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Samacheer Kalvi 11th Physics Nature of Physical World and Measurement TextBook Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choice Questions

Samacheer Kalvi 11th Physics Solution Chapter 1 Question 1.
One of the combinations from the fundamental physical constants is \(\frac{h c}{\mathrm{G}}\). The unit of this expression is
(a) Kg²
(b) m³
(c) S^-1
(d) m
Answer:
(a) Kg²

Nature Of Physical World And Measurement Question 2.
If the error in the measurement of radius is 2%, then the error in the determination of volume of the sphere will be …….
(a) 8%
(b) 2%
(c) 4%
(d) 6%
Answer:
(d) 6%

Nature Of Physical World And Measurement Questions And Answers Question 3.
If the length and time period of an oscillating pendulum have errors of 1 % and 3% respectively
then the error in measurement of acceleration due to gravity is …… [Related to AMPMT 2008]
(a) 4%
(b) 5%
(c) 6%
(d) 7%
Answer:
(d) 7%

11th Physics Nature Of Physical World And Measurement Question 4.
The length of a body is measured as 3.51 m, if the accuracy is 0.01mm, then the percentage error in the measurement is ……
(a) 351%
(b) 1%
(c) 0.28%
(d) 0.035%
Answer:
(c) 0.28%

Nature Of Physical World And Measurement Class 11 Question 5.
Which of the following has the highest number of significant figures?
(a) 0.007 m²
(b) 2.64 × 1024 kg
(c) 0.0006032 m²
(d) 6.3200 J
Answer:
(d) 6.3200 J

Nature Of Physical World And Measurement Class 11 Notes Question 6.
If π = 3.14, then the value of π² is …..
(a) 9.8596
(b) 9.860
(c) 9.86
(d) 9.9
Answer:
(c) 9.86

Samacheer Kalvi Guru 11th Physics Question 7.
Which of the following pairs of physical quantities have same dimension?
(a) force and power
(b) torque and energy
(c) torque and power
(d) force and torque
Answer:
(b) torque and energy

11th Physics Unit 1 Question 8.
The dimensional formula of Planck’s constant h is [AMU, Main, JEE, NEET]
(a) [ML²T^-1]
(b) [ML²T^-3]
(c) [MLTT^-1]
(d) [MLTT^3-3]
Answer:
(a) [ML²T^-1]

Physics Class 11 Chapter 1 Question 9.
The velocity of a particle v at an instant t is given by v = at + bt². The dimensions of b is ……
(a) [L]
(b) [LT^-1]
(c) [LT^-2]
(d) [LT^-3]
Answer:
(d) [LT^-3]

Samacheerkalvi.Guru 11th Physics Question 10.
The dimensional formula for gravitational constant G is [Related to AIPMT 2004]
(a) [ML^-3T^-2]
(b) [M^-1L³T^-2]
(c) [M^-1L^-3T^-2]
(d) [ML^-3T²]
Answer:
(b) [M^-1L³T^-2]

Nature Of Physical World And Measurement In Tamil Question 11.
The density of a material in CGS system of units is 4 g cm^-3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, then the value of density of material will be ……
(a) 0.04
(b) 0.4
(c) 40
(d) 400
Answer:
(c) 40

Samacheer Kalvi 11th Physics Question 12.
If the force is proportional to square of velocity, then the dimension of proportionality constant
is [JEE-2000] ……
(a) [MLT⁰]
(b) [MLT^-1]
(c) [ML^-2T]
(d) [ML^-1T⁰]
Answer:
(d) [ML^-1T⁰]

Samacheer Kalvi 11th Physics Solution Book Question 13.
[MainAIPMT2011]
(a) length
(b) time
(c) velocity
(d) force
Answer:
(c) velocity

11 Th Physics Samacheer Kalvi Guide Question 14.
Planck’s constant (h), speed of light in vaccum (c) and Newton’s gravitational constant (G) are taken as three fundamental constants. Which of the following combinations of these has the dimension of length? [NEET 2016 (Phase II)]

Answer:

Samacheer Kalvi 11th Physics Guide Question 15.
A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (k_B), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression for l is dimensionally correct?. [JEE (advanced) 2016]

Answer:

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions

11th Physics Chapter 1 Numerical Problems Question 1.
Briefly explain the types of physical quantities.
Answer:
Physical quantities are classified into two types. There are fundamental and derived quantities. Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities. These are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
Quantities that can be expressed in terms of fundamental quantities are called derived quantities. For example, area, volume, velocity, acceleration, force.

11th Physics Samacheer Kalvi Question 2.
How will you measure the diameter of the Moon using parallax method?
Answer:
Let θ is the angular diameter of moon
d – is the distance of moon from earth, from figure, θ = \(\frac{\mathrm{D}}{d}\)

Diameter of moon D = d.θ
by knowing θ, d, diameter of moon can be calculated

Samacheer Kalvi Physics 11th Question 3.
Write the rules for determining significant figures.
Answer:
Rules for counting significant figures:

Note: 1 Multiplying or dividing factors, which are neither rounded numbers nor numbers representing measured values, are exact and they have infinite numbers of significant figures as per the situation.
For example, circumference of circle S = 2πr, Here the factor 2 is exact number. It can be written as 2.0, 2.00 or 2.000 as required.
Note: 2 The power of 10 is irrelevant to the determination of significant figures.
For example x = 5.70 m = 5.70 × 10² cm = 5.70 × 10³ mm = 5.70 × 10^-3 km.
In each case the number of significant figures is three.

Physics Class 11 Samacheer Kalvi Question 4.
What are the limitations of dimensional analysis?
Answer:
Limitations of Dimensional analysis
1. This method gives no information about the dimensionless constants in the formula like 1, 2, ……… π, e, etc.
This method cannot decide whether the given quantity is a vector or a scalar.
This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
It cannot be applied to an equation involving more than three physical quantities.
It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation. For example, using dimensional analysis, is dimensionally correct whereas the correct relation is

Samacheer Kalvi Guru 11 Physics Question 5.
Define precision and accuracy. Explain with one example.
Answer:
The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision of a measurement is a closeness of two or more measured values to each other.
The true value of a certain length is near 5.678 cm. In one experiment, using a measuring instrument of resolution 0.1 cm, the measured value is found to be 5.5 cm. In another experiment using a measuring instrument of greater resolution, say 0.01 cm, the length is found to be 5.38 cm. We find that the first measurement is more accurate as it is closer to the true value, but it has lesser precision. On the contrary, the second measurement is less accurate, but it is more precise.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Long Answer Questions

Question 1.
(i) Explain the use of screw gauge and vernier caliper in measuring smaller distances.
Answer:
(i) Measurement of small distances: screw gauge and vernier caliper Screw gauge:
The screw gauge is an instrument used for measuring accurately the dimensions of objects up to a maximum of about 50 mm. The principle of the instrument is the magnification of linear motion using the circular motion of a screw. The least count of the screw gauge is 0.01 mm. Vernier caliper: A vernier caliper is a versatile instrument for measuring the dimensions of an object namely diameter of a hole, or a depth of a hole. The least count of the vernier caliper is 0.1 mm.

(ii) Write a note on triangulation method and radar method to measure larger distances. Triangulation method for the height of an accessible object;
Let AB = h be the height of the tree or tower to be measured. Let C be the point of observation at distance x from B. Place a range finder at C and measure the angle of elevation, ACB =
θ as shown in figure.
From right angled triangle ABC,


(or) height h = x tan θ
Knowing the distance x, the height h can be determined.
RADAR method
The word RADAR stands for radio detection and ranging. A radar can be used to measure accurately the distance of a nearby planet such as Mars. In this method, radio waves are sent from transmitters which, after reflection from the planet, are detected by the receiver. By measuring, the time interval (t) between the instants the radio waves are sent and received, the distance of the planet can be determined as where v is the speed of the radio wave. As the time taken (t) is for the distance covered during the forward and backward path of the radio waves, it is divided by 2 to get the actual distance of the object. This method can also be used to determine the height, at which an aeroplane flies from the ground.

Question 2.
Explain in detail the various types of errors.
Answer:
The uncertainty in a measurement is called an error. Random error, systematic error and gross error are the three possible errors.
(i) Systematic errors: Systematic errors are reproducible inaccuracies that are consistently *, in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows
(1) Instrumental errors: When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.
(2) Imperfections in experimental technique or procedure: These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied
(3) Personal errors: These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.
(4) Errors due to external causes: The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity, or pressure during measurements may affect the result of the measurement.
(5) Least count error: Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument’s
resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.
(ii) Random errors: Random errors may arise due to random and unpredictable variations in
experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called “chance error”. When different readings are obtained by a person every time he repeats the experiment, personal error occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.

Question 3.
What do you mean by propagation of errors? Explain the propagation of errors in addition and multiplication.
Answer:
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for taking readings. Then we may have to look at the errors in measuring various quantities, collectively.
The error in the final result depends on
(i) The errors in the individual measurements
(ii) On the nature of mathematical operations performed to get the final result. So we should know the rules to combine the errors.
The various possibilities of the propagation or combination of errors in different mathematical operations are discussed below:
(i) Error in the sum of two quantities
Let ∆A and ∆B be the absolute errors in the two quantities A and B respectively. Then, Measured value of A = A ± ∆A
Measured value of B = B ± ∆B
Consider the sum, Z = A + B
The error ∆Z in Z is then given by

The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.
Error in the product of two quantities: Let ∆A and ∆B be the absolute errors in the two quantities A, and B, respectively. Consider the product Z = AB
The error ∆Z in Z is given by Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= (AB) ± (A ∆ B) ± (B ∆ A) ± (∆A • ∆B)
Dividing L.H.S by Z and R.H.S by AB, we get,

As ∆A/A, ∆B/B are both small quantities, their product term can be neglected.
The maximum fractional error in Z is

Question 4.
Write short note on the following:
(a) Unit
(b) Rounding – off
(c) Dimensionless quantities
Answer:
(a) Unit: An arbitrarily chosen standard of measurement of a quantity, which is accepted internationally is called unit of the quantity.
The units in which the fundamental quantities are measured are called fundamental or base units and the units of measurement of all other physical quantities, which can be obtained by a suitable multiplication or division of powers of fundamental units, are called derived units.
(b) Rounding – off: In no case should the result have more significant figures than die figures involved in the data used for calculation. The result of calculation with numbers containing more than one uncertain digit should be rounded off. The rules for rounding off are given below.

(c) Dimensionless quantities: On the basis of dimension, dimensionless quantities are classified into two categories.
(i) Dimensionless variables:
Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.
(ii) Dimensionless Constant:
Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are π, e, numbers etc.

Question 5.
Explain the principle of homogeniety of dimensions. What are its uses? Give example.
Answer:
The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v² = u² + 2as, the dimensions of v², u² and 2 as are the same and equal to [L²T^-2].
(i) To convert a physical quantity from one system of units to another: This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant, i.e, = constant n₁[u₁] = constant (or) n₁[u₁] = n₂[u₂].
Consider a physical quantity which has dimension ‘a’ in mass, ‘b’ in length and ‘c’ in time. If the fundamental units in one system are M₁, L₁ and T₁ and the other system are M₂, L₂ and T₂ respectively, then we can write,
We have thus converted the numerical value of physical quantity from one system of units into the other system.
Example: Convert 76 cm of mercury pressure into Nm^-2 using the method of dimensions. Solution: In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm^-2
The dimensional formula of pressure P is [ML^-1T^-2]

(ii) To check the dimensional correctness of a given physical equation:
Example: The equation \(\frac{1}{2} m v^{2}\) = mgh can be checked by using this method as follows.
Solution: Dimensional formula for

Dimensional formula for

Both sides are dimensionally the same, hence the equations is dimensionally correct.
(iii) To establish the relation among various physical quantities:
Example: An expression for the time period T of a simple pendulum can be obtained by using this method as follows.
Let true period T depend upon
(i) mass m of the bob
(ii) length l of the pendulum and
(iii) acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is K = 2π.
Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, a – 0, b + c = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numerical Problems

Question 1.
In a submarine equipped with sonar, the time delay between the generation of a pulse and its echo after reflection from an enemy submarine is observed to be 80 sec. If the speed of sound in water is 1460 ms^-1. What is the distance of enemy submarine?
Answer:
Given:
Speed of sound in water = 1460 ms^-1
Time delay = 80s
Distance of enemy ship = ?
Solution:
Total distance covered = speed × time
= 1460 ms^-1 × 80s = 116800 m
Time taken is for forward and backward path of sound waves.

= 58400 m (or) 58.4 km

Question 2.
The radius of the circle is 3.12 m. Calculate the area of the circle with regard to significant figures.
Answer:
Given: radius : 3.12 m (Three significant figures)
Solution:
Area of the circle = πr² = 3.14 × (3.12 m)² = 30.566
If the result is rounded off into three significant figure, area of the circle = 30.6 m²

Question 3.
Assuming that the frequency v of a vibrating string may depend upon
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m), prove that using dimensional analysis. [Related to JIPMER 2001]
Answer:
Given: The frequency v of a vibrating string depends
(i) applied force (F)
(ii) length (l)
(iii) mass per unit length (m)
Solution:

Substitute the dimensional formulae of the above quantities

Comparing the powers of M, L, T on both sides,
x + z = 0, x + y – z = 0, -2x = -1
Solving for x, y, z, we get

Substitute x, y, z values in equ(1)

Question 4.
Jupiter is at a distance of 824.7 million km from the Earth. Its angular diameter is measured to be 35.72″. Calculate the diameter of Jupiter.
Answer:
Given,
Given Distance of Jupiter = 824.7 × 10⁶ km = 8.247 × 10¹¹ m
angular diameter = 35.72 × 4.85 × 10^-6rad = 173.242 × 10^-6 rad
= 1.73 × 10^-4 rad
∴ Diameter of Jupiter D = D × d = 1.73 × 10^-4 rad × 8.247 × 10¹¹ m
= 14.267 × 1o⁷ m = 1.427 × 10⁸ m (or) 1.427 × 10^{5</sup km}

Question 5.
The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage of accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement.
Answer:
Given,
Length of simple pendulum (l) = 20 cm
absolute error in length (∆l) = 2 mm = 0.2 cm
Time taken for 50 oscillation (t) = 40 s
error in time ∆T = 1 s
Solution: Time period for one oscillation (T)

Hence, the percentage error in g is

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Conceptual Questions

Question 1.
Why is it convenient to express the distance of stars in terms of light year (or) parsec rather than in km?
Answer:
A parsec is 206, 265 AU and is roughly the distance to the nearest stars. If we were to view a giant star with a diameter of 1 AU at a distance of one parsec, it would appear to be just 1/3600^th of a degree in angular size. For comparison, the sun and moon are both half a degree in angular size when viewed from Earth.

Question 2.
Show that a screw gauge of pitch 1 mm and 100 divisions is more precise than a vernier caliper with 20 divisions on the sliding scale.
Answer:

As shown, the least count of screw gauge is lesser then vernier caliper, hence screw gauge is more precise.

Question 4.
Having all units in atomic standards is more useful. Explain.
Answer:
An atomic mass unit (symbolized AMU or amu) is defined as precisely 1/12 the mass of an atom of carbon-12. The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus.
In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass. This is approximately 1.67377 × 10^-27 kilogram (kg), or 1.67377 × 10^-24 gram (g). The mass of an atom in AMU is roughly equal to the sum of the number of protons and neutrons in the nucleus.
The AMU is used to express the relative masses of, and thereby differentiate between, various isotopes of elements. Thus, for example, uranium-235 (U-235) has an AMU of approximately 235, while uranium-238 (U-238) is slightly more massive. The difference results from the fact that U-238, the most abundant naturally occurring isotope of uranium, has three more neutrons than U-235, an isotope that has been used in nuclear reactors and atomic bombs.

Question 5.
Why dimensional methods are applicable only up to three quantities?
Answer:
Understanding dimensions is of utmost importance as it helps us in studying the nature of physical quantities mathematically. The basic concept of dimensions is that we can add or subtract only those quantities which have same dimensions. Also, two physical quantities are equal if they have same dimensions. these basic ideas help us in deriving the new relation between physical quantities, it is just like units.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Additional Questions Solved

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Multiple Choose Questions

Question 1.
The unit of surface tension ……
(a) MT^-2
(b) Nm^-2
(c) Nm
(d) Nm^-1
Answer:
(d) Nm^-1

Question 2.
One atomus equal to ……

Answer:
(c) 160 ms

Question 3.
One light year is ……
(a) 3.153 × 10⁷ m
(b) 1.496 × 10⁷ m
(c) 9.46 × 10¹² km
(d) 3.26 × 10¹⁵ km
Answer:
(c) 9.46 × 10¹² km

Question 4.
One Astronomical unit is
(a) 3.153 × 10⁷ m
(b) 1.496 × 10⁷ m
(c) 9.46 × 10¹² m
(d) 3.26 × 10¹⁵ m
Answer:
(b) 1.496 × 10⁷ m

Question 5.
One parsec is …..
(a) 3.153 × 10⁷ m
(b) 3.26 × 10¹⁵ m
(c) 30.84 × 10¹⁵ m
(d) 9.46 × 10¹⁵ m
Answer:
(c) 30.84 × 10¹⁵ m

Question 6.
One Fermi is …..
(a) 10^-9 m
(b) 10^-10 m
(c) 10^-12 m
(d) 10^-15 m
Answer:
(d) 10^-15 m

Question 7.
One Angstrom is ………
(a) 10^-9 m
(b) 10^-10m
(c) 10^-12 m
(d) 10^-15 m
Answer:
(b) 10^-10 m

Question 8.
One solar mass is ….
(a) 2 × 10³⁰ kg
(b) 2 × 10³⁰ g
(c) 2 × 10³⁰ mg
(d) 2 × 10³⁰ tonne
Answer:
(d) 2 × 10³⁰ tonne

Question 9.
\(\frac{1}{12}\) of the mass of carbon 12 atom is …..
(a) 1 TMC
(b) mass of neutron
(c) 1 amu
(d) mass of hydrogen
Answer:
(d) mass of hydrogen

Question 10.
The word physics is derived from the word …..
(a) scientist
(b) fusis
(c) fission
(d) fusion
Answer:
(b) fusis

Question 11.
The study of forces acting on bodies whether at rest or in motion is …..
(a) classical mechanics
(b) quantum mechanics
(c) thermodynamics
(d) condensed matter physics
Answer:
(a) classical mechanics

Question 12.
Mass of observable universe …..
(a) 10³¹ kg
(b) 10⁴¹ kg
(c) 10⁵⁵ kg
(d) 9.11 × 10³¹ kg
Answer:
(c) 10⁵⁵ kg

Question 13.
Mass of an electron …….

