble rises from the bottom of a lake, where the temperature and pressure are 8° C and 6.4 atm. to the water surface, where the temperature ¡s 25°C and pressure is 1 atm. Calculate the final volume in (ml) of the bubble, if its initial volume is 2.1 ml.
Answer:
T₁ = 8°C = 8 + 273 = 281K
P₁ = 6.4atm V₁ = 2.1 mol
T₂ = 25°C = 25 + 273 = 298 K
P₂ = 1 atm
V₂ = ?
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)

V₂ = 14.25 ml

Question 6.
(a) A mixture of He and O₂ were used ¡n the ‘air’ tanks of underwater divers for deep dives. For a particular dive 12 dm³ of O₂ at 298 K, I atm. and 46 dm3 of He, at 298 K, 1 aim. were both pumped into a 5 dm³ tank. Calculate the partial pressure of each gas and the total pressure In the tank at 298 K
Answer:

P = 1 atm
V_total = 5 dm³
P_O2 = X_O2 x P_total

P_He = 3.54 atm

Question 6.
(b) A sample of solid KClO₃ (potassium chlorate) was heated in a test tube to obtain O₂ according to the reaction 2KClO₃ → 2KCl_(s) + 3O₂ The oxygen gas was collected by downward displacement of water at 295 K. The total pressure of the mixture is 772 mm of Hg. The vapour pressure of water is 26.7 mn of Hg at 300K. What is the partial pressure of the oxygen gas?
Answer:
2KCl_3(s) → 2KCl_(s) 3O_3(g)
P_total = 772 mm Hg
P_H2O = 26.7 mm Hg
P_total = P_O2 + P_H2O
P_O2 = P_total – P_H2O
P₁ = 26.7 mm Hg
T₁ = 300 K
T₂ = 295 K
P₂ = ?
\(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

P₂ = 26.26 mm Hg
∴ P_O2 = 772 – 26.26
= 745.74 mm Hg

Question 7.
A flammable hydrocarbon gas of particular volume is found to diffuse through a small hole in 1.5 minutes. Under the same conditions of temperature and pressure an equal volume of bromine vapour takes 4.73 min to diffuse through the same hole. Calculate the molar mass of the unknown gas and suggest what this gas might be, (Given that molar mass of bromine = 159.8 g/ mole)
Answer:
t₁ = 1.5 minutes (gas)_hydrocarbon
t₂ = 4.73 minutes (gas)_Bromi

n (12) + (2n + 2) 1 = 16 (general formula for hydrocarbon C_nH_2n+2)
12n + 2n + 2 = 16
14n = 16 – 2
14n = 14
n = 1
The hydrocarbon is C₁H_{2(1) + 2} = CH₄

Question 8.
Critical temperature of H₂O, NH₃ and CO₂ are 647.4, 405.5 and 304.2 K, respectively. When we start cooling from a temperature of 700 K which will liquefy first and which will liquefy finally?
Answer:
Critical temperature of a gas is defined as the temperature above which it cannot be liquefied even at high pressures.
∴ When cooling starts from 700 K, H₂O vill liquefied first, then followed by ammonia and finally carbon dioxide will liquefied.

In-Text Example Problems

Question 9.
In the below figure, let us find the missing parameters [volume in (b) and pressure in (c)]
P₁ = 1 atm P₂ = 2 atm P₃ = ? atm
V₁ I dm3  V₂ =? dm3 V₃ = 0.25 dm³
T₁ = 298 K T₂ = 298 K T₃ = 298 K
Solution:
According to Boyle’s law, at constant temperature for a given mass of gas at constant temperature,
P₁V₁ = P₂V₂ = P₃V₃
I atm x 1 dm³ = 2 atm x V₂ = P₃x 0.25 dm³
∴ 2 atm x V₂ = 1 atm x 1 dm³

V₂ = 0.5 dm³
and P₃ x 0.25 dm³ = 1 atm x 1 dm³

P₃ = 4atm

Question 10.
In the below figure, let us find the missing parameters [volume in (b) and temperature in (c)]
P₁ = 1 atm P₂ = 1 atm P₃ = 1 atm
V₁ = 0.3dm³ V₂ = ?dm³ V₃ = 0.15dm³
T₁ = 200K T₂ = 300 K T₃ = ? K
Answer:

According to Charles law,


T₃ = 100k

Question 11.
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm³ at 70°C assuming It is an ideal gas.
Answer:
We will use the ideal gas equation for this calculation as below:
P = \(\frac {nRT}{V}\)

= 9.39 atm.

Question 12.
A mixture of gases contains 4.76 mole of Ne, 0.74 mole of Ar and 2.5 mole of Xe. Calculate the partial pressure of gases, if the total pressure is 2 atm. at a fixed temperature. Solve this problem using Dalton’s law.
Answer:
P_Ne = X_Ne P_Total

P_Ne = X_Ne P_Total = 0.595 x 2 = 1.19 atm.
P_Ne = X_Ne P_Total = 0.093 x 2 = 0. 186 atm.
P_Ne = X_Ne P_Total = 0.312 x 2 = 0.624 atm.

Question 13.
An unknown gas diffuses at a rate of 0.5 time that of nitrogen at the same temperature and pressure. Calculate the molar mass of the unknown gas.
Answer:

Question 14.
If a scuba diver takes a breath at the surface filling his lungs with 5.82 dm3 of air what volume will the air in his lungs occupy when he drives to a depth, where the pressure ¡s 1.92 atm. (assume temperature is constant and the pressure at the surface is exactly 1 atm.)
Solution :
Temperature = Constant
Pressure at the surface = 1 atm – P₁
Pressure at the depth = 1.92 atm – P₂
Vdlume of air breathing at the surface of the air = 5.82 dm³ – V₁
Volume of air breathing at the depth = V₂ = ?

V₁ = 3.03 dm³
The volume of air scuba diver’s lung occupy = 3.03 dm³

Question 15.
Inside a certain automobile engine, the volume of air in a cylinder is 0.475 dm³, when the pressure is 1.05 atm. When the gas is compressed, the pressure increased to 5.65 atm. at the same temperature. What is the volume of compressed air?
Solution:
Volume of air in the cylinder 0.475 dm³ – V₁
Pressure of air P₁ = 1.05 atm
Increased pressure P₂ = 5.65 atm
Volume of air compressed V₂ = ?
P₁V₁ = P₂V₂

V₂ = 0.08827 dm³
Compressed volume of air = 0.08 827 dm³

Samacheer Kalvi 11th Chemistry Gaseous State Additional Questions Solved

I. Choose the correct answer.

Question 1.
For one mole of a gas, the ideal gas equation is ………..
(a) PV = \(\frac {1}{2}\) RT
(b) PV = RT
(c) PV = \(\frac {3}{2}\)RT
(d) PV = \(\frac {5}{2}\)RT
Answer:
(b) PV= RT

Question 2.
The average kinetic energy of the gas molecule is …………
(a) inversely proportional to its absolute temperature
(b) directly proportional to its absolute temperature
(c) equal to the square of its absolute temperature
(d) All of the above
Answer:
(b) directly proportional to Its absolute temperature

Question 3.
Which of the following is the correct mathematical relation for Charles’ law at constant pressure?
(a) V ∝ T
(b) V ∝ t
(c) V ∝ – \(\frac {1}{T}\)
(d) all of above
Answer:
(a) V ∝ T

Question 4.
At constant temperature, the pressure of the gas is reduced to one-third, the volume
(a) reduce to one-third
(b) increases by three times
(c) remaining the same
(d) cannot be predicted
Answer:
(b) increases by three times

Question 5.
With rise in temperature, the surface tension of a liquid …………
(a) decreases
(b) increases
(c) remaining the same
(d) none of the above
Answer:
(a) decreases

Question 6.
Viscosity of a liquid is a measure of ……………
(a) repulsive forces between the liquid molecules
(b) frictional resistance
(c) intermolecular forces between the molecules
(d) none of the above
Answer:
(b) frictional resistance

Question 7.
The cleansing action of soaps and detergents is due to …………..
(a) internal friction
(b) high hydrogen bonding
(c) viscosity
(d) surface tensions
Answer:
(d) surface tensions

Question 8.
In Vander Waals equation of state for a non-ideal gas the net force of attraction among the molecules is given by ………..
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(b) P + \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(c) P – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
(d) – \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)
Answer:
(a) \(\frac{\mathrm{an}^{2}}{\mathrm{V}^{2}}\)

Question 9.
The compressibility factor, z for an ideal gas is ………….
(a) zero
(b) less than one
(c) greater than one
(d) equal to one
Answer:
(d) equal to one

Question 10.
Which of the following gases will have the lowest rate of diffusion?
(a) H₂
(b) N₂
(c) F₂
(d) O₂
Answer:
(c) F₂

Question 11.
Which of the following is a mono atomic gas in nature?
(a) Oxygen
(b) Hydrogen
(c) Helium
(d) Ozone
Answer:
(c) Helium

Question 12.
Which of the following is a diatomic gas in nature?
(a) Oxygen
(b) Ozone
(c) Helium
(d) Radon
Answer:
(a) Oxygen

Question 13.
Which one of the following is not a monoatomic gas?
(a) Neon
(b) Xenon
(c) Argon
(d) Oxygen
Answer:
(d) Oxygen

Question 14.
Among the following groups which contains monoatomic gases?
(a) Group 17
(b) Group 18
(c) Group 1
(d) Group 15
Answer:
(b) Group 18

Question 15.
Which of the following is a tri atomic gas at room temperature?
(a) Oxygen
(b) Helium
(c) Ozone
(d) Nitrogen
Answer:
(c) Ozone

Question 16.
Which of the following gas is essential for our survival?
(a) N₂
(b) H₂
(c) O₂
(d) He
Answer:
(c) O₂

Question 17.
Among the following, which is deadly poison?
(a) CO₂
(b) HCN
(c) HCl
(d) NH₃
Answer:
(b) HCN

Question 18.
Which of the following is not chemically inert?
(a) Helium
(b) Oxygen
(c) Argon
(d) Krypton
Answer:
(b) Oxygen

Question 19.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 20.
Pressure of a gas is equal to ………..
(a) \(\frac {F}{a}\)
(b) F x a
(c) \(\frac {a}{F}\)
(d) F – a
Answer:
(a) \(\frac {F}{a}\)

Question 21.
The SI unit of pressure is ………..
(a) Nm^-2  Kg^-1
(b) Pascal
(c) bar
(d) atmosphere
Answer:
(b) Pascal

Question 22.
Statement-I: The pressure cooker takes more time for cooking at high altitude.
Statement-II: Air is subjected to Earth’s gravitational force. The pressure of air gradually decreases from the surface of the Earth to higher altitude.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(e) Statement-I is wrong but Statement-II is correct
(ð) Statement-I is correct but Statement-II is wrong
Answer:
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I

Question 23.
The instrument used for measuring the atmospheric pressure is …………….
(a) lactometer
(b) barometer
(c) electrometer
(d) ammeter
Answer:
(b) barometer

Question 24.
The standard atmospheric pressure at sea level at 0°C is equal to ……………..
(a) 1 mm Hg
(b) 76 mm Hg
(c) 760 mm Hg
(d) 680 mm Hg
Answer:
(c) 760 mm Hg

Question 25.
Mathematical expression of Boyle’s law is ………….
(a) P₁V₁ = P₂V₂
(b) \(\frac {P}{V}\) = Constant
(c) \(\frac {V}{T}\) = Constant
(J) \(\frac {P}{T}\) = Constant
Answer:
(a) P₁V₁ = P₂V₂

Question 26.
Statement-I: If the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
Statement-II: If the volume is halved, the density of the gas is doubled.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(a) Statement-I and liare correct and Statement-II is the correct explanation of Statement-I

Question 27.
Which one of the following represents the Charles’ law?
(a) PV = Constant
(b) \(\frac {V}{T}\) = Constant
(c) VT Constant
(d) \(\frac {T}{V}\) = R
Answer:
(b) \(\frac {V}{T}\) = Constant

Question 28.
Which one of the following is absolute zero?
(a) 293 K
(b) 273 K
(c) – 273.15°C
(d) 0°C
Answer:
(c) – 273.15°C

Question 29.
\(\frac {P}{T}\) = Constant is known as …………..
(a) Boyle’s law
(b) Charles’ law
(c) Gay Lussac’s law
(d) Dalton’s law
Answer:
(c) Gay Lussac’s law

Question 30.
The ideal gas equation is …………..
(a) PV = RT for 1 mole
(b) P₁ V₁ = P₂V₂
(c) \(\frac {P}{T}\) = R
(d) P = P₂ + P₂ + P₂
Answer:
(a) PV = RT for 1 mole

Question 31.
The value of Universal gas constant in a ideal gas equation is equal to ………….
(a) 8.3 14 KJ
(b) 0.082057 dm³ atm mol^-1 K^-1
(c) 1 Pascal
(d) 8.314 x 10^-2Pascal
Answer:
(b) 0.082057 dm³ atm mol^-1 K^-1

Question 32.
Mathematical expression of Graham’ s law is ……………….