Answer:
(b) 9.11 × 10^-31 kg

Question 14.
The study of production and propagation of sound waves …..
(a) Astrophysics
(b) Acoustics
(c) Relativity
(d) Atomic physics
Answer:
(b) Acoustics

Question 15.
The study of the discrete nature of phenomena at the atomic and subatomic levels.
(a) Quantum mechanics
(b) High energy physics
(c) Acoustics
(d) Classical mechanics
Answer:
(a) Quantum mechanics

Question 16.
The techniQuestion used to study the crystal structure of various rocks are …….
(a) diffraction
(b) interference
(c) total internal reflection
(d) refraction
Answer:
(a) diffraction

Question 17.
The astronomers used to observe distant points of the universe by …….
(a) Electron telescope
(b) Astronomical telescope
(c) Radio telescope
(d) Radar
Answer:
(c) Radio telescope

Question 18.
The comparison of any physical quantity with its standard unit is known as ……..
(a) fundamental quantities
(b) measurement
(c) dualism
(d) derived quantities
Answer:
(b) measurement

Question 19.
Fundamental quantities can also be known as …… quantities.
(a) original
(b) physical
(c) negative
(d) base
Answer:
(d) base

Question 20.
Which one of the following is not a fundamental quantity?
(a) length
(b) luminous intensity
(c) temperature
(d) water current
Answer:
(d) water current

Question 21.
The system of unit not only based on length, mass, and time is
(a) FPS
(b) CGS
(c) MKS
(d) SI
Answer:
(d) SI

Question 22.
The coherent system of units …..
(a) CGS
(b) SI
(c) FPS
(d) MKS
Answer
(b) SI

Question 23.
The triple point temperature of water is ……
(a) -273.16 K
(b) 0K
(c) 273.16 K
(d) 100 K
Answer:
(d) 100 K

Question 24.
Which of the following is a unit of distance?
(a) Light year
(b) Leap year
(c) Dyne-sec
(d) Pauli
Answer:
(a) Light year

Question 25.
The unit of moment of force ……
(a) Nm²
(b) Nm
(c) N
(d) NJ rad
Answer:
(b) Nm

Question 26.
1 radian is ……
(a) 2.91 × 10^-4 m
(b) 57.27°
(c) 180°
(d) \(\frac{\pi}{180}\)
Answer:
(b) 57.27°

Question 27.
One degree of arc is …….
(a) 1″
(b) 60″
(c) 60′
(d) 60°
Answer:
(c) 60′

Question 28.
One degree of arc is equal to …….
(a) 1.457 × 10² rad
(b) 1.457 × 10^-2 rad
(c) 1.745 × 10² rad
(d) 1.745 × 10^-2 rad
Answer:
(b) 1.457 × 10^-2 rad

Question 29.
1 minute of arc is equal to …….
(a) 1.745 × 10^-2 rad
(b) 2.91 × 10^-4 rad
(c) 2.91 × 10⁴ rad
(d) 4.85 × 10^-6 rad
Answer:
(b) 2.91 × 10^-4 rad

Question 30.
1 second of arc is equal to ………

Answer:

Question 31.
1 second of arc is equal to ….
(a) 0.00027°
(b) 1.745 × 10^-2 rad
(c) 2.91 × 10^-4 rad
(d) 4.85 × 10^-6 rad
Answer:
(a) 0.00027°

Question 32.
Unit of impulse ….
(a) NS²
(b) NS
(c) Nm
(d) Kgms^-2
Answer:
(b) NS

Question 33.
The ratio of energy and temperature is known as ……
(a) Stefen’s constant
(b) Boltzmann constant
(c) Plank’s constant
(d) Kinetic constant
Answer:
(b) Boltzmann constant

Question 34.
The range of distance can be measured by using direct methods is …..
(a) 10^-2 to 10^-5 m
(b) 10^-2 to 10² m
(c) 10² to 1(T5 m {d) 10″2 to 105 m
Answer:
(b) 10^-2 to 10² m

Question 35.
Which of the following is in increased order?
(a) exa, tera, hecto
(b) tera, exa, hecto
(c) giga, tera, exa
(d) hecto, exa, giga
Answer:
(c) giga, tera, exa

Question 36.
10^-18 is called as ……
(a) nano
(b) pico
(c) femto
(d) atto
Answer:
(d) atto

Question 37.
A radio signal sent towards the distant planet, returns after “t” s. If “c” is the speed of radio
waves then the distance of the planet and from the earth is …….

Answer:

Question 38.
Find odd one out ….
(a) Newton
(b) metre
(c) candela
(d) Kelvin
Answer:
(a) Newton

Question 39.
The shift in the position of an object when viewed with two eyes, keeping one eye closed at a
time is known as …
(a) basis
(b) fundamental
(c) parallax
(d) pendulum
Answer:
(c) parallax

Question 40.
Chandrasekar limit is ….. times the mass of the sun.
(a) 1.2
(b) 1.4
(c) 1.6
(d) 1.8
Answer:
(b) 1.4

Question 41.
The smallest physical unit of time is
(a) second
(b) minute
(c) microsecond
(d) shake
Answer:
(d) shake

Question 42.
Size of atomic nucleus is …..
(a) 10^-10 m
(b) 10^-12 m
(c) 10^-14 m
(d) 10^-18 m
Answer:
(c) 10^-14 m

Question 43.
Time interval between two successive heart beat is in the order of …….
(a) 10° s
(b) 10 s
(c) 10² s
(d) 10^-3 s
Answer:
(a) 10° s

Question 44.
Half life time of a free neutron is in the order of ……
(a) 10°
(b) 10¹ s
(c) 10² s
(d) 10³ s
Answer:
(d) 10³ s

Question 45.
The uncertainty contained in any measurement is ……
(a) rounding off
(b) error
(c) parallax
(d) gross
Answer:
(b) error

Question 46.
Zero error of an instrument is a ……
(a) Systematic error
(b) Random error
(c) Gross error
(d) Both (a) and (b)
Answer:
(a) Systematic error

Question 47.
Error in the measurement of radius of a sphere is 2%. Then error in the measurement of surface
area is ….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Question 48.
Imperfections in experimental procedure gives ….. error.
(a) random
(b) gross
(c) systematic
(d) personal
Answer:
(c) Systematic

Question 49.
Random error can also be called as ….
(a) personal error
(b) chance error
(c) gross error
(d) system error
Answer:
(b) chance error

Question 50.
To get the best possible true value of the quantity has to be taken.
(a) rms value
(b) net value
(c) arithmetic mean
(d) mode
Answer:
(c) arithmetic mean

Question 51.
The error caused due to the shear carelessness of an observer is called as …… error.
(a) Systematise
(b) Gross
(c) Random
(d) Personal
Answer:
(b) Gross

Question 52.
The uncertainty in a measurement is called as ….
(a) error
(b) systematic
(c) random error
(d) gross error
Answer:
(a) error

Question 53.
The difference between the true value and the measured value of a quantity is known as …..
(a) Absolute error
(b) Relative error
(c) Percentage error
(d) Systemmatic error
Answer:
(a) Absolute error

Question 54.
If a₁, a₂, a₃ …. a_n are the measured value of a physical quantity “a” and a_m is the true value then absolute error …..

Answer:
(d) \(\Delta a_{n}=a_{m}-a_{n}\)

Question 55.
If ‘a_m‘ and ‘∆a_m ‘ are true value and mean absolute error respectively, then the magnitude of the quantity may lie between …..

Answer:
(b) \(a_{m}-\Delta a_{m} \text { to } a_{m}+\Delta a_{m}\)

Question 56.
The ratio of the mean absolute error to the mean value is called as ……
(a) absolute error
(b) random error
(c) relative error
(d) percentage error
Answer:
(c) Relative error

Question 57.
Relative error can also be called as ……
(a) fractional error
(b) absolute error
(c) percentage error
(d) systematic error
Answer:
(a) fractional error

Question 58.
A measured value to be close to targeted value, percentage error must be close to
(a) 0
(b) 10
(c) 100
(d) ∝
Answer:
(a) 0

Question 59.
The maximum possible error in the sum of two quantities is equal to …….
(a) Z = A + B
(b) ∆Z = ∆A + ∆B
(c) ∆Z = ∆A/∆B
(d) ∆Z = ∆A – ∆B
Answer:
(b) ∆Z = ∆A + ∆B

Question 60.
The maximum possible error in the difference of two quantities is ……

Answer:

Question 61.
The maximum fractional error in the division of two quantities is ….

Answer:

Question 62.
The fractional error in the n^th power of a quantity is …..

Answer:

Question 63.
A physical quantity is given as y = \(\frac{a b^{3}}{c^{2}}\). If ∆a, ∆b, ∆c are absolute errors, the possible fractional error in y is …..

Answer:

Question 64.
Number of significant digits in 3256 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 65.
Number of significant digits in 32005 ……
(a) 1
(b) 2
(c) 5
(d) 2
Answer:
(c) 5

Question 66.
Number of significant digits in 2030 …
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 67.
Number of significant digits in 2030N …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 68.
Number of significant digits in 0.0342 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c)3

Question 69.
Number of significant digit in 20.00 …..
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 70.
Number of significant digit in 0.030400
(a) 6
(b) 5
(c) 4
(d) 3
Answer:
(b) 5

Question 71.
The force acting on a body is measured as 4.25 N. Round it off with two significant figure ..
(a) 4.3
(b) 4.2
(c) both
(a) or (b)
(d) 4.25
Answer:
(b) 4.2

Question 72.
The quantities a, b, c are measured as 3.21, 4.253, 7.2346. The sum (a + b + c) with proper
significant digits is ……
(a) 14.6976
(b) 14.697
(c) 14.69
(d) 14.6
Answer:
(c) 14.69

Question 73.
The dimensions of gravitational constant G are …

Answer:
(b) \(\mathbf{M}^{-1} \mathbf{L}^{3} \mathbf{T}^{-2}\)

Question 74.
The ratio of one nanometer to one micron is
(a) 10^-3
(b) 10³
(c) 10^-9
(d) 10^-6
Answer:
(b) 10³

Question 75.
Which of the following pairs does not have same dimension?
(a) Moment of inertia and moment of force
(b) Work and torque
(c) Impulse and momentum
(d) Angular momentum and Plank’s constant
Answer:
(a) Moment of inertia and moment of force

Question 76.
Two quantities A and B have different dimensions. Which of the following is physically meaningful?
(a) A + B
(b) A – B
(c) A /B
(d) None
Answer:
(c) A /B

Question 77.
The dimensional formula for moment of inertia ……

Answer:
(d) \(\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{\mathbf{0}}\)

Question 78.
Which of the following is having same dimensional formula?
(a) Work and power
(b) Radius of gyration and displacement
(c) Impulse and force
(d) Frequencies and wavelength
Answer:
(b) Radius of gyration and displacement

Question 79.
Which of the following quantities is expressed as force per unit area?
(a) Pressure
(b) Stress
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

Question 80.
In equation of motion the dimensional formula for K is …..

Answer:
(b) \(\left[\mathbf{L} \mathbf{T}^{-2}\right]\)

Question 81.
The dimensional formula for heat capacity ……

Answer:
(d) \(\left[\mathbf{M} \mathbf{L}^{2} \mathbf{T}^{2} \mathbf{K}^{-1}\right]\)

Question 82.
The product of Avogadro constant and elementary charge is known as …… constant.
(a) Planck’s
(b) Avagadro
(c) Boltzmann
(d) Faraday
Answer:
(d) Faraday

Question 83.
The force F is given by F = at + bt² where t is time. The dimensions of ‘a’ and ‘b’ respectively are

Answer:
(d) \(\left[\mathrm{MLT}^{-2}\right]\) and \(\left[\mathrm{MLT}^{-0}\right]\)

Question 84.
Dimensions of impulse are …..

Answer:
(c) \(\left[\mathrm{MLT}^{-1}\right]\)

Question 85.
If speed of light (c), acceleration due to gravity (g) and pressure (P) are taken as fundamental
units, the possible relation to gravitational constant (G) is ….

Answer:
(c) \(c^{0} g^{2} p^{-1}\)

Question 86.
Equivalent of one joule is ……
(a) Nm²
(b) kg m² s^-2
(c) kg m s^-1
(d) N kg m²
Answer:
(b) kg m² s^-2

Question 87.
Pick out the dimensionless quantity …..
(a) force
(b) specific gravity
(c) planck’s constant
(d) velocity
Answer:
(b) specific gravity

Question 88.
Odd one out ………
(a) strain
(b) refractive index
(c) numbers
(d) stress
Answer:
(d) stress

Question 89.
A wire has a mass 0.3 ± 0.003g, radius 0.5 ± 0.005 mm and length 6 + 0.06 cm. The maximum percentage error in the measurement of its density is …….
(a) 1%
(b) 2%
(c) 3%
(d) 4%
Answer:
(d) 4%

Question 90.
The dimensions of planck constant equals to that of …..
(a) energy
(b) momentum
(c) angular momentum
(d) power
Answer:
(c) angular momentum

Question 91

(a) The unit of λ is same as that of x and A
(b) The unit of λ is same as that of x but not of A
(c) The unit of c is same as that of 2π/λ
(d) The unit of (ct – x)is same as that 2π/λ
Answer:
(a) The unit of λ is same as that of x and A

Question 92.
The number of significant figures in 0.06900 is …….
(a) 2
(b) 2
(c) 4
(d) 5
Answer:
(c) 4

Question 93.
The numbers 3.665 and 3.635 on rounding off to 3 significant figures will give
(a) 3.66 and 3.63
(b) 3.66 and 3.64
(c) 3.67 and 3.63
(d) 3.67 and 3.64
Answer:
(b) 3.66 and 3.64

Question 94.
Which of the following measurements is most precise?
(a) 4.00 mm
(b) 4.00 cm
(c) 4.00 m
(d) 4.00 km
Answer:
(a) 4.00 mm

Question 95.
The mean radius of a wire is 2 mm. Which of the following measurements is most accurate? (a) 1.9 mm
(b) 2.25 mm
(c) 2.3 mm
(d) 1.83 mm
Answer:
(a) 1.9 mm

Question 96.
If error in measurement of radius of sphere is 1%. What will be the error in measurement of volume?
(a) 1%
(b) \(\frac{1}{3}\)%
(c) 3%
(d) 10%
Answer:
(c) 3%

Question 97.
Dimensions [M L^-1 T^-1] are related to …….
(a) torque
(b) work
(c) energy
(d) Coefficient of viscosity
Answer:
(d) Coefficient of viscosity

Question 98.
Heat produced by a current is obtained a relation H = I²RT. If the errors in measuring these quantities current, resistance, time are 1%, 2%, 1% respectively then total error in calculating the energy produced is
(a) 2%
(b) 4%
(c) 5%
(d) 6%
Answer:
(c) 5%

Question 99.
Length cannot be measured by ….
(a) fermi
(b) angstrom
(c) parsec
(d) debye
Answer:
(d) debye

Question 100.
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate by using the formula p = \([/late\frac{\mathrm{F}}{l^{2}}x]. If the maximum errors in the measurement
of force and length are 4% and 2% respectively, then the maximum error in the measurement of pressure is ……..
(a) 1%
(b) 2%
(c) 8%
(d) 10%
Aswer:
(c) 8%

Question 101.
Which of the following cannot be verified by using dimensional analysis?
1 mv²

Answer:
(b) y = a sin wt

Question 102.
Percentage errors in the measurement of mass and speed are 3% and 2% respectively. The error in the calculation of kinetic energy is …….
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 103.
More number of readings will reduce …….
(a) random error
(b) systematic error
(c) both (a) and (b)
(d) neither (a) nor (b)
Answer:
(a) random error

Question 104.
If the percentage error in the measurement of mass and momentum of a body are 3% and
2%respectively, then maximum possible error in kinetic energy is
(a) 2%
(b) 3%
(c) 5%
(d) 7%
Answer:
(d) 7%

Question 105.
In a vernier caliper, n divisions of vernier scale coincides with (n – 1) divisions of main scale. The least count of the instrument is ………

Answer:

Question 106.
The period of a simple pendulum is recorded as 2.56s, 2.42s, 2.71s and 2.80s respectively.
The average absolute error is
(a) 0.1s
(b) 0.2s
(c) 1.0s
(d) 0.11s
Answer:
(d) 0.11s

Question 107.
In a system of units, if force (F), acceleration (A) and time (T) are taken as fundamental units
then the dimensional formula of energy is
(a) [FA²T]
(b) [FAT²]
(c) [F²AT]
(d) [FAT]
Answer:
(b) [FAT²]

Question 108.
The random error in the arithmetic mean of 50 observations is ‘a’, then the random error in the
arithmetic mean of 200 observations a would be

Answer:
(c) [latex]\frac{a}{4}\)

Question 109.
Which of the following is not dimensionless?
(a) Relative permittivity
(b) Refractive index
(c) Relative density
(d) Relative velocity
Answer:
(d) Relative velocity

Question 110.
If V-velocity, K – kinetic energy and T – time are chosen as the fundamental units, then what is the dimensional formula for surface tension?

Answer:
(a) [K V^-2 T^-2]

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (1 Mark)

Question 1.
A new unit of length is chosen such that the speed of light in vaccum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes 8 min and 20 s to cover this distance.
Answer:
Speed of light in vacuum, c = 1 new unit of length s^-1
t = 8 min. 20 sec, = 500 s
x = ct= 1 new unit of length s^-1 × 500s
x = 500 new unit of length

Question 2.
If x = a + bt + ct², where x is in metre and t in seconds, what is the unit of c ?
Answer:
The unit of left hand side is metre so the units of ct² should also be metre.
Since t² has unit of s², so the unit of c is m/s².

Question 3.
What is the difference between mN, Nm and nm ?
Answer:
mN means milli newton, 1 mN = 10^-3 N, Nm means Newton meter, nm means nano meter.

Question 4.
The radius of atom is of the order of 1 A° & radius of nucleus is of the order of fermi. How many magnitudes higher is the volume of the atom as compared to the volume of nucleus ?
Answer:

Question 5.
How many kg make 1 unified atomic mass unit ?
Answer:
1u = 1.66 × 10^-27 kg

Question 6.
Name some physical quantities that have same dimension.
Answer:
Work, energy and torque.

Question 7.
Name the physical quantities that have dimensional formula [ML ^-1T^-2].
Answer:
Stress, pressure, modulus of elasticity.

Question 8.
Give two examples of dimensionless variables.
Answer:
Strain, refractive index.

Question 9.
State the number of significant figures in
(i) 0.007 m²
(ii) 2.64 × 1024 kg
(iii) 0.2370 g cm^-3
(iv) 0.2300m
(v) 86400
(vi) 86400 m
Answer:
(i) 1,
(ii) 3,
(iii) 4,
(iv) 4,
(v) 3,
(vi) 5 since it comes from a measurement the last two zeros become significant.

Question 10.
Given relative error in the measurement of length is 0.02, what is the percentage error ?
Answer:
2%.

Question 11.
A physical quantity P is related to four observables a, b, c and d as follows :

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2% respectively. What is the percentage error in the quantity P?
Answer:
Relative error in P is given by

Question 12.
A boy recalls the relation for relativistic mass (m) in terms of rest mass (m₀) velocity of particle V, but forgets to put the constant c (velocity of light). He writes correct the equation by putting the missing ‘c’.
Answer:
Since quantities of similar nature can only be added or subtracted, v² cannot be subtracted from 1 but v²/c² can be subtracted from 1.

Question 13.
Name the technique used in locating.
(a) an under water obstacle
(b) position of an aeroplane in space.
Answer:
(a) SONAR ➝ Sound Navigation and Ranging.
(b) RADAR ➝ Radio Detection and Ranging.

Question 14.
Deduce dimensional formulae of—
(i) Boltzmann’s constant
(ii) mechanical equivalent of heat.
Answer:
(i) Boltzmann Constant:

Question 15.
Give examples of dimensional constants and dimensionless constants.
Answer:
Dimensional Constants : Gravitational constant, Plank’s constant. Dimensionless Constants : it, e.

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Short Answer Questions (2 Marks)

Question 16.
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm. Calculate the minimum inaccuracy in the measurement of distance.
Answer:
Minimum inaccuracy = Vernier constant

Question 17.
If the unit of force is 100N, unit of length is 10m and unit of time is 100s. What is the unit of Mass in this system of units ?
Answer:

Question 18.
State the principle of homogeneity. Test the dimensional homogeneity of equations

Answer:

as Dimensions of L.H.S. = Dimensions of R.H.S.
∴ The equation to dimensionally homogeneous.

(ii) S_n = Distance travelled in n^th sec that is (S_n – S_{n – 1})

Hence this is dimensionally correct.

Question 19.

Answer:
Since dimensionally similar quantities can only be added

Question 20.
Magnitude of force experienced by an object moving with speed v is given by F = kv². Find dimensions of k.
Answer:

Question 21.
A book with printing error contains four different formulae for displacement. Choose the
correct formula/formulae

Answer:
The arguments of sine and cosine function must be dimensionless so (a) is the probable correct formulae. Since

Samacheer Kalvi 11th Physics Nature of Physical World and Measurement Numericals Questions

Question 22.
Determine the number of light years in one metre.
Answer:

Question 23.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces 20.15 g and 20.17 g are added to the box.
(i) What is the total mass of the box ?
(ii) The difference in masses of the pieces to correct significant figures.
Answer:
(i) Mass of box = 2.3 kg
Mass of gold pieces = 20.15 + 20.17 = 40.32 g = 0.04032 kg.
Total mass = 2.3 + 0.04032 = 2.34032 kg
In correct significant figure mass = 2.3 kg (as least decimal)
(ii) Difference in mass of gold pieces = 0.02 g
In correct significant figure (2 significant fig. minimum decimal) will be 0.02 g.

Question 24.
5.74 g of a substance occupies 1.2 cm³. Express its density to correct significant figures.
Answer:

Here least significant figure is 2, so density = 4.8 g/cm³.

Question 25.
If displacement of a body s = (200 ± 5) m and time taken by it t = (20 + 0.2) s, then find the percentage error in the calculation of velocity.
Answer:

Question 26.
If the error in measurement of mass of a body be 3% and in the measurement of velocity be 2%. What will be maximum possible error in calculation of kinetic energy.
Answer:

Question 27.
The length of a rod as measured in an experiment was found to be 2.48 m, 2.46 m, 2.49 m, 2.50 m and 2.48 m. Find the average length, absolute error and percentage error. Express the result with error limit.
Answer:

Question 28.
A physical quantity is measured as Q = (2.1 ± 0.5) units. Calculate the percentage error in (1) Q² (2) 2Q.
Answer:

Question 29.
When the planet Jupiter is at a distance of 824.7 million km from the earth, its angular diameter is measured to be 35.72″ of arc. Calculate diameter of Jupiter.
Answer:

Question 30.
A laser light beamed at the moon takes 2.56s and to return after reflection at the moon’s surface. What will be the radius of lunar orbit?
Answer:
t = 2.54 s

Question 31.
Convert
(i) 3 m.s^-2 to km h^-2
(ii) G = 6.67 × 10^-11 N m² kg^-2 to cm³ g^-1 s^-2
Answer:

Question 32.
A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m²s^-2. Suppose we employ a system of units in which unit of mass is α kg, unit of length is β m, unit of time γs. What will be magnitude of calorie in terms of this new system.
Answer:

Question 33.
The escape velocity v of a body depends on—
(i) the acceleration due to gravity ‘g’ of the planet,
(ii) the radius R of the planet. Establish dimensionally the relation for the escape velocity.
Solution:

Question 34.
The frequency of vibration of a string depends of on,
(i) tension in the string
(ii) mass per unit length of string,
(iii) vibrating length of the string. Establish dimensionally the relation for frequency.
Answer:

Question 35.
One mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is the ratio so large? Take radius of hydrogen molecule to be 1°A.
Answer:

This ratio is large because actual size of gas molecule is negligible in comparison to the inter molecular separation.

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Tamilnadu Samacheer Kalvi 10th Social Science Geography Solutions Chapter 6 Physical Geography of Tamil Nadu

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Physical Geography of Tamil Nadu Textual Exercise

I. Choose the correct answer.

Physical Geography Of Tamil Nadu Question 1.
The latitudinal extent of Tamil Nadu is …………..
(a) 8°4’N to 13°35’N
(b) 8°5’S to 13°35’S
(c) 8°0′ to 13°5’N
(d) 8°0’S to 13°05’S
Answer:
(a) 8°4’N to 13°35’N

State The Boundaries Of Tamil Nadu Class 10 Question 2.
The longitudinal extent of Tamil Nadu is
(a) 76° 18’E to 80°20’E
(b) 76° 18’W to 80°20’W
(c) 86°18’E to 10°20’E
(d) 86°18’W to 10°20’W
Answer:
(a) 76° 18’E to 80°20’E

Question 3.
The highest peak in Tamil Nadu is ……………..
(a) Anaimudi
(b) Doddabetta
(c) Mahendragiri
(d) Servarayan
Answer:
(b) Doddabetta

Question 4.
Which of the following passes is not located in the Western Ghats of Tamil Nadu?
(a) Palghat
(b) Shencottah
(c) Bhorghat
(d) Achankoil
Answer:
(c) Bhorghat

Question 5.
Which one of the following river flows into the Arabian Sea?
(a) Periyar
(b) Cauvery
(c) Chittar
(d) Bhavani
Answer:
(a) Periyar

Question 6.
The district with largest mangrove forest cover in Tamil Nadu is:
(a) Ramanathapuram
(b) Nagapattinam
(c) Cuddalore
(d) Theni
Answer:
(c) Cuddalore

Question 7.
The forest cover of Tamil Nadu as per 2017 Indian Forest Report is ………………
(a) 20.21%
(b) 20.31%
(c) 21.20%
(d) 21.30%
Answer:
(a) 20.21%

Question 8.
Retreating monsoon wind picks up moisture from:
(a) Arabian sea
(b) Bay of Bengal
(c) Indian Ocean
(d) Timor sea
Answer:
(b) Bay of Bengal

Question 9.
Which of the following district is affected by sand dunes to a large extent?
(a) Theni
(b) Madurai
(c) Thanjavur
(d) Ramanathapuram
Answer:
(d) Ramanathapuram

Question 10.
The district which has the largest forest cover in Tamil Nadu is:
(a) Dharmapuri
(b) Vellore
(c) Dindigul
(d) Erode
Answer:
(a) Dharmapuri

II. Fill in the blanks.

1. The plateau which lies between the Nilgiris and Dharmapuri districts is ……………..
2. ……………. is the highest peak in the southernmost part of the Eastern Ghats.
3. The riverine Island of Srirangam is located between …………… and ………….. branches of cauvery.
4. ……………… soil is suitable for the cultivation of tea and coffee plants.
5. ……………. is the Tamil Nadu state animal which is found in
Answers:
1. Bharamahal
2. Solaikaradu
3. Northern, Southern
4. Laterite
5. Nilgiri Tahr, Nilgiri Hills

III. Match the following.

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

IV. Assertion type Question.

Question 1.
Assertion (A): Tamil Nadu does not receive much rainfall from the southwest monsoon. Reasoning (R): It is situated in the rain shadow area of the Western Ghats.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both (A) and (R) are true but, (R) does not explain (A).
(c) (A) is true but, (R) is false.
(d) (R) is true but, (A) is false.
Answer:
(a) Both (A) and (R) are true and (R) explains (A).

Question 2.
Assertion (A): Red soil is rich in iron oxides Reasoning (R): It is formed by leaching
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both (A) and (R) are true but, (R) does not explain (A).
(c) (A) is true but, (R) is false.
(d) (R) is true but, (A) is false.
Answer:
(b) Both (A) and (R) are true but, (R) does not explain (A).

V. Answer the following questions briefly.

Question 1.
State the boundaries of Tamil Nadu.
Answer:
Tamil Nadu is bounded by Bay of Bengal in the East, Kerala in the West, Andhra Pradesh in the North Karnataka in the North West and Indian Ocean in the South.

Question 2.
List out the districts of Tamil Nadu which are partly/fully located on Eastern and Western Ghats separately.
Answer:
Eastern Ghats:
Parangimalai, Chennai, Javadhu hills, Vellore, Salem, Sirumalai, Dindigul, Thiruvannamalai, Bhubaneswar, Namakkal, Perambalur.

Western Ghats:
Kanyakumari, Tirunelveli, Theni, Coimbatore, Nilgiri.

Question 3.
What is ‘Teri’?
Answer:
The sand dunes formed along the coast of Ramanathapuram and Thoothukudi districts are called “Teri”.

Question 4.
How is coastal plain formed?
Answer:
It is formed by the rivers that flow towards east and drain in the Bay of Bengal.

Question 5.
Name the major islands of Tamil Nadu.
Answer:
Pamban, Hare, Kurusadai, Nallathanni Theevu, Pullivasal, Srirangam, Upputanni, Island grounds, Kattupalli Island, Quibble island and Vivekananda rock memorial are some major islands of Tamil Nadu.

Question 6.
Name the tributaries of river Thamirabarani.
Answer:
Karaiyar, Servalar, Manimuthar, Gadananathi, Pachaiyar, Chittar and Ramanathi are its main tributaries.

Question 7.
Define Disaster Risk Reduction.
Answer:
Disaster Risk Reduction (UNDRR) According to United Nations Office for Disaster Risk Reduction is “the concept and practice of reducing disaster risks through systematic efforts to analyses and reduce the causal factors of disasters”. This includes reducing exposure to hazards, lessening the vulnerability of people and property, wise management of land and environment and early warning for adverse events.

Question 8.
During cyclone, how does the Meterological department warn the fishermen?
Answer:
During disturbed weather over the seas, the ports likely to be affected are warned by concerned ACWCs / CWCs by advising the port authorities through port warnings to hoist appropriate Storm Warning Signals. The Department also issues “Fleet forecast” for Indian Navy. Coastal Bulletins for Indian coastal areas covering up to 75 km from the coastline and sea area bulletins for the sea areas beyond 75 km. The special warnings are issued for fishermen four times a day in normal weather and every three hourly in accordance with the four-stage warning in case of disturbed weather.

The general public, the coastal residents and fishermen are warned through state government officials and broadcast of warnings through All India Radio and National Television telecast programmes in national and regional hook up. A sytem of warning dissemination for fishermen through World Space Digital Based radio receiver is being planned.

VI. Distinguish between the following.

Question 1.
Western Ghats and Eastern Ghats.
Answer:
Western Ghats Eastern Ghats

Question 2.
Southwest Monsoon and Northeast Monsoon.
Answer:
Southwest Monsoon Northeast Monsoon

Question 3.
Tropical evergreen and Tropical deciduous forests.
Answer:
Tropical evergreen forest Tropical deciduous forests

VII. Give reasons for the following:

Question 1.
Eastern Ghats are not a continuous range.
Answer:
The Eastern Ghats are not a continuous range because it is dissected at many places by the rivers flowing towards the east that drain into the Bay of Bengal.

Question 2.
Tamil Nadu receives low rainfall during southwest monsoon.
Answer:
During southwest monsoon, Tamil Nadu is located in the rain shadow region for the wind, which blows from the Arabian sea. As a result, Tamil Nadu receives only a meagre rainfall from this monsoon. Rainfall during this season decreases from west to east.

Question 3.
Alluvial soil is fertile.
Answer:
Alluvial soil is formed by the deposition of rivers and are rich in minerals like lime, potassium magnesium, nitrogen and phosphoric acid. Hence it is fertile in nature.

Question 4.
Cuddalore is a multiprone disaster zone.
Answer:
Cuddalore districts is prone to natural calamities having experienced landfalls of major cyclones formed in the Bay of Bengal region. Apart from the cyclones, 2004 Tsunami caused massive damages to life and property in Cuddalore and its adjacent Nagapattinam district. Cyclone Thane which made landfall here caused major loss to life and property.

VIII. Answer the following in a paragraph.

Question 1.
Describe the nature of the plateau region of Tamil Nadu.
Answer:

1. Location: Plateaus of Tamil Nadu are located between the Western Ghats and the Eastern Ghats.
2. Area and Shape: It covers an area of about 60,000 sq. km and is roughly triangular in shape.
3. Slope and height: It is broader in the North and very narrow in the South. It’s height increases from East to West ranging between 150 and 600 metres.
4. Sub-divisions: Baramahal plateau, Coimbatore plateau and Madurai plateau.

Question 2.
Write an account on river Cauvery.
Answer:
The main river of Tamil Nadu is cauvery which originates at Talacauvery in the Brahmragiri hills of Kodagu (coorg) district of Karnataka in the Western Ghats.

About 416 km of its course falls in Tamil Nadu. It serves as the boundary between Karnataka and Tamil Nadu for a distance of 64 km. A tributary called Bhavani joins Cauvery on the right bank about 45 km from the Mettur Reservoir.

Thereafter, it takes easterly course to enter into the plains of Tamil Nadu. Cauvery and its distributaries in its lower course drain the districts of Nagapattinam, Thanjavur, Thivarur and Thiruchirapalli. The cauvery, kollidam and the vellar jointly drain central part of the Tamil . Nadu. The head of the cauvery delta is near the islands of Srirangam. Kollidam branches off from cauvery at Grand Anaicut, also called as kallanai was built across the river cauvery.

After kallanai, the river breaks into a large number of distributaries and forms a network all over the delta. The network of tributaries within the delta of cauvery in the coast is called as the ‘Garden of Southern India’. It merges into Bay of Bengal to the south of cuddalore. Cauvery along with its tributaries Bhavani, Noyyal, Mayar and Amaravathi is the most important source of canal irrigation.

Question 3.
Explain the characteristic features of summer and winter seasons of Tamil Nadu.
Answer:

1. The State of Tamil Nadu lies to the South of Tropic of cancer, which is near the Equator.
2. As it receives vertical Sunrays, the temperature the State is relatively high throughout the year.
3. But the climate of the coastal regions are influenced by the surrounding seas Bay of Bengal and the Indian Ocean enjoys tropical maritime climate.

Characteristic features of Summer Season:

1. There is a steady rise in temperature from South to North.
2. Generally the temperature various from 30°C to more than 40°C.
3. Summer falls in the months of March, April and May. Hottest month is the month of May.
4. In this season particularly in the month of May Southern , part of the. State receives some rainfall from pre-monsoon showers and some parts experience convectional rainfall.

Winter Season:

1. During January and February Tamil Nadu receives the slanting rays of the sun.
2. So the weather is slightly cooler. The difference between summer and winter temperature is not very high.
3. Winter temperature in the State varies from 15°C to 25°C. However in hill stations it ranges from 5°C to 0°C forming thick mist and frost.
4. The season is generally dry.

Question 4.
What is desertification and write about the areas affected by it in Tamil Nadu.
Answer:
Desertification is a type of land degradation in which a relatively dry areas of land becomes a desert, typically losing its bodies of water as well as vegetation and wild life. Theni, the Nilgiris and Kanniyakumari are the worst affected districts. About 12,000 hectares (120 Sq.km) were affected by sand deposition in Theni and Rajapalayam.

Question 5.
Bring out the types and distribution of soils in Tamil Nadu.
Answer:

Question 6.
Name the areas which are affected by landslides. What will you do before, during and after landslides?
Answer:

1. A collapse of a mass of Earth or rock from a mountain or cliff is called landslide.
2. One of the most vulnerable district affected by landslides is Nilgiris.
3. The other regions which are prone to landslides are Coimbatore and Palani hill of Dindigul district.

Risk Reduction Measures:
Before:

1. Create awareness
2. Stay alert and awake
3. Monitor the News update
4. Make evacuation plan
5. Listen for any unusual sounds indication such as.
6. Moving debris – trees cracking, boulders knocking.
7. Consider leaving the place of landslides if it is safe to do so.

During:

(a) If Indoors:

• Find cover in the section of the building that is farthest away from the approaching landslide.
• Take shelter under a strong table or bench.
• Hold on firmly and stay until all movement has ceased.

(b) If outdoors:

• Move quickly away from its likely path.
• Keeping clear of embankments, trees, power lines and poles.
• Avoid crossing roads and bridges and stay away from landslides.
• The slope may experience additional failures for hours todays afterwards.

(c) After:

• Stay away from the slide area.
• Listen to local radio or television for the latest emergency information.
• Watch for flooding which may occur after a landslide or debris flow.
• Check for injured and trapped persons near the slide, without entering the direct slide area.