Answer:

Question 33.
Which law is used in the isotopic separation of deuterium and protium?
(a) Boyle’s law
(b) Charles’ law
(c) Graham’ s law
(d) Gay Lussac’s law
Answer:
(c) Graham’ s law

Question 34.
The value of compression factor Z is equal to ………….
(a) \(\frac {nRT}{PV}\)
(b) \(\frac {PV}{RT}\)
(c) PV x nRT
(d) \(\frac {PV}{nRT}\)
Answer:
(d) \(\frac {PV}{nRT}\)

Question 35.
The value of critical volume is equal in terms of Vander Waals constant is ……….
(a) 3b
(b) \(\frac{8a}{27 Rb}\)
(c) \(\frac{a}{27 b^{2}}\)
(d) \(\frac{2a}{Rb}\)
Answer:
(a) 3b

Question 36.
The value of critical temperature of carbon dioxide is …………
(a) 273 K
(b) 303.98 K
(c) 373 K
(d) – 80°C
Answer:
(b) 303.98 K

Question 37.
Match the List-I and List-II using the correct code given below the list.


Answer:

Question 38.
The value of critical pressure of CO₂ is ……………….
(a) 173 atm
(b) 73 atm
(c) 1 atm
(d) 22.4 atm
Answer:
(b) 73 atm

Question 39.
The temperature below which a gas obey Joule Thomson effect is called …………..
(a) critical temperature
(b) standard temperature
(c) inversion temperature
(d) normal temperature
Answer:
(c) Inversion temperature

Question 40.
The substance used in adiabatic process of liquefaction is ……………
(a) liquid helium
(b) gadolinium sulphate
(c) iron sulphate
(d) liquid ammonia
Answer:
(b) Gadolinium sulphate

Question 41.
The temperature produced in adiabatic process of liquefaction is …………..
(a) zero kelvin
(b) -273 K
(c) 10^-4 K
(d) 10⁴ K
Answer:
(c) 10^-4 K

Question 42.
The molecules of a gas A travel four times faster than the molecules of gas B at same temperature. The ratio of molecular weight M_A/ M_B is ………….
(a) 1/16
(b) 4
(c) 1/4
(d) 16
Answer:
(a) 1/16

Question 43.
The compressibility factor for an ideal gas is …………….
(a) 1.5
(b) 2
(c) 1
(d) x
Answer:
(c) 1

Question 44.
Which of the following pair will diffuse at the same rate?
(a) CO₂ and N₂O
(b) CO₂ and NO
(c) CO₂ and CO
(d) N₂O and NO
Answer:
(a) CO₂ and N₂O

Question 45.
The value of Vander Waals constant “a” is maximum for ……………….
(a) helium
(b) nitrogen
(c) methane
(d) ammonia
Answer:
(d) Ammonia

Question 46.
A person living in Shimla observed that cooking food with using pressure cooker takes more time. The reason for this observation is that at high altitude …………..
(a) pressure increases
(b) temperature decreases
(c) pressure decreases
(a) temperature decreases
Answer:
(c) pressure decreases

Question 47.
Statement-I : At constant temperature PV vs V plot for real gases is not a straight line.
Statement-II : At high pressure, all gases have Z >1, but at intermediate pressure most gases have Z <1.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-Il is wrong
(d) Statement-I is wrong but Statement-Il is correct .
Answer:
(a) Statement-I and liare correct and Statement-II is the correct explanation of Statement-I

Question 48.
Statement-I: Gases do not liquefy above their critical temperature, even on applying high press ure.
Statement-II: Above critical temperature, the molecular speed is high and intermolecular
attractions cannot hold the molecules together because they escape because of high speed.
(a) Statement-I and II are correct and Statement-II is the correct explanation of Statement-I
(b) Statement-I and II are correct but Statement-II is not the correct explanation of Statement-I
(c) Statement-I is correct but Statement-II is wrong
(d) Statement-I is wrong but Statement-II is correct
Answer:
(c) Statement-I and II are correct and Statement-IIs the correct explanation of Statement-I

Question 49.
The rate of diffusion of a gas is ………….
(a) directly proportional to its density
(b) directly proportional to its molecular mass
(c) directly proportional to its square root of its molecular mass
(d) inversely proportional to its square root of its molecular mass
Answer:
(d) inversely proportional to its square root of its molecular mass

Question 50.
In a closed flask of 5 liters, 1.0 g of H₂ is heated from 300 to 600 K, which statement is not correct’?
(a) pressure of the gas increases
(b) the rate of the collusion increase
(c) the number of moles of gas increases
(d) the energy of gaseous molecules increases
Answer:
(c) the number of moles of gas increases

Question 51.
Match the List-I and List-II using the correct code given below the list.

Answer:

Question 52.
Consider the following statements.
(i) All the gases have higher densities than liquids and solids.
(ii) All gases occupy zero volume at absolute zero.
(iii) At very low pressure all gases exhibit ideal behaviour.
Which of the above statement is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
Answer:
(a) (i) only

Question 53.
Which of the following gas is present maximum in atmospheric air?
(a) O₂
(b) N₂
(c) H₂
(d) radon
Answer:
(b) N₂

Question 54.
Which law is used in the process of enriching the isotope of U²³⁵ from other isotopes?
(a) Boyle’s law
(b) Dalton’s law of partial pressure
(c) Graham’s law of diffusion
(d) Charles’ law
Answer:
(c) Graham’s law of diffusion

Samacheer Kalvi 11th Chemistry Gaseous State 2 – Mark Questions

Question 1.
Identify the elements that are in gaseous state under normal atmospheric conditions.
Answer:

• Hydrogen, nitrogen, oxygen, fluorine and chlorine exist as gaseous diatomic molecules.
• Another form of oxygen namely ozone tr iatomic molecule exist as a gas at room temperature.
• Noble gases namely helium, neon, argon, krypton, xenon and radon are mono atomic gases.

Question 2.
Distinguish between a gas and a vapour.
Answer:

• Gas : A substance that is normally in a gaseous state at ordinary temperature and pressure. .,e.g., Hydrogen.
• Vapou r: The gaseous form of any substance that is a liquid or solid at normal temperature and pressure. e.g., At 298 K and 1 atm, water exist as water vapour.

Question 3.
Define pressure. Give its units.
Answer:

• Pressure is defined as the force exerted by a gas on unit area of the wall. Force F
• Pressure = \(\frac {Force}{Area}\) = \(\frac {F}{a}\)
• The SI unit of pressure is Pascal (Pa)

Question 4.
Define atmospheric pressure. What is its value?
Answer:

• The pressure exerted on a unit area of Earth by the colunm of air above it is called atmospheric pressure.
• The standard atmospheric pressure = 1 atm.
• 1 atm = 760 mm Hg.

Question 5.
Deep sea divers ascend slowly and breath continuously by time they reach the surface. Give reason.
Answer:

• For every 10 m of depth, a diver experiences an additional 1 atm of pressure due to the weight of water surrounding him.
• At 20 m, the diver experiences a total pressure of 3 atm. So the most important rule in diving is never hold breath.
• Divers must ascend slowly and breath continuously allowing the regulator to bring the air pressure in their lungs to 1 atm by the time they reach the surface.

Question 6.
Most aeroplanes cabins are artificially pressurized. Why?
Answer:
The pressure decreases with the increase in altitude because there are fewer molecules per unit volume of air. Above 9200 m (30,000 ft), for example, where most commercial aeroplanes fly, the pressure is so low that one could pass out for lack of oxygen. For this reason most aeroplanes cabins arc artificially pressurized.

Question 7.
What is the reason behind the cause of ear pain while climbing a mountain? How it can be rectified?
Answer:

• When one ascends a mountain in a plain, the external pressure drops while the pressure within the air cavities remains the same. This creates an imbalance.
• The greater internal pressure forces the eardrum to bulge outward causing pain.
• With time and with the help of a yawn or two, the excess air within your ear’s cavities escapes thereby equalizing the internal and external pressure and relieving the pain.

Question 8.
State Charles’ law.
Answer:
Charles’ law:
For a fixed mass of a gas constant pressure, the volume is directly proportional to temperature (K).
Mathematically V – T at constant P and n. (or) \(\frac {V}{T}\) = Constant (or) \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) = Constant

Question 9.
What are the applications of Charles’ law?
Answer:

• A hot air inside the balloon rises because of its decreased density and causes the balloon to float  inside the balloon rises because of its decreased density and causes the balloon to float.
• If you take a helium balloon outside on a chilly day, the balloon will crumble. Once you get back into warm area, the balloon will return to its original shape. This is because, in accordance with Charles’ law, a gas like helium takes up more space when it is warm.

Question 10.
State Avogadro’s hypothesis.
Answer:
Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules. Mathematically V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant

Question 11.
Define Dalton’s law of partial pressure.
Answer:
Dalton’s law of partial pressure:
It states that the total pressure of a mixture of gases is the sum of partial pressures of the gases present.
P_total = P₁ + P₂ + P₃

Question 12.
What are the applications of Dalton’s law of partial pressure?
Answer:
1. Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab, the values are reported as partial pressures.
Gas – Normal range
P_CO2 35 – 45mm of Hg
P_O2  80 – 100 mm of Hg

2. When gas is collected by downward displacement of water, the pressure of dry vapour collected is computed using Dalton’s law

Question 13.
How can you identify a heavy smoker with the help of Dalton’s law?
Answer:
Physicians report the pressure of the patient’s gases in blood, analyzed by hospital lab. The values are reported as partial pressures.

A heavy smoker may be expected to have low O₂ and huge CO₂ partial pressures.

Question 14.
Define Graham’s law of diffusion.
Answer:
Graham’s law of diffusion:
The rate of diffusion or effusion is inversely proportional to the square root of molecular mass of a gas through an orifice.
\(\frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{M}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{A}}}}\)
r_Ar_B = rate of diffusion of gases A, B
M_A, M_B = Molecular mass of gases A, B

Question 15.
Helium diffuses more than air. Give reason.
Answer:
Take two balloons, one is filled with air and another with helium. After one day, the helium balloon was shrunk, because helium being lighter diffuses out faster than the air

Question 16.
Explain about the applications of Graham’s law of diffusion.
Answer:

• Graham’s law of diffusion is useful to determine the molecular mass of the gas if the rate of diffusion is known.
• Graham’s law forms the basis of the process of enriching the isotopes of U²³⁵ from other isotopes and also useful in isotopic separation of deuterium and protium.

Question 17.
What is compression factor?
Answer:
The deviation of real gases from ideal behaviour is measured in terms of a ratio of PV to nRT.
This is termed as compression factor.
Compression factor = Z = \(\frac {PV}{nRT}\)
For ideal gases Z = 1 at all temperature and pressures.

Question 18.

1. Define critical temperature.
2. What is the critical temperature of CO₂ gas?

Answer:

1. The temperature below which a gas can be liquefied by application of pressure is known as critical temperature.
2. The critical temperature of CO₂ gas is 303.98 K.

Question 19.

1. Define critical pressure.
2. What is the critical pressure of CO₂ gas?

Answer:

1. Critical temperature (P_C) of a gas is defined as the minimum pressure required to liquefy.
2. The critical pressure of CO₂ is 73 atm.

Question 20.
CO₂ gas cannot be liquefied at room temperature. Give reason.
Answer:
Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO₂ has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 3O₃ K. At room temperature, (critical temperature) even by applying large amount of pressure CO₂ cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO₂ remains as gas.

Question 21.
What is meant by Joule-Thomson effect?
Answer:
The phenomenon of lowering of temperature when a gas is made to expand adiabatically form a region of high pressure into a region of low pressure is known as Joule-Thomson effect.

Question 22.
Define inversion temperature.
Answer:
The temperature below which a gas obey Joule-Thomson effect is called inversion temperature (T_i).
Ti = \(\frac {2a}{Rb}\)

Question 23.
State and explain Boyle’s law. Represent the law graphically.
Answer:
It states that, the pressure of a fixed mass of a gas is inversely proportional to its volume if temperature is kept constant.
P – \(\frac {1}{V}\)
PV = Constant (n and T are constant)
P₁v₁ = P₂V₂
Graphical Representation:

Question 24.
Give an expression for the van der Waals equation. Give the significance of the constants used in the equation. What are their units?
Answer:
\(\left(P+\frac{n^{2} a}{V^{2}}\right)\) (V- nb) = nRT
Where n is the number of moles present and ‘a’’ b’ are known as van der Waals constants.

Significance of Van der Waals constants:
Van der Waals constant ‘a’:
‘a’ is related to the magnitude of the attractive forces among the molecules of a particular gas. Greater the value of’a’, more will be the attractive forces.
Unit of’a’ = L² mol^-2

Van der Waals constant ‘b’:
‘b’ determines the volume occupied by the gas molecules which depends upon size of molecule.
Unit of ‘b’ = L mol^-1

Question 25.
What are ideal and real gases? Out of CO₂ and NH₃ gases, which is expected to show more deviation from the ideal gas behaviour?
Answer:
Ideal gas:
A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.

Real gases:
Gases which deviate from ideal gas behaviour are known as real gases. NH₃ is expected to show more deviation. Since NH₃ is polar in nature and it can be liquefied easily.

Samacheer Kalvi 11th Chemistry Gaseous State 3 – Mark Questions

Question 1.
Explain the graphical representation of Boyle’s law.
Answer:
Boyle’s law states that at a given temperature the volume occupied by a fixed mass of a gas is inversely proportional to its pressure. V ∝ (T and n are fixed) .
If the pressure of the gas increases, volume will decrease and if the pressure of the gas decreases, the volume will increase. So PV = Constant.

Question 2.
What are the consequences of Boyle’s law?
Answer:
1. if the volume of a fixed mass of a gas is reduced to half at constant temperature the gas pressure doubles.
2. Boyle’s law also helps to relate pressure to density.
P₁V₁ = P₃V₃ (Boyle’s law)
\(P_{1} \frac{m}{d_{1}}=P_{2} \frac{m}{d_{2}}\)
Where ‘m’ is the mass, d₁ and d₂ are the densities of gases at pressure P₁ and P₂. The density of the gas is directly proportional to pressure.

Question 3.
Explain Charles’ law with an experimental illustration.
Answer:
Charles’ law states that for a fixed mass of a gas at constant pressure, the volume is directly proportional to temperature (K).
V ∝ T (or) \(\frac {V}{T}\) = Constant

Volume vs Temperature:
If a balloon is moved from an ice water bath to a boiling water bath, the gas molecules inside move faster due to increased temperature and hence the volume increases.

Question 4.
Explain the graphical representation of Charles’ law.
Answer:
1. Variation of volume of the gas sample with temperature at constant pressure.
2. Each line (iso bar) represents the variation of volume with temperature at certain pressure. The pressure increases from P₁ to P₅.
3.  i.e. P₁<P₂ < P₃  <P₄ < P₅ . When these lines are extrapolated to zero volume, they intersect at a temperature of -273.15°C.
4. All gases are becoming liquids, if they are cooled to sufficiently low temperatures.
5. In other words, all gases occupy zero volume at absolute zero. So the volume of a gas can be measured over only a limited temperature range.

Question 5.
Explain graphicl representation of Gay Lussac’s law
Answer:
Gay Lussac’s law
At constant volume, the pressure of a fixed mass of a gas is directly proportional to temperature.
P – T (or) \(\frac {P}{T}\) = Constant
It can be graphically represented as shown here:
Lines in the pressure vs temperature graph are known as isochores (constant volume) of a gas.

Question 6.
Explain the graphical representation of Avogadro’s hypothesis.
Answer:
Avogadros hypothesis states that equal volumes of all gases under the same conditions of temperature and pressure contain equaLnumbr of molecules.
V ∝ n
\(\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}\) = Constant.
Where V₁ and are the volume and number 10- of moles of a gas and V₂ and n₂ are the different set of values of volume and number of moles of the same gas.

A better example to iflustrate Avogadro’s Number of Moles (n) hypothesis is to observe the effect of pumping more gas into a balloon. When more gas molecules (particularly CO₂) are passed, the volume of the balloon increases. The pressure and temperature stay constant as the balloon inflates, so the increase in volume is due to the increase in the quantity of gas inside the balloon.

Question 7.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant. Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.
Boyle’s law: V ∝ \(\frac{1}{P}\)
Charles’ law: V ∝ T
Avogadro’s law: V ∝ n
We can combine these equations into the following general equation that describes the physical
behaviour of all gases.
V ∝ \(\frac{nT}{P}\)
V = \(\frac{nRT}{P}\)
V = , where R = Proportionately constant.
The above equation can be rearranged to give PV = nRT – Ideal gas equation. Where, R is also known as Universal gas constant.

Question 8.
Derive the various values of R, gas constant.
Answer:
1. For standard conditions in which P is 1 atm, volume 22.4 14 dm³ for 1 mole at 273.15 K.

= 0.08205 7 dm³ atm mol^-1 K^-1

2. Where P = 10⁵ Pascal, V = 22.71 x 10^-3 m³ for I mole of a gas at 273.15 K.

= 8.314 pa m³ K^-1 mol^-1
=8.314x 10^-2bar dm³ K^-1mol^-1
R = 8.314 J K^-1 mol^-1.

Question 9.
What is meant by Boyle temperature (or) Boyle point? How is it related with compression point?
Answer:
(1) Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.
(2) The Boyle point varies with the nattirc of the gas.
(3) Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.
(4) Below the Boyle point, the real gases first show a decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperature, the real gases behave as ideal gases.

Question 10.
Explain the different methods used for liquefaction of gases.
Answer:

• Linde’s method: Joule-Thomson effect is used to get liquid air or any other gas.
• Claude’s process: In addition to Joule-Thomson effect, the gas is allowed to perform mechanical work so that more cooling is produced.
• Adiabatic process: This method of cooling is produced by removing the magnetic property of magnetic material e.g. Gadolinium sulphate. By this method, a temperature of 10^-4 K i.e. as low as zero Kelvin can be achieved.

Samacheer Kalvi 11th Chemistry Gaseous State 5 – Mark Question

Question 1.
How CH₄ He and NH₃ are deviating from ideal behaviour? (or) Explain how real gases deviate from ideal behaviour.
Answer:
1. The gases which obeys gas equation PV = nRT are known as ideal gases. The gases which do not obey PV = nRT are known as real gases.

2. The gas laws and the kinetic theory are based on the assumption that molecules in the gas phase occupy negligible volume (assumption 1) and that they do not exert any force on one another either attractive or repulsive (assumption 2). Gases whose behaviour is consistent with these assumptions are said to exhibit ideal behaviour.

3. The following graph shows RT plotted against P for three real gases and an ideal gas at a given temperature.

4. According to ideal gas equation, PV/RT is equal to n. Plot PV/RT versus P for ‘n’ moles oigas at 0°C. For1 mole of an ideal gas PV/ RT is equal to 1 irrespective of the pressure of the gas.

5. For real gases, we observe various deviations from ideal behaviour at high pressure. At very low pressure, all gases exhibit ideal behaviour, ie. PV/RT values all converge to n as P approaches zero.

6. For real gases, this is true only at moderate low pressures. (≤ 5 atm) significant variation occurs as the pressure increases. Attractive forces operate among molecules at relatively short deviation.

7. At atmospheric pressure, the molecules in a gas are far apart and attractive forces arc negligible. At high pressure, the density of the gas increases and the molecules are much closer to one another. Intermolecular forces can be significant enough to affect the motion of the molecules and the gas will not behave ideally.

Question 2.
Derive Van der Waals equation of state.
Answer:
1. Consider the effect of intermolecular forces on the pressure exerted by a gas form the following explanation.

2. The speed of a molecule that is moving toward the wall of a container is reduced by the attractive forces exerted by its neighbours. Hence, the measured gas pressure is Q lower than the pressure the gas would exert, if it behave ideally.


Where ‘a’ is the proportionality constant and depends on the nature of the gas and n and V are the number of moles and volume of the container and respectively an²/ V² is the correction term.

3. The frequency of encounters increases with the square of the number of molecules per unit volume n²/ V². Therefore an²/ V² represents the intermolecular interaction that causes non-ideal behaviour.

4.  Another correction is concerned with the volume ¿ccupied by the gas molecules. ‘V’ represents the volume of the container. As every individual molecule of a real gas occupies certain volume, the effective volume V- nb which is the actually available for the gas, n is the number of moles and b is a constant of gas.

5.  Hence Van der Waals equation of state for real gases are given as \(\left(P+\frac{a n^{2}}{V^{2}}\right)\) (V – nb) = nRT Where a and b are Van der Waals constants.

Question 3.
Explain about Andrew’s experimental isotherms of CO₂ gas.
Answer:
1. Andrew plotted isotherms of carbon dioxide at different temperatures. it is then proved that many real gases behave in a similar manner like CO₂.

2. At a temperature of 303.98 K, CO₂ remains as a gas. Below this temperature, CO₂, turns into liquid CO₂ at 7 3atm. It is called the critical temperature of CO₂.

3. At 303.98 K and 73 atm pressure, CO₃,, becomes a liquid but remains a gas at higher temperature.

4. Below the critical temperature, the behaviour of CO₂ is different. For example, consider an isotherm of CO₂ at 294.5 K, it is a gas until the point B, is reached. At B, a liquid separates along the line BC, both the liquid and gas co-exist. At C, the gas is completely condensed.

5. If the pressure is higher than at C, only the liquid is compressed so, a steep rise in pressure is observed. Thus, there exist a continuity of state.

6.  A gas below the critical temperature can be liquefied by applying pressures.

Activity – 1
The table below contains the values of pressure measured at different temperatures for 1 moie of an ideal gas. Plot the values in a graph and verify the Gay Lussac’s law. [Lines in the pressure vs temperature graph are known as iso chores (constant volume) of a gas]

Solution:
Gay Lussac’s law at constant volume = \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

If temperature increases, pressure also increases.
So \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\)

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Samacheer Kalvi 10th Science Periodic Classification of Elements Textual Evaluation Solved

I. Choose the best answer.

Ex 3.14 Class 10 Samacheer Question 1.
The number of periods and groups in the periodic table are ____.
(a) 6, 16
(b) 7, 17
(c) 8, 18
(d) 7, 18.
Answer:
(d) 7, 48.