IX. Map study

Question 1.
Mark important rivers, distribution of soil and forest types on different Tamil Nadu maps.
Answer:

In-text Activity

Question 1.
Find out the coastal districts of Tamil Nadu with the help of a map.
Answer:

Question 2.
Group the districts of Tamil Nadu which share their boundary with the states of Andhra Pradesh, Karnataka and Kerala separately.
Answer:

In-Text Find Out

Question 1.
Name the first state of India created on linguistic basis.
Answer:
Andhra State on October 1st, 1953.

Question 2.
Why was the capital of Tamil Nadu renamed?
Answer:
The then local people called the capital of Tamil Nadu as Chennapattinam after the Vassal Chennapa Naicker who ruled this region and in the Sale Deed it was mentioned in the name after the Chenna Kesava temple which was the face of the city during that time. In 1996 Madras was officially renamed as Chennai in Tamil.

Question 3.
What is the meaning of the word ‘Chennai’?
Answer:
According to a suggestion, the city was named after the Chennakesava Perumal temple in the 1600s. The word Chennai in Tamil means ‘face’, with the temple regarded as the face of the city. The city was called Chennapatnam, which became to be abbreviated as Chennai.

Question 4.
Between in which latitude and longitude, is your school located?
Answer:
Do it yourself.

Question 5.
Why are mountain heights measured from mean sea level and not from ground level?
Answer:
Sea level is the base level for measuring elevation and depth on Earth. Because the ocean is one continuous body of water, its surface tends to seek the same level throughout the world.

Question 6.
Name the hill resorts of Western Ghats and Eastern Ghats in Tamil Nadu?
Answer:
Western Ghats: Annaimalai, Cardamon Hills, Palani Hills.
Eastern Ghats: Shevaroy Hills, Kalrayan Hills.

Question 7.
Is Ooty located on Western Ghats?
Answer:
Ooty is located on the Nilgiri hills which is a range of the Western Ghats.

Question 8.
Name the hill station located in Western Ghats and Eastern Ghats of Tamil Nadu?
Answer:
Western Ghats: Nilgiris, Kodaikanal, Annaimalai, Palani, Andipati and Agasthiyar Hills.
Eastern Ghats: Javadhu, Servarayan, Kalrayan, Kollimalai, Pachaimalai.

Question 9.
Why is the Nilgiri hills called as Blue Mountains?
Answer:
From all the directions and point of view these hills appear bluish hue in their natural settings. Hence the Nilgiri hills are called as Blue Mountains.

Question 10.
What is the kind of landform on which you live and what is its height?
Answer:
Do it yourself.

Question 11.
What is Agni Nakshatram?
Answer:

• Temperature of the states increases in the second week of February.
• Gradually it increases in the months of March, May and June.
• The hottest part of the summer season is known as “Agni Nakshatram” or Katthiriveyyil.

Question 12.
Group the districts of Tamil Nadu into low, moderate and heavy rainfall regions.
Answer:

Physical Geography of Tamil Nadu Additional Questions

I. Choose the correct answer.

Question 1.
There are ……………… districts in Tamil Nadu.
(a) 32
(b) 35
(c) 30
Answer:
(b) 35

Question 2.
Tamil Nadu is the ………………. largest State in India regarding its area.
(a) 10th
(b) 12th
(c) 11th
(d) 20th
Answer:
(c) 11th

Question 3.
The state lying on the …………… tip of the Indian Peninsula is Tamil Nadu.
(a) Southern
(b) Western
(c) Eastern
Answer:
(a) Southern

Question 4.
At present there are districts in Tamil Nadu.
(a) 25
(b) 35
(c) 14
(d) 29
Answer:
(b) 35

Question 5.
Geographically Tamil Nadu is divided into five ……………..
(a) Plain region
(b) Hilly region
(c) Physical division
Answer:
(c) Physical division

Question 6.
The topography of the Tamil Nadu State slopes towards:
(a) East
(b) West
(c) North
(d) South
Answer:
(a) East

Question 7.
………….. is located in the border of Tamil Nadu and Kerala.
(a) Nilgiri Hills
(b) Anaimalai
(c) Cardamom Hills
Answer:
(b) Anaimalai

Question 8.
Cardamom hills are also known as :
(a) Pachaimalai
(b) Servarayan
(c) Yelamalai
(d) Kalvarayan
Answer:
(c) Yelamalai

Question 9.
The trees of ……………. forest shed their leaves during dry season.
(a) Tidel
(b) Deciduous
(c) Evergreen
Answer:
(b) Deciduous

Question 10.
Kolli hills is a small mountain range located in district.
(a) Madurai
(b) Dindigal
(c) Namakkal
(d) Salem
Answer:
(c) Namakkal

Question 11.
…………… determines thickness of soil profile.
(a) Climate
(b) Time
(c) Wind
Answer:
(b) Time

Question 12.
The rivers of Tamil Nadu originates from :
(a) Eastern Ghats
(b) Western Ghats
(c) River plains
(d) Plateaus
Answer:
(b) Western Ghats

Question 13.
………….. is the most common trigger of a landslide.
(a) Air
(b) Fire
(c) Water
Answer:
(c) Water

Question 14.
………………. is also known as retreating monsoon.
(a) South west monsoon
(b) Tropical cyclones
(c) North east monsoon
(d) Pre-monsoon shower
Answer:
(c) North east monsoon

Question 15.
…………. dams are located at the foothills of Anaimalai.
(a) Aliyar and Tirumurthy
(b) Vaigai
(c) Mettur
Answer:
(a) Aliyar and Tirumurthy

II. Match the following.

a.

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

b.

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

III. Answer briefly:

Question 1.
Name the local bodies of Tamil Nadu:
Answer:
Tamil Nadu consists of:

• 35 – Districts
• 39 – Lok Sabha Constituencies
• 234 – Assembly Constituencies
• 76 – Revenue Division
• 125 – Municipalities
• 561 – Town Panchayats
• 12,618 – Village Panchayats

Question 2.
Name the passes that are found in the Western Ghats.
Answer:
The passes in the Western Ghats are Palghat, Shencottah, Aralvoimozhi and Achankoil.

Question 3.
Name the water bodies bordering Tamil Nadu.
Answer:

• Bay of Bengal on the east.
• Indian Ocean on the south.

Question 4.
Write a short note on Anaimalai.
Answer:
Anaimalai is located in the border of Tamil Nadu and Kerala. It is located in the South of Palghat gap. Valparai hill station and Kadamparai hydroelectric Power Plant are located on this hill. Anaimalai Tiger Reserve and Aliyar Reserve Forest is here. The dams Aliyar and Tirumurthy are located at the foothills of this range.

Question 5.
What are called Cholamandalam Plains?
Answer:
The coastal plains of Tiruvallur, Kanchipuram, Cuddalore and Villupuram are together known as the Cholamandalam Plains.

Question 6.
Name the districts that one located in the plains of Cauvery.
Answer:
The fertile plains of Cauvery is found in Salem, Erode, Karur, Tiruchirappalli, Pudukottai, Thanjavur, Thiruvarur and Nagapattinam.

Question 7.
Name the types of forests found in Tamil Nadu.
Answer:

1. Tropical Evergreen Forests
2. Montane Temperate Forests
3. Tropical Deciduous Forests
4. Mangrove Forests and
5. Tropical Thom Forests

Question 8.
Name the Wettest place in Tamil Nadu.
Answer:
Chinnakallar near Valparai is the Wettest place in Tamil Nadu. It is the third Wettest place in India.

Question 9.
What are the factors that determine the formation of soil?
Answer:

• Parent Rock
• Climate
• Relief
• Time factor
• Flora, fauna and micro organisms

Question 10.
Name of factors that are responsible for soil erosion.
Answer:
Soil erosion occurs due to deforestation, overgrazing, urbanisation and heavy rainfall.

IV. Distinguish Between:

Question 1.
Marina Beach and Rameshwaram Beach
Answer:

Question 2.
River Cauvery and River Vaigai
Answer:

Question 3.
Thorn forests and Mangrove forests
Answer:

V. Give Reasons:

Question 1.
Does Thanjavur has less than 1% of forest cover?
Answer:

• Thanjavur has alluvial soil
• It is suitable for agriculture
• So there is less then 1% of forest cover.

Question 2.
Why do Nilgiris and Kanyakumari get more rainfall during the southwest Monsoon season?
Answer:

1. The south west monsoon occurs between June and September.
2. Nilgiris, Kanyakumari, the western part of Coimbatore, Dharmapuri and Salem get benefitted during southwest monsoon season.
3. As the southwest monsoon starts its downpour of rain in the western ghats, the western parts of Tamil Nadu receive about 150 cm of rain fall, on an average.

Question 3.
East Coast of Tamil Nadu does not receive much rainfall during Southwest Monsoon
Answer:

• This occurs due to the southwesterly direction of Monsoon winds.
• Most of the eastern and central parts of Tamil Nadu become rain shadow regions for this season.

VI. Answer in a paragraph.

Question 1.
Describe the geographical location of TamilNadu.
Answer:

• Tamil Nadu, a state in Southern India, is bordered by the states of Kerala, Karnataka and Andra pradesh.
• Bay of Bengal on the east, Indian ocean on the South, Kerala and Karnataka on the West and Andra Pradesh on the North.
• Gulf of Mannar and Palk strait separate Tamil Nadu from the Island of Srilanka, which lies to the South east of India.
• It is the 11th largest state in India.
• The state extends from 8° 4’N to 13° 35’N latitudes and from 76° 18’Eto 80° 20’E longitudes.
• Its Eastern and Western extremities are defined by the point Calimere and the hills of Anaimalai respectively.
• The Northern extremity of the state is marked by Pulicat lake and the Southern most point is Cape Comorin.
• The state of Tamil Nadu is a triangular land mass at the south eastern end of the continent.
• It has 1,076 km long coast line, the second longest in India.

Question 2.
Write about the rainfall regions of Tamil Nadu.
Answer:
Rainfall regions of Tamil Nadu Distribution of rain fall.

Question 3.
Write a paragraph on the Nilgiris hills.
Answer:

1. The Nilgiri hills is located in the Northwestern part of Tamil Nadu.
2. It consists of 24 peaks with more than 2000 metres height.
3. Doddabetta is the highest peak (2,637 metres) followed by Mukkuruthi (2,554 metres).
4. Ooty and Coonoor are the major hill stations.
5. It has more than 2,700 species of flowering plants.
6. The state animal Nilgiri Tahr is found in this hill.
7. By extensive tea plantations and cattle grazing destroyed and disturbed much of the Nilgiris montane grasslands and shrublands.

We think the data given here clarify all your queries of Chapter 6 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science Geography Chapter 6 Physical Geography of Tamil Nadu Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.1

9th Maths Trigonometry Exercise 6.1 Question 1.
From the given figure find all the trigonometric ratios of angle B.
Solution:

9th maths trigonometry exercise 6.1 Question 2.
From the given figure, find the values of
(i) sin B
(ii) sec B
(iii) cot B
(iv) cos C
(v) tan C
(vi) cosec C
Solution:

trigonometry exercise 6.1 Question 3.
If 2 cos θ = \(\sqrt{3}\), then find all the trigonometric ratios of angle θ.
Solution:

Trigonometry solutions exercise 6.1 Question 4.
If cos A = \(\frac{3}{5}\), then find the value of \(\frac{\sin A-\cos A}{2 \tan A}\)
Solution:

trignometry solution of 9th class Question 5.
If cos A = \(\frac{2 x}{1+x^{2}}\) then find the values of sin A and tan A in terms of x.
Solution:

9th Maths Exercise 6.1 Samacheer Kalvi Question 6.

Solution:

9th Maths Exercise 6.1 Question 7.
If 3 cot A = 2, then find the value of \(\frac{4 \sin A-3 \cos A}{2 \sin A+3 \cos A}\)
Solution:

9th Std Maths Trigonometry Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8 \cos \theta-2 \sin \theta}{4 \cos \theta+2 \sin \theta}\)
Solution:

9th Maths Exercise 6.1 Solution Question 9.
From the given figure, prove that θ + ϕ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ϕ
Solution:


(∴ By Pythagoras theorem, in a right angled triangle square of hypotenuse is equal to sum of the squares of other two side)
And also in the figure, ∆ADC, ∆DBC are two other triangles.
As per the data given,
9²+ 12² = 81 + 144 = 225 = 15²
∴ ∆ ADC is a right angled triangle, then 12² + 16² = 144 + 256 = 400 = 20²
∴ ∆ DBC is also a right angled triangle

9th Maths Trigonometry Question 10.
A boy standing at point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios)
Solution:
In the figure,
∆OPM, ∆OQN are similar triangles. In similar triangles the sides are in the same proportional.

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 6 Trigonometry Ex 6.1

10th Maths Exercise 6.1 Samacheer Kalvi Question 1.
Prove the following identities.
(i) cot θ + tan θ = sec θ cosec θ
(ii) tan⁴θ + tan²θ = sec⁴θ – sec²θ
Solution:

10th Maths Trigonometry Exercise 6.1 Question 2.
Prove the following identities

Solution:

Exercise 6.1 Class 10 Samacheer Kalvi Question 3.
Prove the following identities

Solution:

10th Trigonometry Exercise 6.1 Question 4.
Prove the following identities
(i) sec⁶θ = tan⁶θ + 3tan²θ sec²θ + 1
(ii) (sinθ + secθ)² + (cosθ + cosecθ)²
= 1 + (secθ + cosecθ)²
(i) L.H.S = sec⁶θ = (sec²θ)³ = (1 + tan²θ )³ = (tan²θ + 1)³
(a + b)³ = a³ + 3a²b + 3ab² + b³
= (tan²θ)³ + 3(tan²θ)² × 1 + 3 × tan²θ × 1² + 1
= tan⁶θ + 3tan²θ × (sec²θ – 1) + 3tan²θ + 1
= tan⁶θ + 3tan²θsec²θ – 3tan²θ + 3tan²θ +1
= tan⁶θ + 3tan²θ sec²θ + 1 = R.H.S

(ii) L.H.S = (sinθ + secθ )² + (cosθ + cosecθ)²
= sin²θ + 2sinθ secθ + sec²θ + cos²θ + 2cosθ cosecθ+ cosec²θ

Ex 6.1 Class 10 Samacheer Question 5.
Prove the following identities

Solution:

10th Maths Exercise 6.1 Question 6.
Prove the following identities

Solution:

Trigonometry Exercise 6.1 Question 7.

Solution:

10th Maths 6.1 Question 8.

Solution:
(i) LHS:

10th Maths Exercise 6.1 In Tamil Question 9.
(i) If sinθ + cosθ = p and secθ + cosecθ = q then prove that q(p² – 1) = 2p
(ii) If sinθ(1 + sin²θ) = cos²θ, then prove that cos⁶θ – 4cos⁴θ + 8cos²θ = 4
Solution:

(ii) Given sinθ(1 + sin²θ) = cos²θ
Sustitute sin²θ = 1 – co²θ and take cos θ = c
squaring (1) on bothsides we get
sin²θ(1 + sin²θ)² = cos⁴θ
(1 – c²)(1 + 1 – c²) = c⁴
(1 – c²)(2 – c²)² = c⁴
(1 – c²)(4 + c⁴– 4c²) = c⁴
4 + c⁴– 4c² – 4c² – c⁶ + 4c⁴ = c⁴
 -c^{6 +} 4c⁴ – 8c² = -4
c⁶ – 4c⁴ + 8c² = -4
ie cos 6θ – 4cos 4θ + 8cos²θ = 4 = RHS
∴ Hence proved

10th Samacheer Kalvi Maths Trigonometry Question 10.

Solution:

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 8 Statistics and Probability Ex 8.4

10th Maths Exercise 8.4 Samacheer Kalvi Question 1.
If P(A) = \(\frac{2}{3}\), P(B) = \(\frac{2}{5}\), P(A∪B) = 13 then find P(A∩B).
Solution:
P(A) = \(\frac{2}{3}\), P(B) = \(\frac{2}{5}\), P(A∪B) = \(\frac{1}{3}\)
P(A ∩ B) = P(A) + P(B) – P(A∪B)

Ex 8.4 Class 10 Samacheer Question 2.
A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A ∩ B) = 0.16. Find
(i) P(not A)
(it) P(not B)
(iii) P(A or B)
Answer:
(i) P(not A) = 1 – P (A) = 1 – 0.42 = 0.58
(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) P(A or B) = P(A) + P(B) – P(A ∩ B) = 0.42 + 0.48 – 0.16 = 0.90 – 0.16 = 0.74

Exercise 8.4 Class 10 Samacheer Question 3.
If A and B are two mutually exclusive events of a random experiment and P(not A) = 0.45, P(A∪B) = 0.65, then find P(B).
Solution:
A and B are two mutually exclusive events of a random experiment.
P(not A) = 0.45,
P(A) = 1 – P(not A)
P(A∪B) = 0.65 = 1 – 0.45 = 0.55
P(A∪B) = P(A) + P(B) = 0.65
0.55 + P(B) = 0.65
P(B) = 0.65 – 0.55
= 0.10

10th Maths Exercise 8.4 Question 4.
The probability that atleast one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(\(\bar{A}\)) + P(\(\bar{B}\)).
Answer:
Here P (A ∪ B) = 0.6, P (A ∩ B) = 0.2
P (A ∪ B) = P (A) + P (B) – P (A ∩ B)
0.6 = P (A) + P (B) – 0.2
P(A) + P(B) = 0.8
P(\(\bar{A}\)) + P(\(\bar{B}\)) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)]
= 2 – 0.8
= 1.2

10th Maths 8.4 Solutions Question 5.
The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability that neither A nor B happen.
Solution:
P(A) = 0.5 Since A and B are mutually inclusive events
P(B) = 0.3 events.
P(\(\overline{\mathbf{A}}\))∪P(\(\overline{\mathbf{B}}\)) = 1 – [P(A) + P(B)]
= 1 – [0.5 + 0.3] = 0.2

10th Maths Probability Exercise 8.4 Question 6.
Two dice are rolled once. Find the probability of getting an even number on the first die or a total of face sum 8.
Solution:
Two dice rolled once.

10th Maths Ex 8.4 Question 7.
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of it being either a red king or a black queen.
Solution:
n(S) = 52
No. of Red cards = 26,
Red king cards = 2
No. of Black cards = 26,
Black queen cards = 2
No. of red king cards = n(K) = 2

∴ The probability of being either a red king or a black queen = \(\frac{1}{13}\).