10th Maths Exercise 3.14 Samacheer Kalvi Question 2.
The basis of modern periodic law is:
(a) atomic number
(b) atomic mass
(c) isotopic mass
(d) number of neutrons
Answer:
(a) atomic number

Exercise 3.14 Class 10 Samacheer Question 3.
_____ group contains the member of the halogen family.
(a) 17th
(b) 15th
(c) 18th
(d) 16th.
Answer:
(a) 17th

10th Maths Exercise 3.14 Question 4.
……….is a relative periodic property.
(a) atomic radii
(b) ionic radii
(c) electron affinity
(d) electronegativity
Answer:
(d) electronegativity

10th Maths Exercise 3.14 Solutions Question 5.
Chemical formula of rust is ____.
(a) FeO.xH₂O
(b) FeO₄.xH₂O
(c) Fe₂O₃.xH₂O
(d) FeO.
Answer:
(c) Fe₂O₃.xH₂O

10th Maths Exercise 3.14 Solution Question 6.
In the aluminothermic process the role of Al is ____.
(a) oxidizing agent
(b) reducing agent
(c) hydrogenating agent
(d) sulphurising agent.
Answer:
(b) reducing agent

Maths Book For Class 10 Samacheer Kalvi Question 7.
The process of coating the surface of metal with a thin layer of zinc is called:
(a) painting
(b) thinning
(c) galvanization
(d) electroplating
Answer:
(c) galvanization

10th Samacheer Maths Question 8.
Which of the following have inert gases 2 electrons in the outermost shell?
(a) He
(b) Ne
(c) Ar
(d) Kr.
Answer:
(a) He

Samacheer Kalvi 10th Guide Maths Question 9.
Neon shows zero electron affinity due to:
(a) stable arrangement of neutrons
(b) stable configuration of electrons
(c) reduced size
(d) increased density
Answer:
(b) stable configuration of electrons

10th Maths Solution Samacheer Kalvi Question 10.
____ is an important metal to form an amalgam.
(a) Ag
(b) Hg
(c) Mg
(d) Al.
Answer:
(b) Hg

II. Fill in the blanks.

10th Standard Maths Algebra Question 1.
If the electronegativity difference between two bonded atoms in a molecule is greater than 1.7, the nature of bonding is ____.
Answer:
Ionic bond.

10th Maths Samacheer Question 2.
____ is the longest period in the periodical table.
Answer:
Sixth period.

Maths Book Class 10 Samacheer Kalvi Question 3.
______ forms the basis of modem periodic table.
Answer:
Atomic number.

Question 4.
If the distance between two Cl atoms in Cl₂ molecule is 1.98 Å, then the radius of the Cl atom is ____.
Answer:
0.99 Å.

Question 5.
Among the given species A^–, A⁺, and A, the smallest one in size is ______.
Answer:
A^+.

Question 6.
The scientist who propounded the modem periodic law is ______.
Answer:
Henry Moseley.

Question 7.
Across the period, ionic radii ______ (increases,decreases)
Answer:
Decreases.

Question 8.
_______ and ______ are called inner transition elements.
Answer:
Lanthanides and Actinides.

Question 9.
The chief ore of Aluminium is ______.
Answer:
Bauxite (Al₂O₃.2H₂O).

Question 10.
The chemical name of rust is ______.
Answer:
Hydrated ferric oxide.

III. Match the following.

Question 1.

1. Galvanisation	(a) Noble gas elements
2. Calcination	(b) Coating with Zn
3. Redox reaction	(c) Silver-tin amalgam
4. Dental filling	(d) Alumino thermic process
5. Group 18 elements	(e) Heating in the absence of air

Answer:
1 – (b), 2 – (e), 3 – (d), 4 – (c), 5 – (a).

IV. True or False: (If false give the correct statement)

Question 1.
Moseley’s periodic table is based on atomic mass.
Answer:
False.
Correct Statement: Moseley’s periodic table is based on atomic number.

Question 2.
Ionic radius increases across the period from left to right.
Answer:
False.
Correct Statement: Ionic radius decreases across the period from left to right.
Question 3.
All ores are minerals, but all minerals cannot be called as ores.
Answer:
True.

Question 4.
Al wires are used as electric cables due to their silvery – white colour.
Answer:
False.
Correct Statement: Al wires are used as electric cables due to their good conductor of electricity.

Question 5.
An alloy is a heterogeneous mixture of metals.
Answer:
False.
Correct Statement: An alloy is a homogenous mixture of metals.

V. Assertion and Reason:

Answer the following questions using the data given below.
(i) A and R are correct, R explains the A.
(ii) A is correct, R is wrong.
(iii) A is wrong, R is correct.
(iv) A and R are correct, R doesn’t explain A.

Question 1.
Assertion: The nature of bond in HF molecule is ionic
Reason: The electronegativity difference between H and F is 1.9.
Answer:
(i) A and R are correct, R explains the A.

Question 2.
Assertion: Magnesium is used to protect steel from rusting.
Reason: Magnesium is more reactive than iron.
Answer:
(i) A and R are correct, R explains the A.

Question 3.
Assertion: An uncleaned copper vessel is covered with a greenish layer.
Reason: copper is not attacked by alkali
Answer:
(iv) A and R are correct, R doesn’t explain A.

VI. Short Answer Questions.

Question 1.
A is a reddish-brown metal, which combines with O₂ at < 1370 K gives B, a black coloured compound. At a temperature > 1370 K, A gives C which is red in colour. Find A, B and C with reaction.
Answer:
(i) Reddish brown metal (A) is copper.

(ii) (A) reacts with O₂ at bleow 1370 K gives Copper (II) oxide (B), which is black in colour.
2Cu + O₂ \(\xrightarrow [ 1370k ]{ below }\) 2CuO (Copper (II) oxide) (B).

(iii) (A) reacts with O₂ at above 1370 K gives Copper (I) oxide (C), which is red in colour.

Question 2.
A is a silvery – white metal. A combines with O₂ to form B at 800°C, the alloy of A is used in making the aircraft. Find A and B.
Answer:
(i) Silver – white metal (A) is Aluminium.

(ii) (A) combines with O₂ to form aluminium oxide at 800°C.

(iii) Duralumin is the alloy of Al, which is used to make aircraft

Question 3.
What is rust? Give the equation for formation of rust.
Answer:
Rust is hydrated ferric oxide, Fe₂O₃.xH₂O. It is formed when iron is exposed to moist air.
4 Fe + 3O₂ + xH₂O > 2Fe₂O₃.xH₂O

Question 4.
State two conditions necessary for rusting of iron.
Answer:
Conditions for rusting of iron:

• The presence of water and oxygen is essential for the rusting of iron.
• Impurities in the iron, the presence of water vapour, acids, salts and carbon dioxide hasten to rust.
• Pure iron does not rust in dry and CO₂ free air. It also does not rust in pure water, free from dissolved salts.

VII. Long Answer Questions

Question 1.
(a) State the reason for the addition of caustic alkali to bauxite ore during purification of bauxite.
(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture. Name the substance and give one reason for the addition.
Answer:
(a) Naturally, Bauxite is not soluble in normal solvents. Therefore the addition of caustic alkali to bauxite plays an important role while extraction of aluminium. Caustic alkali dissolves bauxite forming soluble sodium meta aluminate while impurities remain insoluble and precipitate as red mud.

(b) Along with cryolite and alumina, another substance is added to the electrolyte mixture is Fluorspar. Adding of fluorspar lowers the fusion temperature of the electrolyte.

Question 2.
The electronic configuration of metal A is 2, 8, 18, 1. The metal A when exposed to air and moisture forms B a green layered compound. A with conc. H₂SO₄ forms C and D along with water. D is a gaseous compound. Find A, B, C and D.
Answer:
(i) The electronic configuration of metal (A) is 2, 8, 18, 1. A is copper (Z = 29)

(ii) (A) Copper exposed to air and moisture forms green layered compound (B) that is a copper carbonate.

(iii) Copper (A) reacts with conc.H₂SO₄ to give copper sulphate (C) and Sulphur dioxide (D).

Question 3.
Explain the smelting process.
Answer:
Smelting (in a Blast Furnace): The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top. There are three important regions in the furnace.

(a) The Lower Region (Combustion Zone): The temperature is at 1500°C. In this region, coke bums with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

(b) The Middle Region (Fusion Zone): The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

(c) The Upper Region (Reduction Zone): The temperature prevails at 400°C . In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.
\(\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{Fe}+3 \mathrm{CO}_{2}\)
The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

VIII. HOT Questions.

Question 1.
Metal A belongs to period 3 and group 13. A in red hot condition reacts with steam to form B. A with strong alkali forms C. Find A, B and C with reactions.
Answer:
(i) Metal (A) belongs to period 3 and group 13, is Aluminium (Al).

(ii) (A) Al in red hot condition reacts with steam to form Aluminium oxide (B).

(iii) Aluminium (A) reacts with strong alkali forms of sodium meta aluminate (C).

Question 2.
Name the acid that renders aluminium passive. Why?
Answer:
Dil. or cone. HNO₃ does not attack aluminium, but renders Al passive due to the formation of oxide film on its surface.

Question 3.
(a) Identify the bond between H and F in the HF molecule.
(b) What property forms the basis of identification?
(c) How does the property7 vary in periods and in groups?
Answer:
(a) The nature of the bond in the HF molecule is ionic.
(b) Electronegativity.
(c) Along the period, from left to right in the periodic table the electronegativity increases because of the increase in the nuclear charge which in turn attracts the electrons more strongly. On moving down a group, the electronegativity of the elements decreases because of the increased number of energy levels.

Samacheer Kalvi 10th Science Periodic Classification of Elements Additional Questions Solved

I. Choose the best answer.

Question 1.
Which period contains only two elements?
(a) Second
(b) First
(c) Third
(d) Fifth.
Answer:
(b) First

Question 2.
The sixth period contains inner transition elements.
(a) 18
(b) 14
(c) 10
(d) 8
Answer:
(b) 14

Question 3.
Lanthanides and Actinides are called as _____.
(a) Alkali metals
(b) Inner transition elements
(c) Transition elements
(d) Representative elements.
Answer:
(b) Inner transition elements

Question 4.
The number of valence electrons present in Halogens is:
(a) One
(b) Seven
(c) Zero
(d) Two
Answer:
(b) Seven

Question 5.
The distance between the two hydrogen nuclei of the molecule is 0.74 Å. So its covalent radius is _____.
(a) 0.74 Å
(b) 0.99 Å
(c) 0.37 Å
(d) 7.4 Å.
Answer:
(c) 0.37 Å

Question 6.
Along the groups, atomic radius _____.
(a) decreases
(b) increases
(c) decreases then increase
(d) no change.
Answer:
(b) increases.

Question 7.
Which one of the following elements will have the highest electronegativity?
(a) chlorine
(b) nitrogen
(c) caesium
(d) fluorine
Answer:
(d) fluorine

Question 8.
Electron affinity is measured in _____.
(a) kJ^-1
(b) mol^-1
(c) kJ/mol
(d) kJ/mol².
Answer:
(c) kJ/mol

Question 9.
Noble gases have ______ electron affinity.
(a) positive
(b) negative
(c) zero
(d) high.
Answer:
(c) zero

Question 10.
The element with positive electron gain enthalpy is:
(a) Hydrogen
(b) Sodium
(c) Argon
(d) Fluorine
Answer:
(b) Sodium

Question 11.
The mineral from which a metal can be readily and economically extracted on a large scale is said to be a / an _____.
(a) Ore
(b) Flux
(c) Slag
(d) Gangue.
Answer:
(a) Ore

Question 12.
Flux + Gangue → _____?
(a) Mineral
(b) Matrix
(c) Slag
(d) Smog.
Answer:
(c) Slag

Question 13.
………… is used for anodizing process.
(a) Zinc
(b) iron
(c) Aluminium
(d) Tin
Answer:
(c) Aluminium

Question 14.
The ore which can be purified by gravity separation method is _____.
(a) Haematite
(b) oxide ores
(c) sulphide ores
(d) both (a) and (b).
Answer:
(d) both (a) and (b).

Question 15.
The calcium silicate slag is formed in ………. zone.
(a) combustion zone
(b) fusion zone
(c) reduction
(d) molten
Answer:
(b) fusion zone

Question 16.
Zinc blende is purified by _____.
(a) Hydraulic method
(b) Magnetic separation method
(c) Froth floatation method
(d) Chemical method.
Answer:
(c) Froth floatation method

Question 17.
Bauxite ore is purified by _____.
(a) Leaching process
(b) Hydraulic method
(c) Froth floatation method
(d) Magnetic separation method.
Answer:
(a) Leaching process

Question 18.
When copper reacts with dil.HNO₃ it liberates …………. gas.
(a) SO₂
(b) NO₂
(C) CO₂
(d) NO
Answer:
(d) NO

Question 19.
Which metal process low melting point?
(a) Gallium
(b) Caesium
(c) Aluminium
(d) Copper.
Answer:
(a) Gallium

Question 20.
Which one of the following is not an ore of aluminium?
(a) Bauxite
(b) Haematite
(c) Cryolite
(d) Corundum.
Answer:
(b) Haematite

Question 21.
……….. is stored in Kerosene.
(a) Iron
(b) Silver
(c) Sodium
(d) Aluminium
Answer:
(c) Sodium

Question 22.
Electrolytic reduction of alumina into aluminium is ______.
(a) Hall’s process
(b) Alumino thermic process
(c) Baeyer’s process
(d) Bessemerisation process.
Answer:
(a) Hall’s process

Question 23.
In Hall’s process, cathode used is ______.
(a) Iron tank
(b) Graphite
(c) Pure alumina
(d) Iron tank linked with graphite.
Answer:
(d) Iron tank linked with graphite.

Question 24.
The metal oxides are usually……….. in nature.
(a) basic
(b) acidic
(c) amphoteric
(d) neither acidic nor basic
Answer:
(a) basic

Question 25.
A silvery-white metal is ______.
(a) Aluminium
(b) Copper
(c) Iron
(d) Zinc.
Answer:
(a) Aluminium

Question 26.
Aluminium reacts with NaOH to give ______.
(a) Al₂O₃
(b) AlCl₃
(c) NaAlO₂
(d) Al(OH)₃
Answer:
(c) NaAlO₂

Question 27.
Atoms of the elements belonging to the same group of periodic table will have:
(a) same number of protons
(b) same number of electrons in the valence shell
(c) same number of neutrons
(d) same number of electrons
Answer:
(b) same number of electrons in the valence shell

Question 28.
Chief ore of copper is ______.
(a) CuFeS₂
(b) Cu₂O
(c) Cu₂S
(d) CuSO₄
Answer:
(a) CuFeS₂

Question 29.
Molecular formula for copper pyrites ______.
(a) Cu₂O
(b) Cu₂S
(c) CuCO₃
(d) CuFeS₂
Answer:
(d) CuFeS₂

Question 30.
The electronegativity of fluorine is:
(a) 4.0
(b) 3
(c) 2.8
(d) 2.1
Answer:
(a) 4.0

Question 31.
The second most abundant metal available next to aluminium is ______.
(a) Cu
(b) Ag
(c) Au
(d) Fe.
Answer:
(d) Fe.

Question 32.
Most important ore of iron is ______.
(a) Haematite
(b) Magnetite
(c) Iron pyrite
(d) Cryolite.
Answer:
(a) Haematite

Question 33.
The volatile impurities present in haematite are:
(a) Carbon and Nitrogen
(b) Oxygen and Nitrogen
(c) Helium and Oxygen
(d) Sulphur and Phosphorus
Answer:
(d) Sulphur and Phosphorus

Question 34.
The solute present in brass is:
(a) copper
(b) zinc
(c) tin
(d) magnesium
Answer:
(b) zinc

Question 35.
Which one of the following is used for making pressure cookers?
(a) Brass
(b) Magnalium
(c) Duralumin
(d) Nickel steel
Answer:
(c) Duralumin

Question 36.
Which is used as propeller?
(a) Stainless steel
(b) Nickel steel
(c) Brass
(d) Magnalium.
Answer:
(b) Nickel steel

Question 37.
Gold does not occur in the combined form. It does not react with air or water. It is in the ______ state.
(a) native
(b) combined
(c) complex
(d) molten.
Answer:
(a) native

Question 38.
Which of the following metal is not found in a free state?
(a) Ag
(b) Au
(c) Pt
(d) Al.
Answer:
(d) Al.

Question 39.
Which one of the following does not react with copper?
(a) Oxygen
(b) Conc.H₂SO₄
(c) NaOH
(d) Conc.HNO₃
Answer:
(c) NaOH

Question 40.
An element which is an essential constituent of all organic compounds belongs to ______ group.
(a) 14th
(b) 15th
(c) 16th
(d) 17th.
Answer:
(a) 14th

Question 41.
The highest ionization energy is exhibited by ______.
(a) Halogens
(b) Alkaline earth metals
(c) Transition metals
(d) Nobel gases.
Answer:
(d) Nobel gases.

Question 42.
Which two elements of the following belongs to the same period? (Al, Si, Ba, O).
(a) Si, Ba
(b) Al, Ba
(c) Al, Si
(d) Al, O.
Answer:
(c) Al, Si

Question 43.
98% pure copper and 2% impurities is called ______.
(a) Matte
(b) Copper pyrites
(c) blister copper
(d) cuprite.
Answer:
(c) blister copper

Question 44.
_____ is used in making anchors and electromagnets.
(a) Steel
(b) Pig iron
(c) Cast iron
(d) Wrought iron.
Answer:
(d) Wrought iron.

Question 45.
Which reagent does not react with iron?
(a) Conc. HNO₃
(b) Conc.H₂SO₄
(c) Steam
(d) Dil. HNO₃
Answer:
(a) Conc. HNO₃

II. Fill in the blanks.

Question 1.
Matte is a mixture of ______.
Answer:
Cu₂S + FeS.

Question 2.
Second group elements are called ______.
Answer:
Alkaline earth metals.

Question 3.
Ionisation energy is measured in _____ unit.
Answer:
kJ/mol

Question 4.
The ionisation energy ______ along the period.
Answer:
Increases.

Question 5.
______ property, which predicts the nature of bonding between the atoms in a molecule.
Answer:
Electronegativity.

Question 6.
If the difference in electronegativity between two elements is 1.7, the bond has ____ and _____.
Answer:
50% ionic character, 50% covalent character.

Question 7.
If the difference in electronegativity between two elements is less than 1.7, the bond is considered to be ______.
Answer:
Covalent.

Question 8.
If the difference in electronegativity between two elements is greater than 1.7, the bond is considered to be ______.
Answer:
Ionic.

Question 9.
The process of extracting the ores from the earth’s crust is called ______.
Answer:
Mining.

Question 10.
______ is the main principle behind in Hydraulic method.
Answer:
Specific gravity.

Question 11.
Froth floatation process is preferable for ______ ores.
Answer:
Lighter
(or)
Sulphide.

Question 12.
On heating in air, iron forms ______.
Answer:
Fe₃O₄

Question 13.
Iron reacts with Chlorine to form _____ compound.
Answer:
Ferric chloride.

Question 14.
The corrosive action in the absence of moisture is called ______.
Answer:
Dry corrosion.

Question 15.
______ technique used to renovate the Pamban bridge.
Answer:
Protective coating.

Question 16.
The atomic number is the number of ____ in the nucleus or number of ______ revolving around the nucleus in an atom.
Answer:
Protons, electrons.

Question 17.
The long form of periodic table is based upon the _____ of elements.
Answer:
Electronic configuration.

Question 18.
In the periodic table, the horizontal rows are called _____ and vertical columns are called ______.
Answer:
Periods, groups.

Question 19.
The modem periodic table has been divided into ______ blocks known as _______ blocks.
Answer:
Four, s, p, d, f

Question 20.
The ______ of the elements in a period decreases from left to right and the atomic radii of the elements present in a group downwards.
Answer:
Atomic size, increases.

Question 21.
_______ period is the longest period and it contains ______ elements.
Answer:
Seventh, 32.

Question 22.
In the periodic table, there are ______ groups and _____ periods.
Answer:
18, 7.

Question 23.
Metals like Ti, Cr, Mn, Zr find their application in the manufacturing of defence equipment called ______.
Answer:
Strategic metals.

Question 24.
The metal ______ plays a vital role in nuclear reactions releasing nuclear energy and used in nuclear weapons.
Answer:
Uranium.

Question 25.
Copper, silver and gold are called _______ as they are used in making ____ and ______.
Answer:
Coinage metals, coins and jewellery.

Question 26.
Purity of gold is expressed in ______ and ______ is pure gold.
Answer:
Carats, 24 – carat gold.

Question 27.
_______ is an ore of aluminium and _____ is its mineral.
Answer:
Bauxite, clay.

Question 28.
All ______ cannot be called as ores but all ______ are minerals.
Answer:
Minerals, ores.

Question 29.
The process of extracting the ores from the earth’s crust is called ______.
Answer:
Mining.

Question 30.
The rocky impurity associated with the ore is called ____ or ______.
Answer:
Gangue, matrix.

Question 31.
_____ is a substance added to the ore to reduce the fusion temperature and to remove impurities.
Answer:
Flux.

Question 32.
____ is the process of reducing the roasted metallic oxide to metal.
Answer:
Smelting.

Question 33.
Slag is the fusible product formed when ____ reacts with ____ during the extraction of metals.
Answer:
Flux, gangue

Question 34.
The temperature applied in Hall’s process is _____ and the voltage used in ______.
Answer:
900 – 950°C, 5 – 6V

Question 35.
_____ is used in making manhole covers and drain pipes and _______ is used in making transmission cables and T.V. towers.
Answer:
Pig iron, steel.

Question 36.
______ is defined as the slow and steady destruction of a metal by the environment.
Answer:
Corrosion.

Question 37.
_______ is a process of coating zinc on iron sheets by using electric current.
Answer:
Galvanization.

III. Match the following.

Question 1.

i. Boron family	(a) Group 17
ii. Carbon family	(b) Group 16
iii. Nitrogen family	(c) Group 13
iv. Chalcogen family	(d) Group 15
v. Halogen family	(e) Group 14

Answer:
i – c, ii – e, iii – d, iv – b , v – a.

Question 2.

i. Alkali metals	(a) Lanthanides & Actinides
ii. Alkaline earth metals	(b) Groups 3 – 12
iii. Transition elements	(c) Group 2
iv. Inner transition elements	(d) Group 1

Answer:
i – d, ii – c, iii – b, iv – a.

Question 3.

i. Group 18	(a) Main group elements
ii. Group 3 – 12	(b) Noble gases
iii. Group 13 – 18	(c) Halogens
iv. Group 17	(d) Transition elements

Answer:
i – b, ii – d, iii – a, iv – c.

Question 4.

i. Copper	(a) Lustrous greyish white metal
ii. Iron	(b) Cn 112
iii. Aluminium	(c) Reddish-brown metal
iv. Copernicium	(d) Silvery white metal

Answer:
i – c, ii – a, iii – d, iv – b.

Question 5.

Alloys	Composition
i. Brass	(a) Al, Mg, Mn, Cu
ii. Duralumin	(b) Cu, Zn
iii. Bronze	(c) Al, Mg
iv. Magnalium	(d) Cu, Sn

Answer:
i – b, ii – a, iii – d, iv – c.

Question 6.

Elements	Electronegative value
i. F	(a) 2.5
ii. Cl	(b) 2.8
iii. Br	(c) 3.0
iv. I	(d) 4.0

Answer:
i – d, ii – c, iii – b, iv – a.

Question 7.

Process	Ores
i. Hydraulic process	(a) ZnS
ii. Magnetic separation	(b) Fe2O3
iii. Froth floatation process	(c) SnO2
iv. Leaching process	(d) Al2O3. 2H2O

Answer:
i – b, ii – c, iii – a, iv – d.

Question 8.

i. Cuprite	(a) Halide ore
ii. Marble	(b) Oxide ore
iii. Fluorspar	(c) Sulphide ore
iv. Galena	(d) Carbonate ore

Answer:
i – b, ii – d, iii – a, iv – c.

Question 9.

Alloy	Metals present	Uses
1. Brass	Fe, C, Ni	Statues, Coins
2. Bronze	Al, Mg, Mn, Cu	Aircraft, Pressure cookers
3. Duralumin	Fe, C, Ni, Cr	Cables, Propeller
4. Magnalium	Cu, Sn	Automobile parts, Utensils
5. Stainless steel	Cu, Zn	Scientific instruments, Air craft
6. Nickel steel	Al, Mg	Medals, decorative items

Answer:

Alloy	Metals present	Uses
1. Brass	Cu, Zn	Medals, decorative items
2. Bronze	Cu, Sn	Statues, Coins
3. Duralumin	Al, Mg, Mn, Cu	Aircraft, Pressure cookers
4. Magnalium	Al, Mg	Scientific instruments, Air craft
5. Stainless steel	Fe, C, Ni, Cr	Automobile parts, Utensils
6. Nickel steel	Fe, C, Ni	Cables, Propeller

Question 10.