10th Maths Exercise 8.4 Solution Question 8.
A box contains cards numbered 3, 5, 7, 9,… 35, 37. A card is drawn at random from the box. Find the probability that the drawn card has either multiples of 7 or a prime number.
Solution:
S = {3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37}
n(S) = 18
Multiplies of seven cards (A) = {7, 21, 35}
= n(A) = 3

Let the prime number cards B
B = {3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37}
n(B) = 11

10th Maths Exercise 8.4 In Tamil Question 9.
Three unbiased coins are tossed once. Find the probability of getting at most 2 tails or at least 2 heads.
Solution:
When we toss three coins, the sample space S = {HHH, TTT, HTT, THH, HHT, TTH, HTH, THT}
n(S) = 8
Event of getting at most 2 tails be A.
∴ A = { HHH, HTT, THH, HHT, TTH, HTH, THT}

10th Maths Statistics And Probability Question 10.
The probability that a person will get an electrification contract is the probability that he will not get plumbing contract is \(\frac{3}{5}\). The probability of getting at least one contract is \(\frac{5}{8}\). What is the probability that he will get both?
Solution:

10th Maths 8.4 Question 11.
In a town of 8000 people, 1300 are over 50 years and 3000 are females. It is known that 30% of the females are over 50 years. What is the probability that a chosen individual from the town is either a female or over 50 years ?
Solution:

10th Exercise 8.4 Question 12.
A coin is tossed thrice. Find the probability of getting exactly two heads or at least one tail or two consecutive heads.
Solution:
Three coins tossed simultaneously.
S = { HHH, TTT, HHT, TTH, HTH, THT, HTT, THH}
n(S) = 8
Happening of getting exactly two heads be A.
A= {HHT, HTH, THH}
n(A) = 3

Event of getting at least one tail be B.
∴ B = {TTT, HHT, TTH, HTH, THT, HTT, THH}

Samacheer Kalvi Guru 10th Maths Question 13.
If A, B, C are any three events such that probability of B is twice as that of probability of A and probability of C is thrice as that of probability of A and if P(A∩B) = \(\frac{1}{6}\), P(B∩C) = \(\frac{1}{4}\), P(A∩C) = \(\frac{1}{8}\), \(\mathbf{P}(\mathbf{A} \cup \mathbf{B} \cup \mathbf{C})=\frac{9}{10}, \mathbf{P}(\mathbf{A} \cap \mathbf{B} \cap \mathbf{C})=\frac{1}{15}\), then find P(A), P(B) and P(C)?
Solution:
P(B) = 2P(A)
P(C) = 3P(A)

10th Maths Chapter 8 Exercise 8.4 Question 14.
In a class of 35, students are numbered from 1 to 35. The ratio of boys to girls is 4:3. The roll numbers of students begin with boys and end with girls. Find the probability that a student selected is either a boy with a prime roll number or a girl with a composite roll number or an even roll number.
Solution:
n(S) = 35

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 5.5 தொகைநிலை, தொகாநிலைத் தொடர்கள் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 5.5 தொகைநிலை, தொகாநிலைத் தொடர்கள்

கற்பவை கற்றபின்

Question 1.
பாடப்பகுதியில் இடம்பெற்றுள்ள தொகைநிலைத் தொடர், தொகாநிலைத் தொடர்களைக் கண்டறிந்து தனித்தனியே தொகுக்க.
Answer:
தொகைநிலைத் தொடர்கள்

தொகா நிலைத்தொடர்கள்

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
சொற்களுக்கு இடையே வேற்றுமை உருபு மறைந்து வருவது ……………………….
அ) வேற்றுமைத் தொகை
ஆ) உம்மைத் தொகை
இ) உவமைத் தொகை
ஈ) அன்மொழித் தொகை
Answer:
அ) வேற்றுமைத் தொகை

Question 2.
‘செம்மரம்’ என்னும் சொல் …………………. த்தொகை.
அ) வினை
ஆ) பண்பு
இ) அன்மொழி
ஈ) உம்மை
Answer:
ஆ) பண்பு

Question 3.
‘கண்ணா வா!’ – என்பது ……………….. த் தொடர்.
அ) எழுவாய்
ஆ) விளி
இ) வினைமுற்று
ஈ) வேற்றுமை
Answer:
ஆ) விளி

பொருத்துக

1. பெயரெச்சத் தொடர் – அ) கார்குழலி படித்தாள்.
2. வினையெச்சத் தொடர் – ஆ) புலவரே வருக.
3. வினைமுற்றுத் தொடர் – இ) பாடி முடித்தான்.
4. எழுவாய்த் தொடர் – ஈ) எழுதிய பாடல்.
5. விளித் தொடர் – உ) வென்றான் சோழன்.
Answer:
1. ஈ
2. இ
3. உ
4. அ
5. ஆ

சிறுவினா

Question 1.
தொகைநிலைத் தொடர்கள் எத்தனை வகைப்படும்? அவை யாவை?
Answer:
தொகைநிலைத் தொடர்கள் ஆறு வகைப்படும். அவையாவன:

• வேற்றுமைத்தொகை
• உவமைத்தொகை
• வினைத்தொகை
• உம்மைத்தொகை
• பண்புத்தொகை
• அன்மொழித்தொகை

Question 2.
இரவு பகல் என்பது எவ்வகைத் தொடர் என விளக்குக.
Answer:

• ‘இரவு பகல்’ இத்தொடர், ‘இரவும் பகலும்’ என விரிந்து பொருள் தருகின்றது.
• இதில் சொல்லின் இடையிலும் இறுதியிலும் `உம்’ என்னும் இடைச் சொல் மறைந்து நின்று பொருள் தருகிறது.
• இவ்வாறு சொற்களுக்கு இடையிலும் இறுதியிலும் ‘உம்’ என்னும் இடைச் சொல் மறைந்து நின்று பொருள் தருவதை உம்மைத்தொகை என்பர்.

Question 3.
அன்மொழித்தொகையை எடுத்துக்காட்டுடன் விளக்குக.
Answer:
வேற்றுமை, வினை, பண்பு, உவமை, உம்மை ஆகிய தொகைநிலைத் தொடர்களில் அவை அல்லாத வேறு பிற சொற்களும் மறைந்து வருவது அன்மொழித்தொகை எனப்படும்.

சான்று : பொற்றொடி வந்தாள்.

இத்தொடர் ‘பொன்னாலாகிய வளையலை அணிந்த பெண் வந்தாள்’ என்னும் பொருள் தருகிறது. இதில் ‘ஆல்’ என்னும் மூன்றாம் வேற்றுமை உருபும் ‘ஆகிய என்னும் அதன் பயனும் மறைந்து வந்துள்ளது. ‘வந்தாள்’ என்னும் சொல்லால் பெண் என்பதையும் குறிப்பதால், இது மூன்றாம் வேற்றுமைப் புறத்துப் பிறந்த அன்மொழித் தொகை ஆகும்.

மொழியை ஆள்வோம்

கீழ்க்காணும் தலைப்பில் இரண்டு நிமிடம் பேசுக.

1. கைத்தொழில் ஒன்றைக் கற்றுக்கொள்

பெருமை மிகுந்த சான்றோர் சபைக்கு என் முதற்கண் வணக்கத்தைத் தெரிவித்துக் கொள்கின்றேன். இப்பெருமைமிகு சபையில் நான் பேச எடுத்துக் கொண்ட தலைப்பு கைத்தொழில் ஒன்றைக் கற்றுக்கொள் என்பதாகும்.

ஏட்டுச் சுரைக்காய் கறிக்கு உதவாது’ என்பார்கள். அதைப்போல ஏட்டுப் படிப்பு படித்தவர்களுக்கு எந்த வேலையும் கிடைப்பதாகத் தெரியவில்லை.

படித்தாலும், படித்துப் பட்டம் பெற்றாலும் கைத்தொழில் ஒன்றையும் நாம் கூடவே, சேர்த்துக் கற்றுக்கொள்ள வேண்டும். படிப்புக்கேற்ற வேலை கிடைக்கும் வரை அந்த வேலைக்காகக் காத்திராமல் கற்ற கைத்தொழில் நமக்கு மிகவும் பயன்படும்.

தையல், ஓவியம், மரவேலை, மின்னணுச் சாதனங்கள் பழுதுபார்ப்பு, தட்டச்சு, கணிப்பொறி, கூடை பின்னுதல், அலங்காரப் பொருட்கள் செய்தல் இவற்றைப் பொழுதுபோக்கிற்காக நாம் பள்ளியில் கற்றாலும், அங்கு ஆழமாக ஆழ்ந்து கற்க வேண்டும். அதுவேதான், பிற்காலத்தில் படிப்புக்கு ஏற்ற வேலை கிடைக்கும் வரையில் நமக்கு நல்ல சம்பாத்தியத்தைக் கொடுக்கும்.

ஏட்டுக்கல்வி கைவிட்டாலும், கைத்தொழில் கல்வி உன்னைக் கைவிடாது என்பதைப் புரிந்து கொள்ள வேண்டும். எனவே, இளைஞர்களாகிய நாம் கைத்தொழில் ஒன்றைக் கற்றுக் கொள்ள வேண்டும் என்று சொல்லி என்னுடைய உரையை நிறைவு செய்கின்றேன்.

2. இதயம் கவரும் இசை

என்னை ஈன்ற தாய் மொழிக்கும், இந்தச் சான்றோர் பேரவைக்கும் முதற்கண் வணக்கத்தைத் தெரிவித்துக்கொண்டு, இதயம் கவரும் இசை என்ற நல்லதொரு தலைப்பில் சில நிமிடங்கள் பேசுகின்றேன். ஒவ்வொருநாளும் ஒவ்வொரு மனிதனுக்கும் ஒவ்வொரு பிரச்சினைகள் இருக்கத்தான் செய்கின்றன. துன்பங்கள் நம்மைத் துரத்தும் போது மன அமைதி தானாக தேடி வருவதில்லை . இசையின் பக்கம் நாம் தான் ஓடி வர – வேண்டும். புல்லாங்குழல் இசையும், வீணை இசையும், நாத முழக்கமும், மத்தளம் இசையும் மனதைப் பண்படுத்தும். இசைக்கச்சேரி கேட்கும் போது இதயமெல்லாம் மென்மையாகிவிடும்.

சங்க காலத்தில் தலைவன் ஒருவன் கள் உண்ட மயக்கத்தில் படுத்து கிடக்கின்றான். தொலைவில் தலைவிதினையைக் காயவைத்துக் கொண்டிருக்கின்றாள். தலைவன் படுத்து இருந்த இடத்தை நோக்கி மத யானை ஒன்று ஓடி வருகின்றது .ஐயோ! தலைவனுக்கு என்ன ஆகுமோ? என்று கவலைப்படாமல், தலைவி அருகிலிருந்த யாழை எடுத்து மீட்டினாள். மதம் பிடித்த யானை யாழ் இசையில் மயங்கி தலைவனை மிதிக்காமல் தெளிந்து சென்றதாம். இசை உயிரையும் காப்பாற்றும்.

குழந்தை பிறந்ததும் தாலாட்டி இசை செய்தான், அவன் வளர்ந்து திருமணம் ஆகும் போதும் மங்கள இசைதான். இப்படி இசை வாழ்வில் எத்தனையோ இடத்திலும் இடம் பெற்றிருக்கிறது. நம் வாழ்க்கையில் ‘இதயம் கவரும் இசை அனைவரையும் கவரும் தசை’ என்று சொல்லி என் உரையை நிறைவு செய்கிறேன்.

நன்றி! வணக்கம்!!

சொல்லக் கேட்டு எழுதுக.

முல்லை நில மக்களாகிய ஆயர்கள் குழல் ஊதுவதில் வல்லவர்கள். இதனைச் சம்பந்தர் திருப்பதிகத்தில் அமைந்த நிகழ்ச்சி ஒன்று விளக்குகிறது. திருவண்ணாமலைச் சாரலில் ஆயர் ஒருவர் ஆநிரைகளையும் எருமையினங்களையும் மேய்த்துக் கொண்டிருந்தார். மாலையில் அவற்றை எல்லாம் ஒன்று திரட்டினார்.

அப்போது எருமை ஒன்று காணாமல் போனதை அறிந்தார். தன் கையிலிருந்த குழலை எடுத்து இனிய இசையை எழுப்பினார். இன்னிசை கேட்ட எருமை அவரை வந்தடைந்தது. இவ்வாறு ஆயர்களின் இசைத் திறத்தைத் திருப்பதிகம் விளக்குகிறது.

கோடிட்ட இடங்களில் பொருத்தமான சொல்லுருபுகளை இட்டு நிரப்புக.

(கொண்டு, இருந்து, உடைய, காட்டிலும், ஆக, நின்று, உடன், விட, பொருட்டு)

1. இடி ………………….. மழை வந்தது.
2. மலர்விழி தேர்வின் …………………. ஆயத்தமானாள்.
3. அருவி மலையில் …………………. வீழ்ந்தது.
4. தமிழைக் ……………….. சுவையான மொழியுண்டோ !
5. யாழ், தமிழர் …………………….. இசைக் கருவிகளுள் ஒன்று.
Answer:
1. உடன்
2. பொருட்டு
3. இருந்து
4. காட்டிலும்
5. உடைய

பின்வரும் இசைக்கருவிகளின் பெயர்களை அகர வரிசைப்படுத்துக.

படகம், தவில், கணப்பறை, பேரியாழ், உறுமி, உடுக்கை, தவண்டை , பிடில், நாகசுவரம், மகுடி.
Answer:
உடுக்கை, உறுமி, கணப்பறை, தவண்டை , தவில், நாகசுவரம், படகம், பிடில், பேரியாழ், மகுடி.

பின்வரும் இணைச்சொற்களை வகைப்படுத்துக.

உற்றார் உறவினர், விருப்பு வெறுப்பு, காலைமாலை, கன்னங்கரேல், ஆடல்பாடல், வாடிவதங்கி, பட்டிதொட்டி, உள்ளும் புறமும், மேடுபள்ளம், நட்ட நடுவில்.

சரியான இணைச்சொற்களை இட்டு நிரப்புக.

(மேடுபள்ளம், ஈடுஇணை, கல்விகேள்வி, போற்றிப்புகழப்பட, வாழ்வு தாழ்வு, ஆடிஅசைந்து)

1. சான்றோர் எனப்படுபவர் …………………………. களில் சிறந்தவர் ஆவார்.
2. ஆற்று வெள்ளம் …………………… பாராமல் ஓடியது.
3. இசைக்கலைஞர்கள் …………………. வேண்டியவர்கள்.
4. தமிழ் இலக்கியங்களின் பெருமைக்கு …………………….. இல்லை.
5. திருவிழாவில் யானை ………………. வந்தது.
Answer:
1. கல்விகேள்வி
2. மேடுபள்ளம்
3. போற்றிப்புகழப்பட
4. ஈடுஇணை
5. ஆடி அசைந்து

கடிதம் எழுதுக

இருப்பிடச் சான்று வேண்டி வட்டாட்சியருக்கு விண்ணப்பம் எழுதுக.

அனுப்புநர்
சா. சுந்தர்,
த/பெ. ஆ. சங்கர்
34, குறிஞ்சி நகர்,
ஈரோடு – 638 001.

பெறுநர்
உயர்திரு. வட்டாட்சியர் அவர்கள்,
வட்டாட்சியர் அலுவலகம்,
ஈரோடு.

மதிப்புக்குரிய அய்யா,

பொருள் : இருப்பிடச் சான்றிதழ் வேண்டுதல் சார்பாக. வணக்கம்.

ஈரோடு, அரசு உயர்நிலைப் பள்ளியில் எட்டாம் வகுப்பு படிக்கின்றேன். 34, குறிஞ்சி நகர், ஈரோடு – 638 001 என்ற முகவரியில் பத்து ஆண்டுகளாக நாங்கள் வசித்து வருகின்றோம். அரசின் கல்வி உதவித்தொகைக்கு விண்ணப்பிக்க இருப்பிடச் சான்றிதழ் தேவைப்படுகின்றது. இத்துடன் குடும்ப அட்டை நகலும் ஆதார் அட்டை நகலும் இணைத்துள்ளேன். ஆகவே, எனக்கு இருப்பிடச் சான்றிதழ் வழங்கும்படி பணிவுடன் கேட்டுக் கொள்கிறேன்.

நன்றி!

இடம் : ஈரோடு
நாள் : 25.06.2020

இப்படிக்கு,
தங்கள் உண்மையுள்ள,
சா. சுந்தர்.

உறைமேல் முகவரி:
பெறுநர்
உயர்திரு. வட்டாட்சியர் அவர்கள்,
வட்டாட்சியர் அலுவலகம்,
ஈரோடு.

மொழியோடு விளையாடு

குறுக்கெழுத்துப்புதிர்.

இடமிருந்து வலம் :
1. முதற்கருவி எனப் பெயர் பெற்றது.
2. யாழிலிருந்து உருவான பிற்காலக் கருவி
7. இயற்கைக் கருவி
12. விலங்கின் உறுப்பைப் பெயராகக் கொண்ட கருவி

வலமிருந்து இடம் :
4. வட்டமான மணி போன்ற கருவி
8. ஐந்து வாய்களைக் கொண்ட கருவி
9. இசைக்கருவிகளை இசைத்துப் பாடல் பாடுவோர்

மேலிருந்து கீழ் :
1. 19 நரம்புகளைக் கொண்ட யாழ்
3. ஒன்றோடு ஒன்று மோதி இசைக்கப்படுபவை
5. சிறிய வகை உடுக்கை
6. பறை ஒரு ……………… கருவி

கீழிருந்து மேல் :
8. மூங்கிலால் செய்யப்படும் காற்றுக்கருவி
10. வீணையில் உள்ள நரம்புகளின் எண்ணிக்கை
11. திருமணத்தின்போது கொட்டும் முரசு

நிற்க அதற்குத் தக

என் பொறுப்புகள்:

1. கைவினைக்கலைகளுள் ஒன்றைக் கற்றுக்கொள்வேன்.
2. இசைக் கலையை வளர்த்த சான்றோர்களைப் பற்றி அறிந்து போற்றுவேன்.

கலைச்சொல் அறிவோம்

1. கைவினைப் பொருள்கள் – Crafts
2. புல்லாங்குழல் – Flute
3. முரசு – Drum .
4. கூடைமுடைதல் – Basketry
5. பின்னுதல் – Knitting
6. கொம்பு – Horn
7. கைவினைஞர் – Artisan
8. சடங்கு – Rite

இணையத்தில் காண்க

இசையின் வகைப்பாடுகள் பற்றிய செய்திகளை இணையத்தில் தேடி எழுதுக.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக..