Metal	Ore	Chemical formula
1. Copper	Bauxite	ZnCO3
2. Aluminium	Haematite	CuFeS2
3. Iron	Copper pyrite	Fe2O3
4. Zinc	Calamine	Al2O3.2H2O

Answer:

Metal	Ore	Chemical formula
1. Copper	Copper pyrite	CuFeS2
2. Aluminium	Bauxite	Al2O3.2H2O
3. Iron	Haematite	Fe2O3
4. Zinc	Calamine	ZnCO3

IV. State whether true or false. If false, give the correct statement.

Question 1.
First period contains only one element.
Answer:
False.
Correct Statement: First period contains two elements.(Hydrogen & Ftelium)

Question 2.
The valency of all alkali metals is one.
Answer:
True.
Question 3.
Noble gases are more reactive.
Answer:
False.
Correct Statement: Noble gases are less reactive.

Question 4.
The atomic radius decreases from Li to B?
Answer:
True.

Question 5.
As the positive charge increases, the size of the cation also increases.
Answer:
False.
Correct Statement: As the positive charge increases, the size of the cation also decreases.

Question 6.
Copper pyrite ore is concentrated by gravity separation method.
False
Correct Statement: Copper pyrite ore is concentrated by froth floatation process.

Question 7.
Aluminium alloyed with gold and silver for making coins and jewels.
Answer:
False.
Correct Statement: Copper alloyed with gold and silver for making coins and jewels.

Question 8.
The corrosive action in the presence of moisture is called wet corrosion.
Answer:
True

Question 9.
The physical and chemical properties of elements are the periodic function of their atomic numbers – modern periodic law.
Answer:
True.

Question 10.
The long form of periodic table consists of horizontal rows called groups and vertical columns called periods.
Answer:
False.
Correct statement: The long form of periodic table consists of horizontal rows called periods and vertical columns called groups.

Question 11.
The first period in the periodic table is the shortest period and contains 8 elements from Lithium to Neon.
Answer:
False.
Correct statement: The first period in the periodic table is the shortest period and contains 2 elements Hydrogen and Helium.
(OR)
The second period in the periodic table is the short period and contains 8 elements from Lithium to Neon.

Question 12.
The sixth period in the periodic table is the longest period and contains 32 elements.
Answer:
True.

Question 13.
Group 1, 2 and 13 – 18 are called normal elements.
(or)
Main group elements.
(or)
Representative elements.
Answer:
True.

Question 14.
The atomic size of the elements in a period increases from left to right.
Answer:
False.
Correct statement: The atomic size of the elements in a period decreases from left to right.

Question 15.
In a period, the metallic character of the element increases while their non-metallic character decreases.
Answer:
False.
Correct statement: In a period, the metallic character of the element decreases while their non-metallic character increases.

Question 16.
The last element authenticated by IUPAC is Cn 112 [Copemicium].
Answer:
True.

Question 17.
Silver was the first metal to be used in making utensils and weapons.
Answer:
False.
Correct statement: Copper was the first metal to be used in making utensils and weapons.

Question 18.
The strategic metals such as copper, silver and gold are used in the manufacturing of defence equipment.
Answer:
False.
Correct statement: The strategic metals such as titanium, chromium manganese, zirconium are used in the manufacturing of defence equipment.

Question 19.
Copper, silver and gold are called coinage metals.
Answer:
True.

Question 20.
For making ornaments, 24 – carat gold is used which is pure gold.
Answer:
False.
Correct statement: For making ornaments, 22 – carat gold is used which contains 22 parts of gold by weight and 2 parts of copper by weight.

Question 21.
The mineral from which a metal can be readily and economically extracted on a large scale is said to be ore.
Answer:
True.

Question 22.
The rocky impurity associated with the ore is called flux.
Answer:
False.
Correct statement: The rocky impurity associated with the ore is called gangue or matrix.

Question 23.
Slag is the fusible product formed when flux reacts with gangue during the extraction of metals.
Answer:
True.

Question 24.
Metals which have high chemical reactivity are found in a free state or in the native state.
Answer:
False.
Correct statement: Metals which have low chemical reactivity are found in a free state or in the native state.

Question 25.
Aluminium is the metal found most abundantly in the earth’s crust.
Answer:
True

Question 26.
Aluminium is a reddish-brown metal and it is a bad conductor of heat and electricity.
Answer:
False.
Correct statement: Aluminium is a silvery – white metal and it is a good conductor of heat and electricity.
Question 27.
Aluminium reacts with strong caustic alkalis forming aluminates.
Answer:
True

Question 28.
Conc. Nitric acid renders aluminium active due to the formation of nitride film on its surface.
Answer:
False.
Correct statement: Cone. Nitric acid renders aluminium passive due to the formation of oxide film on its surface.

Question 29.
Aluminium is a powerful reducing agent.
Answer:
True.

Question 30.
Duralumin alloy is light, having high tensile strength and corrosion-resistant.
Answer:
True.

Question 31.
Fe and Al₂O₃ are used in thermite welding.
Answer:
False.
Correct statement: Al powder and Fe₂O₃ is used in thermite welding.

Question 32.
The chief ore of copper is Ruby copper.
Answer:
False.
Correct statement: The chief ore of copper is copper pyrite.

Question 33.
Iron is a lustrous greyish white metal and can be magnetised.
Answer:
True.

Question 34.
The rust has the chemical formula as Fe₃O₄.
Answer:
False.
Correct statement: The rust has the chemical formula as Fe₂O₃. xH₂O.

V. Assertion and Reason.

Question 1.
Assertion (A): Nobel gas is unreactive.
Reason (R): They have unstable electronic configuration in their valence shells.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(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 2.
Assertion (A): The nature of bond in NaI molecule is covalent.
Reason (R): The electronegativity difference between Na and I is 1.5
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 3.
Assertion (A): Haematite ore was purified by Hydraulic method.
Reason (R): Haematite is oxide ore.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct
Question 4.
Assertion (A): Corundum is a chief ore of aluminium.
Reason (R): Molecular formula of Corundum is Al₂O₃
(a) (A) and (R) are correct, (R) explains the (A)
(b) (A) is correct, (R) is wrong
(c) (A) is wrong, (R) is correct
(d) (A) and (R) are correct, (R) doesn’t explain (A).
Answer:
(c) (A) is wrong, (R) is correct

Question 5.
Assertion (A): The chemical properties of the elements in the same period are not similar.
Reason (R): As the electronic configuration changes across the period, the chemical properties of the elements are not similar.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are not correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 6.
Assertion (A): Copper, Silver and Gold are used in making coins and jewellery. So they are called coinage metals.
Reason (R): These metals release an enormous amount of nuclear energy.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(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 7.
Assertion (A): Metals like Titanium, Chromium, Manganese and Zirconium are called strategic metals!
Reason (R): They find their applications in the manufacturing of defence equipment.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 8.
Assertion (A): Gold, Silver and Platinum are the metals that are found in a free state.
Reason (R): Those metals have low chemical reactivity and are found in a free state or in the native state.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 9.
Assertion (A): Aluminium occurs in the combined state.
Reason (R): It is a reactive metal and so it occurs in combined state.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 10.
Assertion (A): In the aluminothermic process, Iron oxide is reduced to iron by igniting with Aluminium powder.
Reason (R): Aluminium is a powerful reducing agent.
(a) Both (A) and (R) are wrong
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are correct.
Answer:
(d) Both (A) and (R) are correct

Question 11.
Assertion (A): When iron is dipped in conc.HNO₃ it becomes chemically inert (or) passive.
Reason (R): Iron becomes passive when treated with nitric acid is due to the formation of a layer of iron oxide Fe₃O₄ on its surface.
(a) Both (A) and (R) are correct
(b) (A) is correct but (R) is wrong
(c) (A) is wrong but (R) is correct
(d) Both (A) and (R) are wrong.
Answer:
(d) Both (A) and (R) are wrong

Question 12.
Assertion (A): Duralumin is used in making aircraft, tools and pressure cookers.
Reason (R): Duralumin is an alloy that is light, strong, resistant to corrosion.
(a) Both (A) and (R) are wrong
(b) Both (A) and (R) are correct
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct

Question 13.
Assertion (A): Nickel steel is used in making cables, aircraft parts and propeller.
Reason (R): Nickel steel alloy is hard, brittle and polishable.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(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 14.
Assertion (A): Magnesium is used in a sacrificial protection method to prevent corrosion.
Reason (R): Magnesium is more reactive than iron. When it is coated on the articles made of steel, it sacrifices itself to protect steel.
(a) Both (A) and (R) are correct
(b) Both (A) and (R) are wrong
(c) (A) is correct but (R) is wrong
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct

Question 15.
Assertion (A): Electroplating method not only protects but also enhances the metallic appearance.
Reason (R): Electroplating is a method of coating one metal with another by passing current.
(a) (A) is right, (R) is wrong
(b) (A) is right, (R) is not relevant
(c) (A) is right, (R) is relevant
(d) Both (A) and (R) are wrong.
Answer:
(c) (A) is right, (R) is relevant

VI. Short Answer Questions.

Question 1.
State modern periodic law.
Answer:
Modem periodic law states that the physical and chemical properties of elements are the periodic function of their atomic numbers.

Question 2.
Write the flow chart of the long form of the periodic table.
Answer:

Question 3.
The distance between the adjacent copper atoms in solid copper is 2.56 Å. Calculate what is the metallic radius of Cu?
Answer:
Metallic radius of copper = \(\frac{2.56}{2}\) = 1.28 Å

Question 4.
Briefly write any four characteristics of a group in the periodic table.
Answer:

• The elements present in a group have the same valency.
• The elements present in a group have identical chemical properties.
• The physical properties of the elements in the group very gradually.
• The atomic radii of die elements present in a group increase downwards.

Question 5.
What is the principle behind Froth floatation?
Answer:
This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method. Eg: Zinc blende (ZnS).

Question 6.
What are minerals?
Answer:
A mineral may be a single compound or a complex mixture of various compounds of metals found in the earth, e.g. Clay Al₂O₃. 2SiO₂. 2H₂O is the mineral of Aluminium.

Question 7.
Write the increasing order of radii of the following species.
(a) Na, Na⁺, Cl, Cl^–
(b) Li, Na, K, Rb
Answer:
(a) Na⁺, Cl, Na, Cl^–
(b) Li, Na, K, Rb

Question 8.
Differentiate ore and mineral.
Answer:

Ore	Mineral
1. Ores contain a large percentage of metal.	1. Minerals contain a low percentage of metal.
2. Ores can be used for die extraction of metals on a large scale readily and economically.	2. Metals cannot be extracted easily from minerals.
3. Bauxite Al2O3. 2H2O is the ore of aluminium.	3. Clay Al2O3.2SiO.2H2O is the mineral of aluminium.

Question 9.
Define metallurgy.
Answer:
The various steps involved in the extraction of metals from their ores as well as refining of crude metals are collectively known as metallurgy.

Question 10.
Elements a, b, c and d have the following electronic configurations
(a) ls² 2s² 2p⁶
(b) Is², 2s², 2p⁶, 3s², 3p¹
(c) Is², 2s², 2p⁶, 3s², 3p⁶
(d) Is², 2s², Ip¹
Answer:
(a) and (c), (b) and (d). Because the number of valence electrons are the same.

Question 11.
What is slag? Give an example.
Answer:
Slag is a fusible product formed when flux reacts with gangue during the extraction of metals.
Flux + gangue → slag
CaO + SiO₂ → CaSiO₃

Question 12.
Why the electron affinities of noble gases are zero?
Answer:
Noble gases show no tendency to accept electrons because the outer 5 and p orbitals of noble gases are completely filled. No more electrons can be added to them and hence their electron affinities are zero.

Question 13.
Why does gold, silver and platinum occur in free state?
Answer:
Gold, silver and platinum have low chemical reactivity and so they are found in the free state or in a native state.