Question 1.
திருவாசகம் படித்தாள் – இதில் மறைந்து வரும் வேற்றுமை உருபு …………………..
அ) இரண்டாம் வேற்றுமை உருபு
ஆ) மூன்றாம் வேற்றுமை உருபு
இ) நான்காம் வேற்றுமை உருபு
ஈ) ஐந்தாம் வேற்றுமை உருபு
Answer:
அ) இரண்டாம் வேற்றுமை உருபு

Question 2.
கம்பர் பாடல் – இதில் இடம்பெற்றுள்ள வேற்றுமையுருபு ……………………
அ) கு
ஆ) இன்
இ) அது
ஈ) கண்
Answer:
இ) அது

Question 3.
காலம் கரந்த பெயரெச்சம் ………………..
அ) வினைத்தொகை
ஆ) பண்புத்தொகை
இ) வேற்றுமைத்தொகை
ஈ) உவமைத்தொகை
Answer:
அ) வினைத்தொகை

Question 4.
ஆடுகொடி, வளர்தமிழ் – ஆகியன …………………. க்குச் சான்றுகள்.
அ) வேற்றுமைத்தொகை
ஆ) உம்மைத்தொகை
இ) உவமைத்தொகை
ஈ) வினைத்தொகை
Answer:
ஈ) வினைத்தொகை

Question 5.
வெண்ணிலவு, கருங்குவளை ஆகியன ………………….. க்குச் சான்றுகளாகும்.
அ) வினைத்தொகை
ஆ) பண்புத்தொகை
இ) உவமைத்தொகை
ஈ) உம்மைத்தொகை
Answer:
ஆ) பண்புத்தொகை

Question 6.
இருபெயரொட்டு பண்புத்தொகைக்குச் சான்றாக அமையும் ஒன்றினைத் தேர்வு செய்க.
அ) மலர்விழி
ஆ) இரவுபகல்
இ) பொற்றொடி வந்தாள்
ஈ) பனைமரம்
Answer:
ஈ) பனைமரம்

Question 7.
இரவுபகல், தாய்தந்தை ஆகியன ………………………. க்குச் சான்றாகும்.
அ) அன்மொழித்தொகை
ஆ) உவமைத்தொகை
இ) உம்மைத்தொகை
ஈ) இருபெயரொட்டுப் பண்புத்தொகை
Answer:
இ) உம்மைத்தொகை

Question 8.
தொகாநிலைத் தொடர் வகைகள்
அ) 6
ஆ) 8
இ) 9
ஈ) 3
Answer:
இ) 9

Question 9.
எழுவாய்த் தொடர் அமையும் சான்றினைத் தேர்ந்தெடுக்க.
அ) மல்லிகை மலர்ந்தது
ஆ) நண்பா படி
இ) சென்றனர் வீரர்
ஈ) வரைந்த ஓவியம்
Answer:
அ) மல்லிகை மலர்ந்தது

Question 10.
‘நண்பா படி’ என்பது ……………………
அ) எழுவாய்த் தொடர்
ஆ) விளித் தொடர்
இ) வினைமுற்றுத்தொடர்
ஈ) பெயரெச்சத் தொடர்
Answer:
ஆ) விளித்தொடர்

குறுவினா

Question 1.
வேற்றுமைத் தொகை என்றால் என்ன? சான்று தருக.
Answer:

•  இருசொற்களுக்கு இடையில் வேற்றுமை உருபு மறைந்துவந்து பொருள் தந்தால், அதனை வேற்றுமைத் தொகை என்பர். 
• சான்று: திருவாசகம் (ஐ)படித்தான். (இரண்டாம் வேற்றுமைத் தொகை)

Question 2.
உருபும் பயனும் உடன்தொக்கத் தொகை என்றால் என்ன? சான்று தருக.
Answer:

• ஒரு தொடரில் வேற்றுமை உருபும் அதன் பொருளை விளக்கும் சொல்லும் மறைந்து வருவது உருபும் பயனும் உடன் தொக்கத் தொகை எனப்படும்.
• சான்று: பால் குடம்.
• பாலைக் கொண்ட குடம்’ இரண்டாம் வேற்றுமை உருபும் பயனும் உடன் தொக்கத் தொகை.

Question 3.
வினைத்தொகையை சான்றுடன் விளக்குக.
Answer:

• காலம் காட்டும் இடைநிலையும், பெயரெச்சவிகுதியும் மறைந்து வரும் பெயரெச்சம் வினைத்தொகை என்பர்.
• சான்று: வளர்தமிழ். இதில் காலம் காட்டும் இடைநிலைகள் தொக்கி வந்துள்ளன.
• வளர்ந்த தமிழ், வளர்கின்ற தமிழ், வளரும் தமிழ் எனவும் முக்காலத்திற்கும் பொருந்தும் படியாகப் பொருள் தருகின்றன. எனவே, இது வினைத்தொகை ஆகும்.

Question 4.
பண்புத்தொகை என்றால் என்ன? சான்று தருக.
Answer:

• பண்புப்பெயருக்கும் அது தழுவி நிற்கும் பெயர்ச்சொல்லுக்கும் இடையே ‘ஆன்’, ‘ஆகிய’ என்னும் பண்பு உருபுகள் மறைந்து வருவது பண்புத்தொகை எனப்படும்.
• சான்று: வெண்ணிலவு, கருங்குவளை.

Question 5.
‘மலர்விழி’ என்னும் சான்று அமையும் தொகை யாது? விளக்கு.
Answer:
‘மலர்விழி’ என்பது உவமைத்தொகை. மலர் போன்ற விழி என்ற பொருளைத் தருகிறது. ‘மலர்’ என்பது உவமை. ‘விழி’ என்பது உவமேயம். ‘போன்ற’ என்பது உவம உருபு. உவமைக்கும் உவமேயத்துக்கும் இடையில் உவம உருபு மறைந்து வந்தால் அது உவமைத்தொகை எனப்படும்.

Question 6.
எண்ணும்மை என்றால் என்ன? சான்று தருக.
Answer:
ஒன்றுக்கு மேற்பட்ட சொற்களில் ‘உம்’ என்னும் உருபு வெளிப்பட வருவது எண்ணும்மை ஆகும். சான்று: இரவும் பகலும், பசுவும் கன்றும்

சிறு வினா

Question 1.
தொகாநிலைத் தொடர் என்றால் என்ன? அதன் வகைகளை எழுதுக.
Answer:
(i) ஒரு தொடரில் இரு சொற்கள் வந்து அவற்றின் இடையில் சொல்லுருபு மறையாமல் நின்று பொருள் தந்தால் அதனை தொகாநிலைத்தொடர் என்பர்.

(ii) தொகாநிலைத்தொடர் ஒன்பது வகைப்படும். எழுவாய்த்தொடர், விளித்தொடர், வினைமுற்றுத் தொடர், பெயரெச்சத் தொடர், வினையெச்சத் தொடர், வேற்றுமைத் தொகாநிலைத் தொடர், இடைச்சொல் தொடர், உரிச்சொல் தொடர், அடுக்குத்தொடர்.

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Tamilnadu Samacheer Kalvi 11th Physics Solutions Chapter 3 Laws of Motion

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Samacheer Kalvi 11th Physics Chapter 3 Laws of Motion Textual Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

11th Physics Chapter 3 Book Back Answers Question 1.
When a car takes a sudden left turn in the curved road, passengers are pushed towards the right due to
(a) inertia of direction
(b) inertia of motion
(c) inertia of rest
(d) absence of inertia
Answer:
(a) inertia of direction

Samacheer Kalvi 11th Physics Solution Chapter 3 Question 2.
An object of mass m held against a vertical wall by applying horizontal force F as shown in the figure. The minimum value of the force F is [IIT JEE 1994]
(a) Less than mg
(b) Equal to mg
(c) Greater than mg
(d) Cannot determine
Answer:
(c) Greater than mg

11th Physics Lesson 3 Book Back Answers Question 3.
A vehicle is moving along the positive x direction, if sudden brake is applied, then
(a) frictional force acting on the vehicle is along negative x direction
(b) frictional force acting on the vehicle is along positive x direction
(c) no frictional force acts on the vehicle
(d) frictional force acts in downward direction
Answer:
(a) frictional force acting on the vehicle is along negative x direction

11th Physics 3rd Chapter Book Back Answers Question 4.
A book is at rest on the table which exerts a normal force on the book. If this force is considered as reaction force, what is the action force according to Newton’s third law?
(a) Gravitational force exerted by Earth on the book
(b) Gravitational force exerted by the book on Earth
(c) Normal force exerted by the book on the table
(d) None of the above
Answer:
(c) Normal force exerted by the book on the table

Laws Of Motion Class 11 State Board Question 5.
Two masses m₁ and m₂ are experiencing the same force where m₁ < m₂ The ratio of their acceleration \(\frac{a_{1}}{a_{2}}\) is –
(a) 1
(b) less than 1
(c) greater than 1
(d) all the three cases
Answer:
(c) greater than 1

Samacheer Kalvi Guru 11th Physics Question 6.
Choose appropriate free body diagram for the particle experiencing net acceleration along negative y direction. (Each arrow mark represents the force acting on the system).

Answer:

Class 11 Physics Solutions Samacheer Kalvi Question 7.
A particle of mass m sliding on the smooth double inclined plane (shown in figure) will experience –
(a) greater acceleration along the path AB
(b) greater acceleration along the path AC
(c) same acceleration in both the paths
(d) no acceleration in both the paths

Answer:
(a) greater acceleration along the path AC

Samacheer Kalvi 11th Physics Question 8.
Two blocks of masses m and 2m are placed on a smooth horizontal surface as shown. In the first case only a force F₁  is applied from the left. Later only a force F₂  is applied from the right. If the force acting at the interface of the two blocks in the two cases is same, then F₁ : F₂ is [Physics Olympiad 2016]

(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1

11th Physics 3rd Lesson Book Back Answers Question 9.
Force acting on the particle moving with constant speed is –
(a) always zero
(b) need not be zero
(c) always non zero
(d) cannot be concluded
Answer:
(b) need not be zero

11th Physics 3rd Chapter Exercise Question 10.
An object of mass m begins to move on the plane inclined at an angle 0. The coefficient of static friction of inclined surface is lay. The maximum static friction experienced by the mass is –
(a) mg
(b) µ_s mg
(c) µ_s mg sin θ
(d) µ_s mg cos θ
Answer:
(d) µ_s mg cos θ

Class 11 Samacheer Physics Solutions Question 11.
When the object is moving at constant velocity on the rough surface –
(a) net force on the object is zero
(b) no force acts on the object
(c) only external force acts on the object
(d) only kinetic friction acts on the object
Answer:
(a) net force on the object is zero

Samacheer Kalvi 11th Physics Book Back Answers Question 12.
When an object is at rest on the inclined rough surface –
(a) static and kinetic frictions acting on the object is zero
(b) static friction is zero but kinetic friction is not zero
(c) static friction is not zero and kinetic friction is zero
(d) static and kinetic frictions are not zero
Answer:
(c) static friction is not zero and kinetic friction is zero

Class 11 Physics Samacheer Kalvi Question 13.
The centrifugal force appears to exist –
(a) only in inertial frames
(b) only in rotating frames
(c) in any accelerated frame
(d) both in inertial and non-inertial frames
Answer:
(b) only in rotating frames

11th Physics Unit 3 Book Back Answers Question 14.
Choose the correct statement from the following –
(a) Centrifugal and centripetal forces are action reaction pairs
(b) Centripetal forces is a natural force
(c) Centrifugal force arises from gravitational force
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion
Answer:
(d) Centripetal force acts towards the center and centrifugal force appears to act away from the center in a circular motion

Laws Of Motion Class 11 Numericals With Solutions Pdf Question 15.
If a person moving from pole to equator, the centrifugal force acting on him –
(a) increases
(b) decreases
(c) remains the same
(d) increases and then decreases
Answer:
(a) increases

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions

Question 1.
Explain the concept of inertia. Write two examples each for inertia of motion, inertia of rest and inertia of direction.
Answer:
The inability of objects to move on its own or change its state of motion is called inertia. Inertia means resistance to change its state. There are three types of inertia:
1. Inertia of rest:
The inability of an object to change its state of rest is called inertia of rest.
Example:

• When a stationary bus starts to move, the passengers experience a sudden backward push.
• A book lying on the table will remain at rest until it is moved by some external agencies.

2. Inertia of motion:
The inability of an object to change its state of uniform speed (constant speed) on its own is called inertia of motion.
Example:

• When the bus is in motion, and if the brake is applied suddenly, passengers move forward and hit against the front seat.
• An athlete running is a race will continue to run even after reaching the finishing point.

3. Inertia of direction:
The inability of an object to change its direction of motion on its own is called inertia of direction.
Example:

• When a stone attached to a string is in whirling motion, and if the string is cut suddenly, the stone will not continue to move in circular motion but moves tangential to the circle.
• When a bus moving along a straight line takes a turn to the right. The passengers are thrown towards left.

Question 2.
State Newton’s second law.
Answer:
The force acting on an object is equal to the rate of change of its momentum –
\(\overline{\mathrm{F}}\) = \(\frac{d \bar{p}}{d t}\)

Question 3.
Define one newton.
Answer:
One newton is defined as the force which acts on 1 kg of mass to give an acceleration 1 ms^-2 in the direction of the force.

Question 4.
Show that impulse is the change of momentum.
Answer:
According to Newton’s Second Law
F = \(\frac {dp}{dt}\) i.e. dp = Fdt
Integrate it over a time interval from t_i  to t_f

P_i → initial momentum of the object at t_i
P_f → Final momentum of the object at t_f
P_f – P_i = ∆p = change in momentum during the time interval ∆t.
\(\int_{t_{i}}^{t_{f}} \mathrm{F} \cdot d t=\mathrm{J}\) is called the impulse.
If the force is constant over the time interval ∆t, then

Hence the proof.

Question 5.
Using free body diagram, show that it is easy to pull an object than to push it.
Answer:
When a body is pushed at an arbitrary angle θ [0 to \(\frac {π}{2}\)], the applied force F can be resolved into two components as F sin 0 parallel to the surface and F cos 0 perpendicular to the surface as shown in figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to
N_push = mg + F cos θ …………(1)
As a result the maximal static friction also increases and is equal to
\(f_{S}^{\max }\) = \(\mu_{r} \mathrm{N}_{\mathrm{push}}\) = µ_s(mg + F cos θ) ……(2)
Equation (2) shows that a greater force needs to be applied to push the object into motion.

When an object is pulled at an angle θ, the applied force is resolved into two components as shown in figure. The total downward force acting on the object is –
N_pull = mg – F cos θ ………….(3)

Equation (3) shows that the normal force is less than – N_push. From equations (1) and (3), it is easier to pull an object than to push to make it move.

Question 6.
Explain various types of friction. Suggest a few methods to reduce friction.
Answer:
There are two types of Friction:
(1) Static Friction:
Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force f_s lies between
\(0 \leq f_{s} \leq \mu_{s} \mathrm{N}\)
where, µ_s – coefficient of static friction
N – Normal force

(2) Kinetic friction:
The frictional force exerted by the surface when an object slides is called as kinetic friction. Also called as sliding friction or dynamic friction,
f_k – µ_kN
where µ_k – the coefficient of kinetic friction
N – Normal force exerted by the surface on the object

Methods to reduce friction:
Friction can be reduced

• By using lubricants
• By using Ball bearings
• By polishing
• By streamlining

Question 7.
What is the meaning by ‘pseudo force’?
Answer:
Pseudo force is an apparent force which has no origin. It arises due to the non-inertial nature of the frame considered.

Question 8.
State the empirical laws of static and kinetic friction.
Answer:
The empirical laws of friction are:

• Friction is independent of surface of contact.
• Coefficient of kinetic friction is less than coefficient of static friction.
• The direction of frictional force is always opposite to the motion of one body over the other.
• Frictional force always acts on the object parallel to the surface on which the objet is placed,
• The magnitude of frictional force between any two bodies in contact is directly proportional to the normal reaction between them.

Question 9.
State Newton’s third law.
Answer:
Newton’s third law states that for every action there is an equal and opposite reaction.

Question 10.
What are inertial frames?
Answer:
Inertial frame is the one in which if there is no force on the object, the object moves at constant velocity.

Question 11.
Under what condition will a car skid on a leveled circular road?
Answer:
On a leveled circular road, if the static friction is not able to provide enough centripetal ’force to turn, the vehicle will start to skid
\(\mu_{s}<\frac{v^{2}}{r g}\)

Samacheer Kalvi 11th Physics Laws of Motion Long Answer Questions

Question 1.
Prove the law of conservation of linear momentum. Use it to find the recoil velocity of a gun when a bullet is fired from it.
Answer:
In nature, conservation laws play a very important role. The dynamics of motion of bodies can be analysed very effectively using conservation laws. There are three conservation laws in mechanics. Conservation of total energy, conservation of total linear momentum, and conservation of angular momentum. By combining Newton’s second and third laws, we can derive the law of conservation of total linear momentum. When two particles interact with each other, they exert equal and opposite forces on each other.

The particle 1 exerts force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 2 and particle 2 exerts an exactly equal and opposite force \(\overrightarrow{\mathrm{F}}_{12}\) on particle 1 according to Newton’s third law.
\(\overrightarrow{\mathrm{F}}_{12}\) = –\(\overrightarrow{\mathrm{F}}_{12}\) ……..(1)
In terms of momentum of particles, the force on each particle (Newton’s second law) can be written as –
\(\overrightarrow{\mathrm{F}}_{12}\) = \(\frac{d \vec{p}_{1}}{d t}\) and \(\overrightarrow{\mathrm{F}}_{21}\) = \(\frac{d \vec{p}_{2}}{d t}\) ………(2)
Here \(\vec{p}_{1}\) is the momentum of particle 1 which changes due to the force \(\overrightarrow{\mathrm{F}}_{12}\) exerted by particle 2. Further Po is the momentum of particle \(\vec{p}_{2}\) This changes due to \(\overrightarrow{\mathrm{F}}_{21}\) exerted by particle 1.
Substitute equation (2) in equation (1)
\(\frac{d \vec{p}_{1}}{d t}\) = – \(\frac{d \vec{p}_{2}}{d t}\) …………(3)
\(\frac{d \vec{p}_{1}}{d t}\) + \(\frac{d \vec{p}_{2}}{d t}\) = 0 ………(4)
\(\frac {d}{dt}\)(\(\vec{p}_{1}\) + \(\vec{p}_{2}\)) = 0
It implies that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = constant vector (always).
\(\vec{p}_{1}\) + \(\vec{p}_{2}\) is the total linear momentum of the two particles (\(\vec{p}_{tot}\) = \(\vec{p}_{1}\) + \(\vec{p}_{2}\)).It is also called as total linear momentum of the system. Here, the two particles constitute the system. From this result, the law of conservation of linear momentum can be stated as follows.

If there are no external forces acting on the system, then the total linear momentum of the system (\(\vec{p}_{tot}\)) is always a constant vector. In other words, the total linear momentum of the system is conserved in time. Here the word ‘conserve’ means that \(\vec{p}_{1}\) and \(\vec{p}_{2}\) can vary, in such a way that \(\vec{p}_{1}\) + \(\vec{p}_{2}\) is a constant vector.

The forces \(\overrightarrow{\mathrm{F}}_{12}\) and \(\overrightarrow{\mathrm{F}}_{21}\) are called the internal forces of the system, because they act only between the two particles. There is no external force acting on the two particles from outside. In such a case the total linear momentum of the system is a constant vector or is conserved.

Meaning of law of conservation of momentum:
1. The Law of conservation of linear momentum is a vector law. It implies that both the magnitude and direction of total linear momentum are constant. In some cases, this total momentum can also be zero.

2. To analyse the motion of a particle, we can either use Newton’s second law or the law of conservation of linear momentum. Newton’s second law requires us to specify the forces involved in the process. This is difficult to specify in real situations. But conservation of linear momentum does not require any force involved in the process. It is convenient and hence important.

For example, when two particles collide, the forces exerted by these two particles on each other is difficult to specify. But it is easier to apply conservation of linear momentum during the collision process.

Examples:
Consider the firing of a gun. Here the system is Gun+bullet. Initially the gun and bullet are at rest, hence the total linear momentum of the system is zero. Let \(\vec{p}_{1}\) be the momentum of the bullet and \(\vec{p}_{2}\) the momentum of the gun before firing. Since initially both are at rest,

Total momentum before firing the gun is zero, \(\vec{p}_{1}\) + \(\vec{p}_{2}\) = 0.
According to the law of conservation of linear momentum, total linear momentum has to be zero after the firing also.

When the gun is fired, a force is exerted by the gun on the bullet in forward direction. Now the momentum of the bullet changes from \(\vec{p}_{1}\) to \(\vec{p}_{1}\) To conserve the total linear momentum of the system, the momentum of the gun must also change from \(\vec{p}_{2}\) to \(\vec{p}_{2}^{\prime}\). Due to the conservation of linear momentum, \(\vec{p}_{1}\) + \(\vec{p}_{2}^{\prime}\) = 0.

It implies that \(\vec{p}_{1}^{\prime}\)= \(\vec{p}_{2}^{\prime}\), the momentum of the gun is exactly equal, but in the opposite direction to the momentum of the bullet. This is the reason after firing, the gun suddenly moves backward with the momentum (-\(\vec{p}_{2}^{\prime}\)). It is called ‘recoil momentum’. Th is is an example of conservation of total linear momentum.

Question 2.
What are concurrent forces? State Lami’s theorem.
Answer:
Concurrent force:
A collection of forces is said to be concurrent, if the lines of forces act at a common point.

Lami’s Theorem:
If a system of three concurrent and coplanar forces is in equilibrium, then Lami’s theorem states that the magnitude of each force of the system is proportional to sine of the angle between the other two forces.
i.e. |\(\overrightarrow{\mathrm{F}}_{1}\)|∝ sin α, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin β, |\(\overrightarrow{\mathrm{F}}_{2}\)| ∝ sin γ,

Question 3.
Explain the motion of blocks connected by a string in

1. Vertical motion
2. Horizontal motion.

Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in-extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂ T \(\hat{j}\) – m₂ g \(\hat{j}\) = m₂ a \(\hat{j}\) The left hand side of the above equation is the total force that acts on m₂ and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂ g = m₂ a ……….(1)
Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁ g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass m₁ moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)
Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m₂g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration.
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

• Downward gravitational force (m₂g)
• Upward normal force (N) exerted by the surface
• Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

• Downward gravitational force (m₁g)
• Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
Ti = m₁ai (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 4.
Briefly explain the origin of friction. Show that in an inclined plane, angle of friction is pqual to angle of repose.
Answer:
If a very gentle force in the horizontal direction is given to an object at rest on the table it does not move. It is because of the opposing force exerted by the surface on the object which resists its motion. This force is called the frictional force. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But in the twentieth century, the understanding on atoms, electron and protons has changed the perspective.

The frictional force is actually the electromagnetic force between the atoms on the two surfaces. Even well polished surfaces have irregularities on the surface at the microscopic level. The component of force parallel to the inclined plane (mg sin θ) tries to move the object down. The component of force perpendicular to the inclined plane (mg cos θ) is balanced by the Normal force (N).
N = mg cos θ ………(1)

When the object just begins to move, the static friction attains its maximum value
f_s = \(f_{s}^{\max }\) = µ_s N
This friction also satisfies the relation
\(f_{s}^{\max }\) = µ_s mg sin θ ……….(2)
Equating the right hand side of equations (1) and (2),
(\(f_{s}^{\max }\))/N = sin θ / cos θ
From the definition of angle of friction, we also know that
tan θ = µ_s ………..(3)
in which θ is the angle of friction.
Thus the angle of repose is the same as angle of friction. But the difference is that the angle of repose refers to inclined surfaces and the angle of friction is applicable to any type of surface.

Question 5.
State Newton’s three laws and discuss their significance.
Answer:
First Law:
Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.

Second Law:
The force acting on an object is equal to the rate of change of its momentum

Third Law:
For every action there is an equal and opposite reaction.

Significance of Newton’s Laws:
1. Newton’s laws are vector laws. The equation \(\overline{\mathrm{F}}\) = m\(\overline{\mathrm{a}}\) is a vector equation and essentially it is equal to three scalar equations. In Cartesian coordinates, this equation can be written as F_x\(\hat{i}\) + F_y\(\hat{j}\) + F_z\(\hat{j}\) = ma_x\(\hat{i}\) + ma_y\(\hat{j}\) + ma_z\(\hat{j}\)
By comparing both sides, the three scalar equations are

F_x = ma_x The acceleration along the x-direction depends only on the component of force acting along the x – direction.
F_y = ma_y The acceleration along the y direction depends only on the component of force acting along the y – direction.
F_z = ma_z The acceleration along the z direction depends only on the component of force acting along the z – direction.
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along x direction. In the same way, F_z cannot affect a_y and a_x. This understanding is essential for solving problems.

2. The acceleration experienced by the body at time t depends on the force which acts on the body at that instant of time. It does not depend on the force which acted on the body before the time t. This can be expressed as
\(\overline{\mathrm{F}}\)(t) = m\(\overline{\mathrm{a}}\)(t)
Acceleration of the object does not depend on the previous history of the force. For example, when a spin bowler or a fast bowler throws the ball to the batsman, once the ball leaves the hand of the bowler, it experiences only gravitational force and air frictional force. The acceleration of the ball is independent of how the ball was bowled (with a lower or a higher speed).

3. In general, the direction of a force may be different from the direction of motion. Though in some cases, the object may move in the same direction as the direction of the force, it is not always true. A few examples are given below.

Case 1:
Force and motion in the same direction:
When an apple falls towards the Earth, the direction of motion (direction of velocity) of the apple and that of force are in the same downward direction as shown in the Figure.

Case 2:
Force and motion not in the same direction:
The Moon experiences a force towards the Earth. But it actually moves in elliptical orbit. In this case, the direction of the force is different from the direction of motion as shown in Figure.

Case 3:
Force and motion in opposite direction:
If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward as shown in the Figure.

Case 4:
Zero net force, but there is motion:
When a raindrop gets detached from the cloud it experiences both downward gravitational force and upward air drag force. As it descends towards the Earth, the upward air drag force increases and after a certain time, the upward air drag force cancels the downward gravity. From then on the raindrop moves at constant velocity till it touches the surface of the Earth. Hence the raindrop comes with zero net force, therefore with zero acceleration but with non-zero terminal velocity. It is shown in the Figure

4. If multiple forces \(\overrightarrow{\mathrm{F}}_{1}\), \(\overrightarrow{\mathrm{F}}_{2}\), \(\overrightarrow{\mathrm{F}}_{3}\),……. \(\overrightarrow{\mathrm{F}}_{n}\) act on the same body, then the total force (\(\overrightarrow{\mathrm{F}}_{net}\)) is equivalent to the vectorial sum of the individual forces. Their net force provides the acceleration.
\(\overrightarrow{\mathrm{F}}_{net}\) = \(\overrightarrow{\mathrm{F}}_{1}\) + \(\overrightarrow{\mathrm{F}}_{2}\) + \(\overrightarrow{\mathrm{F}}_{3}\) + ……… + \(\overrightarrow{\mathrm{F}}_{n}\)

Newton’s second law for this case is –
\(\overrightarrow{\mathrm{F}}_{net}\) = m\(\overline{\mathrm{a}}\)
In this case the direction of acceleration is in the direction of net force.
Example:
Bow and arrow

5. Newton’s second law can also be written in the following form.
Since the acceleration is the second derivative of position vector of the body \(\left(\vec{a}=\frac{d^{2} \vec{r}}{d t^{2}}\right)\)
the force on the body is –
\(\overline{\mathrm{F}}\) = m\(\frac{d^{2} \vec{r}}{d t^{2}}\)
From this expression, we can infer that Newton’s second law is basically a second order ordinary differential equation and whenever the second derivative of position vector is not zero, there must be a force acting on the body.

6. If no force acts on the body then Newton’s second law, m = \(\frac{d \vec{v}}{d t}\) = 0
It implies that \(\overline{\mathrm{v}}\) = constant. It is essentially Newton’s first law. It implies that the second law is consistent with the first law. However, it should not be thought of as the reduction of second law to the first when no force acts on the object. Newton’s first and second laws are independent laws. They can internally be consistent with each other but cannot be derived from each other.

7. Newton’s second law is cause and effect relation. Force is the cause and acceleration is the effect. Conventionally, the effect should be written on the left and cause on the right hand side of the equation. So the correct way of writing Newton’s second law is –
m\(\overline{\mathrm{a}}\) = \(\overline{\mathrm{F}}\) or \(\frac{d \vec{p}}{d t}\) = \(\overline{\mathrm{F}}\)

Question 6.
Explain the similarities and differences of centripetal and centrifugal forces.
Answer:
Salient features of centripetal and centrifugal forces.
Centripetal Force:

• It is a real force which is exerted on the body by the external agencies like gravitational force, tension in the string, normal force etc.
• Acts in both inertial and non-inertial frames
• It acts towards the axis of rotation or center of the circle in circular motion
  \(\left|\overrightarrow{\mathrm{F}}_{\mathrm{C}_{\mathrm{P}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Real force and has real effects
• Origin of centripetal force is interaction between two objects.
• In inertial frames centripetal force has to be included when free body diagrams are drawn.

Centrifugal Force:

• It is a pseudo force or fictitious force which cannot arise from gravitational force, tension force, normal force etc.
• Acts only in rotating frames (non-inertial frame)
• It acts outwards from the axis of rotation or radially outwards from the center of the circular motion
  \(\left|\overrightarrow{\mathrm{I}}_{\mathrm{C}_{\mathrm{f}}}\right|\) = mω²r = \(\frac{m v^{2}}{r}\)
• Pseudo force but has real effects
• Origin of centrifugal force is inertia. It does not arise from interaction. In an inertial frame the object’s inertial motion appears as centrifugal force in the rotating frame.
• In inertial frames there is no centrifugal force. In rotating frames, both centripetal and centrifugal force have to be included when free body diagrams are drawn.

Question 7.
Briefly explain ‘centrifugal force’ with suitable examples.
Answer:
To use Newton’s first and second laws in the rotational frame of reference, we need to include a Pseudo force called centrifugal force. This centrifugal force appears to act on the object with respect to rotating frames.

Circular motion can be analysed from two different frames of reference. One is the inertial frame (which is either at rest or in uniform motion) where Newton’s laws are obeyed. The other is the rotating frame of reference which is a non – inertial frame of reference as it is accelerating.

When we examine the circular motion from these frames of reference the situations are entirely different. To use Newton’s first and second laws in the rotational frame of reference, we need to include a pseudo force called ‘centrifugal force’. This ‘centrifugal force’ appears to act on the object with respect to rotating frames. To understand the concept of centrifugal force, we can take a specific case and discuss as done below.

Free body diagram of a particle including the centrifugal force Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest). If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.

This implies that in addition to the inward centripetal force – mω²r there must be an equal and opposite force that acts on the stone outward with value +mω²r . So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω² r = 0). This outward force +mω²r is called the centrifugal force. The word ‘centrifugal’ means ‘flee from center’.

Note that the ‘centrifugal force’ appears to act on the particle, only when we analyse the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the rstring. For this reason centrifugal force is called as a ‘pseudo force’. A pseudo force has no origin. It arises due to the non inertial nature of the frame considered. When circular motion problems are solved from a rotating frame of reference, while drawing free body diagram of a particle, the centrifugal force should necessarily be included as shown in the figure.

Question 8.
Briefly explain ‘rolling friction’.
Answer:
The invention of the wheel plays a crucial role in human civilization. One of the important applications is suitcases with rolling on coasters. Rolling wheels makes it easier than carrying luggage. When an object moves on a surface, essentially it is sliding on it. But wheels move on the surface through rolling motion. In rolling motion when a wheel moves on a surface, the point of contact with surface is always at rest. Since Rolling and kinetic friction the point of contact is at rest, there is no relative motion between the wheel and surface.

Hence the frictional fore is very less. At the same time if an object moves without a wheel, there is a relative motion between the object and the surface. As a result frictional force is larger. This makes it difficult to move the object. The figure shows the difference between rolling and kinetic friction. Ideally in pure rolling, motion of the point of contact with the surface should be at rest, but in practice it is not so.

Due to the elastic nature of the surface at the point of contact there will be some deformation on the object at this point on the wheel or surface as shown in figure. Due to this deformation, there will be minimal friction between wheel and surface. It is called ‘rolling friction. In fact, rolling friction’ is much smaller than kinetic friction.

Question 9.
Describe the method of measuring angle of repose.
Answer:
When objects are connected by strings and When objects are connected by strings and a force F is applied either vertically or horizontally or along an inclined plane, it produces a tension T in the string, which affects the acceleration to an extent. Let us discuss various cases for the same.

Case 1:
Vertical motion:
Consider two blocks of masses m₁ and m₂ (m₁> m₂) connected by a light and in extensible string that passes over a pulley as shown in Figure.

Let the tension in the string be T and acceleration a. When the system is released, both the blocks start moving, m₂ vertically upward and m_k, downward with same acceleration a. The gravitational force m₁g on mass m₁ is used in lifting the mass m₂. The upward direction is chosen as y direction. The free body diagrams of both masses are shown in Figure.

Applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₂g\(\hat{j}\) = m₂a\(\hat{j}\) The left hand side of the above equation is the total force that acts on m2 and the right hand side is the product of mass and acceleration of m₂ in y direction.
By comparing the components on both sides, we get
T = m₂g = m₂a ……….(1)

Similarly, applying Newton’s second law for mass m₂
T \(\hat{j}\) – m₁g\(\hat{j}\) = m₁a\(\hat{j}\)
As mass mj moves downward (-\(\hat{j}\)), its acceleration is along (-\(\hat{j}\))
By comparing the components on both sides, we get
T = m₁g = -m₁a
m₁g – T = m₁a ………..(2)

Adding equations (1) and (2), we get
m₁g – m₂g = m₁a + m₂a
(m₁ – m₂)g = (m₁ + m₂)a …………(3)
From equation (3), the acceleration of both the masses is –
a = (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ………..(4)
If both the masses are equal (m₁ = m₂), from equation (4)
a = 0
This shows that if the masses are equal, there is no acceleration and the system as a whole will be at rest.
To find the tension acting on the string, substitute the acceleration from the equation (4) into the equation (1).
T = m₂g = m₂(\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))
T = m₂g + m₂ (\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\))g ……….(5)
By taking m2g common in the RHS of equation (5)

Equation (4) gives only magnitude of acceleration
For mass m₁, the acceleration vector is given by \(\vec{a}\) = –\(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\)\(\hat{j}\)
For mass m₂, the acceleration vector is given by \(\vec{a}\) = \(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\) \(\hat{j}\)

Case 2:
Horizontal motion:
In this case, mass m₂ is kept on a horizontal table and mass m₁, is hanging through a small pulley as shown in figure. Assume that there is no friction on the surface

As both the blocks are connected to the un stretchable string, if m₁ moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are

1. Downward gravitational force (m₂g)
2. Upward normal force (N) exerted by the surface
3. Horizontal tension (T) exerted by the string

The forces acting on mass m₁ are

1. Downward gravitational force (m₁g)
2. Tension (T) acting upwards

The free body diagrams for both the masses is shown in figure 2.

Applying Newton’s second law for m₁
T\(\hat{i}\) – m₁g\(\hat{j}\) = -m₁a\(\hat{j}\) (alongy direction)
By comparing the components on both sides of the above equation,
T – m₁g = -m₁a …………(1)
Applying Newton’s second law for m₂
T\(\hat{i}\) = m₁a\(\hat{i}\) (along x direction)
By comparing the components on both sides of above equation,
T = m₂a ………….(2)
There is no acceleration along y direction for m₂.
N\(\hat{j}\) – m₂g\(\hat{j}\) = 0
By comparing the components on both sides of the above equation
N – m₂g = 0
N = m₂g ……….(3)
By substituting equation (2) in equation (1), we can find the tension T
m₂a – m₁g = -m₁a
m₂a + m₁a = m₁g
a = \(\frac{m_{1}}{m_{1}+m_{2}}\)g …………(4)
Tension in the string can be obtained by substituting equation (4) in equation (2)
T = \(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\)g ………..(5)

Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings. This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Question 10.
Explain the need for banking of tracks.
Answer:
In a leveled circular road, skidding mainly depends on the coefficient of static friction py. The coefficient of static friction depends on the nature of the surface which has a maximum limiting value. To avoid this problem, usually the outer edge of the road is slightly raised compared to inner edge. This is called banking of roads or tracks. This introduces an inclination, and the angle is called banking angle. “Let the surface of the road make angle 9 with horizontal surface. Then the normal force makes the same angle 9 with the vertical. When the car takes a turn.

there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface)
We can resolve the normal force into two components N cos θ and N sin θ. The component balances the downward gravitational force ‘mg’ and component will provide the necessary centripetal acceleration. By using Newton second law.