Question 14.
How does Aluminium reacts with air?
Answer:
Reaction with air: It is not affected by dry air. On heating at 800°C, aluminium bums very brightly forming it’s oxide and nitride.
4Al + 3O₂ → Al₂O₃(Aluminium oxide)
2Al + N₂ → 2 AlN (Aluminium nitride)

Question 15.
How does Aluminium react with caustic soda? Give an equation.
Answer:
Aluminium reacts with caustic soda to give sodium meta aluminate with the liberation of H₂ gas.

Question 16.
Prove that aluminium is a powerful reducing agent.
Answer:
Aluminium is a powerful reducing agent When a mixture of aluminium powder and iron oxide is ignited, iron oxide is reduced to iron. This process is known as the aluminothermic process.

Question 17.
Write a note on Aluminothermic process.
Answer:
As reducing agent: Aluminium is a powerful reducing agent. When a mixture of aluminium powder and iron oxide is ignited, the latter is reduced to metal. This process is known as aluminothermic process.
Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat

Question 18.
What is the action of heat on copper?
Answer:
On heating at different temperatures in the presence of oxygen, copper forms two types of oxides CuO, Cu₂O.

Question 19.
Explain the action of dilute nitric acid with copper.
Answer:
Copper reacts with dil.HNO₃ with the liberation of Nitric oxide gas.

Question 20.
What happens when copper is treated with conc.HNO₃ and with conc.H₂SO₄?
Answer:

Question 21.
What do you mean by Ferrous and Non-Ferrous alloys? Given an example for each.
Answer:
Ferrous alloys : Contain iron as a major component.
Eg: Stainless steel, Nickel steel

Non-Ferrous alloys : These alloys do not contain iron as a major part.
Eg: Aluminium alloy, Copper alloy.

Question 22.
Explain the action of air with iron.
Answer:
3Fe + 2O₂ → Fe₃O₄ [Magnetic oxide (Black)].

Question 23.
What are amalgams? How are they prepared?
Answer:
An amalgam is an alloy of mercury with another metal.
Amalgams are formed by the metallic bonding with the electrostatic force of attraction between the electrons and the positively charged metal ions.

Question 24.
Explain the action of steam with iron.
Answer:

Question 25.
Mention the types of iron on the basis of carbon content.
Answer:

Pig iron	Iron with 2 – 4.5% carbon
Wrought iron	Iron with < 0.25% carbon
Steel	Iron with 0.25 – 2% carbon

Question 26.
What is dry corrosion or chemical corrosion?
Answer:
The corrosive action in the absence of moisture is called dry corrosion. It is the process of a chemical attack on a metal by corrosive liquids or gases such as O₂, N₂, SO₂, H₂S etc… It occurs at high temperature, of all the gases mentioned above O₂ is the most reactive gas to impart the chemical attack.

Question 27.
What is an amalgam? Give one example with its use.
Answer:

• An amalgam is an alloy of mercury with metals such as sodium, gold and silver.
• Dental amalgam is an alloy of mercury with silver and tin and it is used in the dental filling.

Question 28.
Explain sacrificial protection.
Answer:
Magnesium is more reactive than iron. When it is coated on the articles made of steel, it sacrifices itself to protect steel.

Question 29.
Define corrosion.
Answer:
Corrosion is defined as the slow and steady destruction of a metal by the environment. It results in the deterioration of the metal to form metal compounds by means of chemical reactions with the environment.

Question 30.
What is the action of Air and moisture on copper?
Answer:
Copper gets covered with a green layer of basic carbonate in the presence of CO₂ and moisture.
2 Cu + O₂ + CO₂ + H₂O → CUCO₃.CU(OH)₂

Question 31.
Give any two uses of aluminium.
Answer:

• Aluminium metal is a corrosion – resistant and a good conductor of heat. So it is used in making utensils.
• Aluminium is used in welding as thermite and a very good reducing agent.

Question 32.
What is the modern periodic table?
Answer:
The modem periodic table is a tabular arrangement of elements in rows and columns, highlighting the regular repetition of properties of the elements.

Question 33.
Explain the smelting process of iron in the Blast furnace.
Answer:
The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top.

Question 34.
What are the periodic properties?
Answer:
Properties such as atomic radius, ionic radius, ionisation energy, electronegativity, electron affinity, show a regular periodicity and hence they are called periodic properties.

Question 35.
Define Atomic radius.
Answer:
The atomic radius of an atom is defined as the distance between the centre of its nucleus and the outermost shell containing the valence electron.

Question 36.
Write the action of dil.HCl and dil.H₂SO₄ with iron.
Answer:
(i) Fe + 2HCl → FeCl₂ + H₂↑
(ii) Fe + H₂SO₄ → FeSO₄ + H₂↑

Question 37.
What is a covalent radius?
Answer:
It is defined as half the distance between the nuclei of two covalently bonded atoms of the same element in a molecule.

Question 38.
Define Ionisation energy.
Answer:
Ionisation energy is the minimum energy required to remove an electron from a gaseous atom in its ground state to form a cation.

Question 39.
What is an alloy?
Answer:
An alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements. The properties of alloys are often different from the component metals.

Question 40.
What is Electronegativity?
Answer:
Electronegativity of an element is the measure of the tendency of its atom to attract the shared pair of electrons towards itself in a covalent bond.

Question 41.
Write the steps involved in metallurgical process.
Answer:
A metallurgical process involves three main steps as follows:

• Concentration or Separation of the ore: It is the process of removal of impurities from the ore.
• Production of the metal: It is the conversion of the ore into metal.
• Refining of the metal: It is the process of purification of the metal.

Question 42.
Give a single term for each of the following:
(i) The process of extracting the ores from the Earth’s crust is called:
Answer:
Mining

(ii) The rocky impurities associated with an ore is called:
Answer:
Gangue or matrix

(iii) The substance added to the ore to reduce fusion temperature:
Answer:
Flux

(iv) Noble metals occur in this state:
Answer:
Native

Question 43.
How will you convert copper into copper carbonate?
Answer:
Copper reacts with oxygen in the presence of CO₂ and moisture to give copper carbonate.
2Cu + O₂ + CO₂ + H₂O → CuCO₃. Cu(OH)₂ (Copper Carbonate).

Question 44.
Mention the uses of iron.
Answer:
Uses of iron:

1. Pig iron (Iron with 2 – 4.5 % of carbon): It is used in making pipes, stoves, radiators, railings, manhole covers and drain pipes.
2. Steel (Iron with < 0.25 % of carbon): It is used in the construction of buildings, machinery, transmission cables and T.V towers and in making alloys.
3. Wrought iron (Iron with 0.25 – 2 % of wrought carbon): It is used in making springs, anchors and electromagnets.

Question 45.
What are the chemical properties of metals in terms of

1. valence electrons
2. Atomicity.

Answer:

1. Valence electrons: Metals usually have 1, 2 or 3 electrons in their outermost shell.
2. Atomicity: Metals are usually monoatomic in their vapour state.

Question 46.
Why alloys are said to solid solutions?
Answer:
Alloys can be considered solid solutions in which the metal with high concentration is solvent and other metals are solute.
Example: brass is a solid solution of zinc (solute) in copper (solvent).

Question 47.
Write a note on Dry corrosion.
Answer:

• The corrosive action in the absence of moisture is called dry corrosion.
• It is the process of a chemical attack on a metal by corrosive liquids or gases such as O₂, N₂, SO₂, H₂S etc. in which O₂ is more reactive.
• It occurs at high temperature

Question 48.
Explain Wet Corrosion.
Answer:

• The corrosive action in the presence of moisture is called wet corrosion.
• It occurs as a result of the electrochemical reaction of metal with water or an aqueous solution of salt or acids or bases.

Question 49.
What is electroplating?
Answer:
Electroplating is a method of coating one metal over another metal by passing an electric current.

VII. Long Answer Questions.

Question 1.
Explain the variation of ionisation energy along the group and period.
Answer:

• As the atomic size decreases from left to right in a period, more energy is required to remove the electrons. So, the ionisation energy increases throughout the period.
• Down the group, the atomic size increases and hence the valence electrons are loosely bound. They require relatively less energy for the removal. Thus, ionisation energy decreases down the group in the periodic table.

Question 2.
Explain the electrolytic refining of copper.
Answer:
(i) Blister copper contains 98% of pure Cu and 2 % of impurities and is purified by electrolytic refining.
For electrolytic refining of Cu we use:
Cathode: A thin plate of pure Cu.
Anode: A block of impure Cu
Electrolyte: CuSO₄ + dil H₂SO₄

(ii) When electric current is passed through the electrolytic solution, Pure Cu gets deposited at the cathode and the impurities settle at the bottom of the anode in the form of anode mud.

Question 3.
Explain Gravity separation method.
Answer:
Gravity Separation (or) Hydraulic method:
1. Principle: The difference in the densities or specific gravities of the ore and the gangue is the main principle behind this method. Oxide ores are purified by this method,
e.g., Haematite Fe₂O₃ the ore of iron.

2. Method: The ore is poured over a sloping, vibrating corrugated table with grooves and a jet of water is allowed to flow over it. The denser ore particles settle down in the grooves and lighter gangue particles are washed down by the water.

Question 4.
Discuss the magnetic separation methods.
Answer:
Magnetic separation method:
Principle: The magnetic properties of the ores from the basis of separation. When either the ore or the gangue is magnetic, this method is employed, e.g., Tinstone SnO₂, the ore of tin.

Method: The crushed ore is placed over a conveyer belt which rotates around two metal wheels, one of which is magnetic. The magnetic particles are attracted to the magnetic wheel and fall separately apart from the nonmagnetic particles.

Question 5.
Explain the froth floatation process.
Answer:
Froth floatation Process:
Principle: This process depends on the preferential wettability of the ore with oil (pine oil) and the gangue particles by water. Lighter ores, such as sulphide ores, are concentrated by this method, e.g., Zinc blende (ZnS).

Method: The crushed ore is taken in a large tank containing oil and water and agitated with a current of compressed air. The ore is wetted by the oil and gets separated from the gangue in the form of froth. Since the ore is lighter, it comes on the surface with the froth and the impurities are left behind, e.g., Zinc blende (ZnS).

Question 6.
How will you extract aluminium from its ore?
Answer:
The extraction of aluminium from bauxite involves two steps:
(i) Conversion of bauxite into alumina – Baeyer’s Process
The conversion of Bauxite into Alumina involves the following steps:
Bauxite ore is finely ground and heated under pressure with a solution of concentrated caustic soda solution at 150° C to obtain sodium metal aluminate.
On diluting sodium meta aluminate with water, a precipitate of aluminium hydroxide is formed.
The precipitate is filtered, washed, dried and ignited at 1000°C to get alumina.
\(2 \mathrm{Al}(\mathrm{OH})_{3} \stackrel{1000^{\circ} \mathrm{C}}{\longrightarrow} \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{H}_{2} \mathrm{O}\)

(ii) Electrolytic reduction of alumina – Hall’s Process
Aluminium is produced by the electrolytic reduction of fused alumina (Al₂O₃) in the electrolytic cell.

Cathode: Iron tank linked with graphite
Anode: A bunch of graphite rods suspended in a molten electrolyte.
Electrolyte: Pure alumina + molten cryolite + fluorspar (fluorspar lowers the fusion temperature of electrolyte)
Temperature: 900 – 950°C
The voltage used: 5 – 6 V
Aluminium is deposited at the cathode and oxygen gas is liberated at the anode. Oxygen combines with graphite to form CO₂.

Question 7.
Explain the extraction of copper from copper pyrites.
Answer:
Extraction of copper from copper pyrites involves the following steps:
(i) The concentration of ore: The ore is crushed and then concentrated by froth floatation process.

(ii) Roasting: The concentrated ore is roasted in excess of air. During the process of roasting, the moisture and volatile impurities are removed. Sulphur, phosphorus, arsenic and antimony are removed as oxides. Copper pyrite is partly converted into sulphides of copper and iron.
\(2 \mathrm{CuFeS}_{2}+\mathrm{O}_{2} \rightarrow \mathrm{Cu}_{2} \mathrm{S}+2 \mathrm{FeS}+\mathrm{SO}_{2} \uparrow\)

(iii) Smelting: The roasted ore is mixed with powdered coke and sand and is heated in a blast furnace to obtain matte (Cu₂S + FeS) and slag. The slag is removed as waste.

(iv) Bessemerisation: The molten matte is transferred to the Bessemer converter in order to obtain blister copper. Ferrous sulphide from matte is oxidized to ferrous oxide, which is removed as slag using silica.

(v) Refining: Blister copper contains 98% of pure copper and 2% of impurities and is purified A by electrolytic refining. This method is used to get metal of a high degree of purity. For electrolytic refining of copper, we use:
Cathode: A thin plate of pure copper metal.
Anode: A block of impure copper metal.
Electrolyte: Copper sulphate solution acidified with sulphuric acid.
When an electric current is passed through the electrolytic solution, pure copper gets deposited at the cathode and the impurities settle at the bottom of the anode in the form of sludge called anode mud.

Question 8.
Explain the metallurgy of iron.
Answer:
Iron is chiefly extracted from haematite ore (Fe₂O₃):
(i) Concentration by Gravity Separation: The powdered ore is washed with steam of water. As a result, the lighter sand particles and other impurities are washed away and the heavier ore particles settle down.

(ii) Roasting and Calcination: The concentrated ore is strongly heated in a limited supply of air in a reverberatory furnace. As a result, moisture is driven out and sulphur, arsenic and phosphorus impurities are oxidized off.

(iii) Smelting (in a Blast Furnace): The charge consisting of roasted ore, coke and limestone in the ratio 8 : 4 : 1 is smelted in a blast furnace by introducing it through the cup and cone arrangement at the top. There are three important regions in the furnace.

a. The Lower Region (Combustion Zone):
The temperature is at 1500°C. In this region, coke bums with oxygen to form CO₂ when the charge comes in contact with a hot blast of air.

It is an exothermic reaction since heat is liberated.

b. The Middle Region (Fusion Zone):
The temperature prevails at 1000°C. In this region, CO₂ is reduced to CO.

Limestone decomposes to calcium oxide and CO₂

These two reactions are endothermic due to absorption of heat. Calcium oxide combines with silica to form calcium silicate slag.
CaO + SiO₂ → CaSiO₃

c. The Upper Region (Reduction Zone): The temperature prevails at 400°C. In this region carbon monoxide reduces ferric oxide to form a fairly pure spongy iron.
\(\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \stackrel{400^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{Fe}+3 \mathrm{CO}_{2}\)
The molten iron is collected at the bottom of the furnace after removing the slag.
The iron thus formed is called pig iron. It is remelted and cast into different moulds. This iron is called cast iron.

Question 9.
Explain the types of alloys.
Answer:
Based on the presence or absence of Iron, alloys can be classified into:
1. Ferrous alloys: Contain Iron as a major component.
A few examples of ferrous alloys are Stainless Steel, Nickel Steel etc.

2. Non – ferrous alloys: These alloys do not contain Iron as a major component.
For example, Aluminium alloy, Copper alloy etc.
Copper Alloys (Non – ferrous):

Alloys	Uses
Brass (Cu, Zn)	Electrical fittings, medal, decorative items, hardware
Bronze (Cu, Sn)	Statues, coins, bells, gongs

Aluminium Alloys (Non – ferrous):

Alloys	Uses
Duralumin (Al, Mg, Mn, Cu)	Aircraft tools, pressure cookers
Magnalium (Al, Mg)	Aircraft, scientific instruments

Iron Alloys(Ferrous):

Alloys	Uses
Stainless steel (Fe, C, Ni, Cr)	Utensils, cutlery, automobile parts
Nickel steel (Fe, C, Ni)	Cables, aircraft parts, propeller

VIII. HOT Questions.

Question 1.
What would be the atomic number of the next

1. alkali metal
2. alkaline earth metal
3. Halogens
4. inert gas, if discovered in future.

Answer:

1. Alkali metal: 118 + 1 = 119
2. Alkaline earth metal: 120
3. Halogens: 117
4. Inert gas: 118

Question 2.
Explain the mechanism of rusting?
Answer:
Rust is chemically known as hydrated ferric oxide (it is formulated as Fe₂O₃.xH₂O).
Rusting results in the formation of scaling reddish – brown hydrated ferric oxide on the surface of iron and iron-containing materials.

Question 3.
All ores are minerals, but ait minerals are not ores. Why?
Answer:

• The elements or compounds of the metals which occur in nature in the earth’s crust are called minerals whereas ore is a mineral from which the metal is profitably extracted.
• For example, aluminium exists in the two mineral forms, that is clay and bauxite. But aluminium is mainly extracted from bauxite which contains 70 % aluminium oxide. So, bauxite is an ore of aluminium whereas clay is not ore.
• So, all ores are minerals but all minerals need not be ores.

Question 4.
Why the graphite rods acting as anode in the Hall’s process should be replaced periodically?
Answer:
During Hall’s Process, the carbon electrodes get consumed, so they have to be replaced periodically.

Question 5.
Anionic radius Is higher than the corresponding neutral atom. Give reason.
Answer:
When an atom gains one or more electrons it forms an anion. During the formation of anion, the number of orbital electrons becomes greater than the nuclear charge. Hence, the electrons are not strongly attracted by the lesser number of nuclear charges. Hence anionic radius is higher than the corresponding neutral atom.

Question 6.
A reddish – brown metal A, when exposed to moist air, forms a green layer B. When A is heated at different temperatures in the presence of O₂, it forms two types of oxides – C (black) and D (red). Identify A, B, C, D and write the balanced equation.
Answer:
(i) A reddish – brown metal A is a copper (Cu).

(ii) When copper (A) is exposed to moist air it forms a green layer (B) is copper carbonate.

(iii) When copper is heated at different temperature in the presence of oxygen, it forms two types of oxides CuO and Cu₂O. (C and D)

Question 7.
A silvery – white metal on treatment with NaOH and HCl liberates H₂ gas to form B and C respectively. The metal A will not react with acid D due to the formation of a passive film on the surface. Hence it is used for transporting acid D. Identify A, B, C, D and support your answer with balanced equations.
Answer:
(i) A silvery – white metal (A) is Aluminium (Al).
(ii) Aluminium reacts with NaOH to form B which is known as sodium meta aluminate with the liberation of H₂ gas.

(iii) Aluminium reacts with HCl to form Aluminium chloride which is known as C with the liberation of H₂ gas.

(iv) Aluminium does not react with conc. nitric acid (HNO₃) which is known as D, due to the formation of a passive film on the surface.

A	Aluminium	Al
B	Sodium meta aluminate	NaAlO2
C	Aluminium chloride	AlCl3
D	Nitric acid	HNO3

Question 8.
Metal A belongs to period 4 and group 8. A in red hot condition reacts with steam to form B. A reacts with dilute HNO₃ to give C. A again reacts with conc. H₂SO₄ to give D. Find A, B, C and D with suitable reaction.
Answer:
(i) Metal (A) belongs to period 4 and group 8 is iron (Fe).

(ii) Iron (A) reacts with steam to form magnetic oxide (B)

(iii) Iron (A) reacts with dilute HNO₃ in cold condition to give ferrous nitrate (C).

(iv) Iron (A) reacts with conc.H₂SO₄ to form Ferric Sulphate (D).

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.2

10th Maths Exercise 2.2 Samacheer Kalvi Question 1.
For what values of natural number n, 4n can end with the digit 6?
Solution:
4ⁿ = (2 × 2)ⁿ = 2ⁿ × 2ⁿ
2 is a factor of 4ⁿ.
So, 4ⁿ is always even and end with 4 and 6.
When n is an even number say 2, 4, 6, 8 then 4ⁿ can end with the digit 6.
Example:
4² = 16
4³ = 64
4⁴ = 256
4⁵ = 1,024
4⁶ = 4,096
4⁷ = 16,384
4⁸ = 65, 536
4⁹ = 262,144

Ex 2.2 Class 10 Samacheer Question 2.
If m, n are natural numbers, for what values of m, does 2ⁿ × 5^m ends in 5?
Answer:
2ⁿ is always even for any values of n.
[Example. 2² = 4, 2³ = 8, 2⁴ = 16 etc]

5^m is always odd and it ends with 5.
[Example. 5² = 25, 5³ = 125, 5⁴ = 625 etc]
But 2ⁿ × 5^m is always even and end in 0.
[Example. 2³ × 5³ = 8 × 125 = 1000
2² × 5² = 4 × 25 = 100]
∴ 2ⁿ × 5^m cannot end with the digit 5 for any values of m.

Exercise 2.2 Class 10 Maths Samacheer Question 3.
Find the H.C.F. of 252525 and 363636.
Solution:
To find the H.C.F. of 252525 and 363636
Using Euclid’s Division algorithm
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0.
∴ Again by division algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0.
∴ Again by division algorithm.
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0.
∴ Again by division algorithm
30303 = 20202 × 1 + 10101
The remainder 10101 ≠ 0.
∴ Again using division algorithm
20202 = 10101 × 2 + 0
The remainder is 0.
∴ 10101 is the H.C.F. of 363636 and 252525.

10th Maths Exercise 2.2 Question 4.
If 13824 = 2^a × 3^b then find a and b.
Solution:
If 13824 = 2^a × 3^b
Using the prime factorisation tree

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 2⁹ × 3³ = 2^a × 3^b
∴ a = 9, b = 3.

10th Maths Exercise 2.2 In Tamil Question 5.
If p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄ = 113400 where p₁, p₂, p₃, p₄ are primes in ascending order and x₁, x₂, x₃, x₄ are integers, find the value of P₁, P₂, P₃, P₄ and x₁, x₂, x₃, x₄.
Solution:
If p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄ = 113400
p₁, p₂, p₃, P₄ are primes in ascending order, x₁, x₂, x₃, x₄ are integers.
using Prime factorisation tree.

113400 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5 × 7
= 23 × 34 × 52 × 7
= p₁^x₁ × p₂^x₂ × p₃^x₃ × p₄^x₄
∴ p₁= 2, p₂ = 3, p₃ = 5, p₄ = 7, x₁ = 3, x₂ = 4, x₃ = 2, x₄ = 1.

Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of arithmetic.
Solution:
408 and 170.

408 = 2³ × 3¹ × 17¹
170 = 2¹ × 5¹ × 17¹

∴ H.C.F. = 2¹ × 17¹ = 34.
To find L.C.M, we list all prime factors of 408 and 170, and their greatest exponents as follows.

∴ L.C.M. = 2³ × 3¹ × 5¹ × 17¹
= 2040.

10th Maths 2.2 Exercise Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24, 15, 36?
Solution:
To find L.C.M of 24, 15, 36

24 = 2³ × 3
15 = 3 × 5
36 = 2² × 3²

∴ L.C.M = 2³ × 3² × 5¹
= 8 × 9 × 5
= 360
If a number has to be exactly divisible by 24, 15, and 36, then it has to be divisible by 360. Greatest 6 digit number is 999999.
Common multiplies of 24, 15, 36 with 6 digits are 103680, 116640, 115520, …933120, 999720 with six digits.
∴ The greatest number consisting 6 digits which is exactly divisible by 24, 15, 36 is 999720.

10th Maths Ex 2.2 Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Answer:
Find the L.C.M of 35, 56, and 91
35 – 5 × 7 56
56 = 2 × 2 × 2 × 7
91 = 7 × 13
L.C.M = 23 × 5 × 7 × 13
= 3640
Since it leaves remainder 7
The required number = 3640 + 7
= 3647
The smallest number is = 3647

10th Maths 2.2 Question 9.
Find the least number that is divisible by the first ten natural numbers.
Solution:
The least number that is divisible by the first ten natural numbers is 2520.
Hint:
1,2, 3,4, 5, 6, 7, 8,9,10
The least multiple of 2 & 4 is 8
The least multiple of 3 is 9
The least multiple of 7 is 7
The least multiple of 5 is 5
∴ 5 × 7 × 9 × 8 = 2520.
L.C.M is 8 × 9 × 7 × 5
= 40 × 63
= 2520