The banking angle 0 and radius of curvature of the road or track determines the safe speed of the car at the turning. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.

At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Question 11.
Calculate the centripetal acceleration of Moon towards the Earth.
Answer:
The centripetal acceleration is given by a = \(\frac{v^{2}}{r}\) This expression explicitly depends on Moon’s speed which is nontrivial. We can work with the formula
ω²R_m = a_m
a_m is centripetal acceleration of the Moon due to Earth’s gravity, ω is angular velocity
R_m is the distance between Earth and the Moon, which is 60 times the radius of the Earth.
R_m = 60R = 60 x 6.4 x 10⁶ = 384 x 10⁶ m
As we know the angular velocity ω = \(\frac { 2π}{ T }\) and T = 27.3 days = 27.3 x 24 x 60 x 60 second = 2.358 x 10⁶ sec.
By substituting these values in the formula for acceleration
a₆ = \(\frac{\left(4 \pi^{2}\right)\left(384 \times 10^{6}\right)}{\left(2.358 \times 10^{8}\right)^{2}}\) = 0.00272 ms^-2

Samacheer Kalvi 11th Physics Laws of Motion Conceptual Questions

Question 1.
Why it is not possible to push a car from inside?
Answer:
While trying to push a car from outside, he pushes the ground backwards at an angle. The ground offers an equal reaction in the opposite direction, so car can be moved. But the person sits inside means car and the person becomes a single system, and the force given will be a internal force. According to Newton’s third law, total internal force acting on the system is zero and it cannot accelerate the system.

Question 2.
There is a limit beyond which the polishing of a surface increases frictional resistance rather than decreasing it why?
Answer:
Friction is a contact force. Friction is directly proportional to area of contact. In the normal surfaces there are bumps to interlock the surfaces in contact. But the surfaces are polished beyond certain limit. The area of contact will be increased and the molecules come closer to each other. It increases electrostatic force between the molecules. As a result it increases friction.

Question 3.
Can a single isolated force exist in nature? Explain your answer.
Answer:
No. According to Newton’s third law, for every action, there is an equal and opposite reaction. So, whatever case we consider, if there is an action there is always a reaction. So it is impossible.

Question 4.
Why does a parachute descend slowly?
Answer:
A parachute descends slowly because the surface area of parachute is large so that air gives more resistance when it descends down.

Question 5.
When walking on ice one should take short steps. Why?
Answer:
Let R represent the reaction offered by the ground. The vertical component R cos θ will balance the weight of the person and the horizontal component R sin θ will help the person to walk forward.
Now, normal reaction = R cos θ
Friction force = R sin θ
Coefficient of friction, µ = \(\frac {R sin θ}{R cos θ }\) = tan θ
In a long step, θ is more. So tan θ is more. But p has a fixed value. So, there is danger of slipping in a long step.

Question 6.
When a person walks on a surface, the frictional force exerted by the surface on the person is opposite to the direction of motion. True or false?
Answer:
False. In frictional force exerted by the surface on the person is in the direction of his motion. Frictional force acts as an external force to move the person. When the person trying to move, he gives a push to ground on the backward direction and by Newton’s third law he is pushed by the ground in the forward direction. Hence frictional force acts along the direction of motion.

Question 7.
Can the coefficient of friction be more than one?
Answer:
Yes. The coefficient of friction can be more than one in some cases such as silicone rubber. Coefficient of friction is the ratio of frictional force to normal force, i.e. F = μR. If p is greater than one means frictional force is greater than normal force. But in general case the value is usually between 0 and 1.

Question 8.
Can we predict the direction of motion of a body from the direction of force on it?
Answer:
Yes. The direction of motion is always opposite to the force of kinetic friction. By using the principle of equilibrium, the direction of force of static friction can be determined. When the object is in equilibrium, the frictional force must point in the direction which results as a net force is zero.

Question 9.
The momentum of a system of particles is always conserved. True or false?
Answer:
True. The total momentum of a system of particles is always constant i.e. conserved. When no external force acts on it.

Samacheer Kalvi 11th Physics Laws of Motion Numerical Problems

Question 1.
A force of 50 N act on the object of mass 20 kg. shown in the figure. Calculate the acceleration of the object in x and y directions.
Answer:
Given F = 50 N and m = 20 kg
(1) component of force along x – direction
F_x = F cos θ
= 50 x cos 30° = 43.30 N
a_x = \(\frac{F_{x}}{m}\) = \(\frac {43.30}{20}\) =2.165 ms^-2

(2) Component of force along y – direction
F_y = F sin θ = 50 sin 30° = 25 N
a_y = \(\frac{F_{y}}{m}\) = \(\frac {25}{20}\) = 1.25 ms^-2

Question 2.
A spider of mass 50 g is hanging on a string of a cob web as shown in the figure. What is the tension in the string?
Answer:
Given m = 50 g, T = ?
Tension in the string T = mg
= 50 x 10^-2 x 9.8 = 0.49 N

Question 3.
What is the reading shown in spring balance?

Answer:
When a spring balance hung on a rigid support and load is attached at its other end, the weight of the load exerts a force on the rigid support in turn support exerts equal and opposite force on that load, so that balance will be stretched. This is the principle of spring balance. Flence the answer is 4 kg.
Given: m = 2 kg, 0 = 30°.
Resolve the weight into its component as mg sin θ and mg cos θ.
Here mg sin θ acts parallel to the surface
∴ W = mg sin θ
= 2 x 9.8 x sin 30° = 2 x 9.8 x \(\frac {1}{2}\) = 9.8 N

Question 4.
The physics books are stacked on each other in the sequence: +1 volumes 1 and 2; +2 volumes 1 and 2 on a table
(a) Identify the forces acting on each book and draw the free body diagram.
(b) Identify the forces exerted by each book on the other.
Answer:
Let m₁, m₂, m₃, m_4, are the masses of +1 volume I and II and +2 volumes I & II

(a) Force on book m₄

• Downward gravitational force acting downward (m₃g)
• Upward normal force (N₃) exerted by book of mass m₃
  |

(b) Force on book m₃

• Downward gravitational force (m₃g)
• Downward force exerted by m₄ (N₄)
• Upward force exerted by m₂ (N₂)
  

(c) Force on book m₂

• Downward gravitational force (m₂g)
• Downward force exerted by m₃ (N₃)
• Upward force exerted by m₁ (CN₁)
  

(d) Force on book m₁

• Downward gravitational force exerted by earth (m₁g)
• Downward force exerted by m₂ (N₂)
• Upward force exerted by the table (N_table)
  

Question 5.
A bob attached to the string oscillates back and forth. Resolve the forces acting on the bob in to components. What is the acceleration experienced by the bob at an angle θ.
Answer:
The gravitational force (mg) acting downward can be resolved into two components as mg cos θ and mg sin θ
T – tension exerted by the string.
Tangential force F_T = ma_T = mg sin θ
∴ Tangential acceleration a_T = g sin θ
Centripetal force Fc = ma_c = T – mg cos θ
a_c = \(\frac { T – mg cos θ }{ m }\)

Question 6.
Two masses m₁ and m₂ are connected with a string passing over a friction-less pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m₁ with the table is µ_s Calculate the minimum mass m₃ that may be placed on m₁to prevent it from sliding. Check if m₁ = 15 kg, m₂ = 10 kg, m₃ = 25 and µ_s = 0.2.

Solution:
Let m₃ is the mass added on m₁
Maximal static friction
\(f_{s}^{\max }\) = µ_sN = µ_s (m₁ + m₃ )g
Here
N = (m₁ + m₃ )g
Tension acting on string = T = m₂ g
Equate (1) and (2)
µ_s(m₁ + m₃) = m₂g
µ_sm₁ + µ_sm₃ = m₂
m₃ = \(f_{s}^{\max }\) – m₁

(ii) Given,
m₁ = 15 kg, m₂ = 10 kg : m₃ = 25 kg and µ_s = 0.2
m₃ = \(f_{s}^{\max }\) – m₁
m₃ = \(\frac {10}{ 0.2 }\) – 15 = 50 – 15 = 35 kg
The minimum mass m₃ = 35 kg has to be placed on ml to prevent it from sliding. But here m₃ = 25 kg only.
The combined masses (m₁ + m₃) will slide.

Question 7.
Calculate the acceleration of the bicycle of mass 25 kg as shown in Figures 1 and 2.

Answer:
Given:
Mass of bicycle m = 25 kg
Fig. I:
Net force acting in the forward direction, F = 500 – 400 = 100 N
acceleration a = \(\frac { F}{ m }\) = \(\frac { 100 }{25}\) = 4 ms^-2

Fig. II:
Net force acting on bicycle F = 400 – 400 = 0
∴ acceleration a = \(\frac { F}{ m }\) = \(\frac { 0}{25}\) = 0

Question 8.
Apply Lami’s theorem on sling shot and calculate the tension in each string?

Answer:
Given F = 50 N, θ = 30°
Here T is resolved into its components as T sin θ and T cos θ as shown.
According to Lami’s theorem,
\(\frac { F}{sin θ}\) = \(\frac { T}{sin (180 – θ)}\) = \(\frac { T}{sin (180 – θ)}\)
\(\frac { F}{sin θ}\) = \(\frac { T}{sin θ}\)
\(\frac { F}{2 sin θ cos θ}\) = \(\frac { T}{sin θ}\) [T = \(\frac { T}{2 cos θ}\) ]
T = \(\frac {T}{2 cos θ}\) = \(\frac { 50}{ 2 cos 30}\) = 28.868 N

Question 9.
A football player kicks a 0.8 kg ball and imparts it a velocity 12 ms^-1. The contact between the foot and ball is only for one – sixtieth of a second. Find the average kicking force.
Answer:
Given,
Mass of the ball = 0.8 kg
Final velocity (V) =12 ms^-1 and time t =\(\frac {1}{60}\) s
Initial velocity = 0
We know the average kicking force
F = ma = \(\frac {m(v – u)}{t}\) = \(\frac{0.8(12-0)}{\left(\frac{1}{60}\right)}\)
F = 576 N

Question 10.
A stone of mass 2 kg is attached to a string of length 1 meter. The string can withstand maximum tension 200 N. What is the maximum speed that stone can have during the whirling motion?
Solution:
Given,
Mass of a stone = 2 kg,
length of a string = 1 m
Maximum tension = 200 N
The force acting on a stone in the whirling motion is centripetal force. Which is provided by tension of the string.
T_max = F_max = \(\frac{m \mathrm{V}_{\mathrm{max}}^{2}}{r}\)
200 = \(v_{\max }^{2}\) = 100
\(v_{\max }^{2}\) = 10 ms^-1

Question 11.
Imagine that the gravitational force between Earth and Moon is provided by an invisible string that exists between the Moon and Earth. What is the tension that exists in this invisible string due’ to Earth’s centripetal force? (Mass of the Moon = 7.34 x 1022 kg, Distance between Moon and Earth = 3.84 x 108 m).

Solution:
Given,
Mass of the moon = 7.34 x 1022 kg
Distance between moon and earth = 3.84 x 108 m
Centripetal force = F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{7.34 \times 10^{22} \times\left(1.023 \times 10^{3}\right)^{2}}{3.84 \times 10^{8}}\) = 2 x 10²⁰

Question 12.
Two bodies of masses 15 kg and 10 kg are connected with light string kept on a smooth surface. A horizontal force F = 500 N is applied to a 15 kg as shown in the figure. Calculate the tension acting in the string.

Answer:
Given,
m₁  = 15 kg, m₂  = 10 kg, F = 500 N
Tension acting in the string T = \(\frac{m_{2}}{m_{1}+m_{2}}\) F
T = \(\frac {10}{25}\) x 500 = 200 N

Question 13.
People often say “For every action there is an equivalent opposite reaction”. Here they meant ‘action of a human’. Is it correct to apply Newton’s third law to human actions? What is meant by ‘action’ in Newton third law? Give your arguments based on Newton’s laws.
Answer:
Newton’s third law is applicable to only human’s physical actions which involves physical force. Third law is not applicable to human’s psychological actions or thoughts.

Question 14.
A car takes a turn with velocity 50 ms^-1 on the circular road of radius of curvature To m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car?
Answer:
Given,
Mass of a person = 60 kg
Velocity of the car = 50 ms^-1
Radius of curvature = 10 m
Centrifugal force F = \(\frac{m \mathrm{V}^{2}}{r}\) = \(\frac{60 \times(50)^{2}}{10}\) = 15,000 N

Question 15.
A long stick rests on the surface. A person standing 10 m away from the stick. With what minimum speed an object of mass 0.5 kg should he thrown so that it hits the stick. (Assume the coefficient of kinetic friction is 0.7).
Answer:
Given,
Distance (s) = 10 m
Mass of the object (m) = 0.5 kg
Coefficient of kinetic friction (µ) = 0.7
Work done in moving a body in horizontal surface ω = µ_R x s = µmg x s
This work done is equal to initial kinetic energy of the object
\(\frac{1}{2} m v^{2}\) = µ mg s
\(\left|v^{2}\right|\) = 2 µgs = 2 x 0.7 x 9.8 x 10
v² = 14 x 9.8 = 137. 2
v = 11. 71 ms^-1

Samacheer Kalvi 11th Physics Laws of Motion Additional Questions Solved

Samacheer Kalvi 11th Physics Laws of Motion Multiple Choice Questions

Question 1.
The concept “force causes motion” was given by –
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(b) Aristotle

Question 2.
Who decoupled the motion and force?
(a) Galileo
(b) Aristotle
(c) Newton
(d) Joule
Answer:
(a) Galileo

Question 3.
The inability of objects to move on its own or change its state of motion is called as –
(a) force
(b) momentum
(c) inertia
(d) impulse
Answer:
(c) inertia

Question 4.
Inertia means –
(a) inability
(b) resistance to change its state
(c) movement
(d) inertial frame
Answer:
(b) resistance to change its state

Question 5.
When a bus starts to move from rest, the passengers experience a sudden backward push is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(c) Inertia of rest

Question 6.
If the brake is applied in the moving bus suddenly, passengers move forward is an example for –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(a) Inertia of motion

Question 7.
In whirling motion, if the string is cut suddenly, the stone moves tangential to circle is an –
(a) Inertia of motion
(b) Inertia of direction
(c) Inertia of rest
(d) back pull
Answer:
(b) Inertia of direction

Question 8.
Newtons laws are applicable in –
(a) Inertial frame
(b) non inertial frame
(c) in any frame
(d) none
Answer:
(a) Inertial frame

Question 9.
The accelerated train is an example for –
(a) inertial frame
(b) non-inertial frame
(c) both (a) and (b)
(d) none of the above
Answer:
(b) non-inertial frame

Question 10.
Rate of change of momentum of an object is equal to –
(a) acceleration
(b) work done
(c) force
(d) impulse
Answer:
(c) force

Question 11.
The product of mass and velocity is –
(a) force
(b) impulse
(c) momentum
(d) acceleration
Answer:
(c) momentum

Question 12.
Unit of momentum –
(a) kg ms^-2
(b) kg ms^-1
(c) MLT^-2
(d) MLT^-1
Answer:
(b) kg ms^-1

Question 13.
According to Newton’s third law –
(a) F₁₂ = F₂₁
(*) F₁₂ = -F₂₁
(c) F₁₂ + F₂₁ = 0
(d) F₁₂ x F₂₁ = 0
Answer:
(a) F₁₂ = F₂₁

Question 14.
According to Newton’s third law –
(a) \(\overrightarrow{\mathrm{F}_{12}}=\overrightarrow{\mathrm{F}_{21}}\)
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)
(c) \(\mathrm{F}_{12}+\mathrm{F}_{21}\) = 0
(d) \(\mathrm{F}_{12}x\mathrm{F}_{21}\) = 0
Answer:
(b) \(\overrightarrow{\mathrm{F}_{12}}=-\overrightarrow{\mathrm{F}_{21}}\)

Question 15.
The law which is valid in both inertial and non-inertial frame is –
(a) Newton’s first law
(b) Newton’s second law
(c) Newton’s third law
(d) none
Answer:
(c) Newton’s third law

Question 16.
When a force is applied on a body, it can change –
(a) velocity
(b) momentum
(c) direction of motion
(d) all the above
Answer:
(d) all the above

Question 17.
The rate of change of velocity is 1 ms^-2 when a force is applied on the body of mass 75 gm the force is –
(a) 75 N
(b) 0.75 N
(c) 0.075 N
(d) 0.75 x 10^-3 N
Answer:
(c) Force is given by
F = m a
= 75 gm x 1 cm s^-2 = 75 x 10^-3 x 1 = 75 x 10^-3 = 0.075 N

Question 18.
The action and reaction forces acting on –
(a) same body
(b) different bodies
(c) either same or different bodies
(d) none of the above
Answer:
(b) different bodies

Question 19.
Newton’s first law of motion gives the concept of –
(a) velocity
(b) energy
(c) momentum
(d) Inertia
Answer:
(d) Inertia

Question 20.
Inertia of a body has direct dependence on –
(a) velocity
(b) area
(c) mass
(d) volume
Answer:
(c) mass

Question 21.
If a car and a scooter have the same momentum, then which one is having greater speed?
(a) scooter
(b) car
(c) both have same velocity
(d) data insufficient
Answer:
(a) scooter

Question 22.
Newton’s second law gives –
(a) \(\overrightarrow{\mathrm{F}} \propto \frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(b) \(\overrightarrow{\mathrm{F}}=\frac{d \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)
(c) \(\overrightarrow{\mathrm{F}}=m \vec{a}\)
(d) all the above
Answer:
(d) all the above

Question 23.
1 dyne is –
(a) 10⁵N
(b) 10^-5N
(c) 1N
(d) 10^-3N
Answer:
(b) 10^-5N

Question 24.
If same force is acting on two masses m₁ and m₂, and the accelerations of two bodies are a₁ and a₂ respectively, then –
(a) \(\frac{a_{2}}{a_{1}}=\frac{m_{2}}{m_{1}}\)
(b) \(\frac{a_{1}}{a_{2}}=\frac{m_{1}}{m_{2}}\)
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)
(d) m₁ a₁ + m₂a₂ = 0
Answer:
(c) \(\frac{a_{1}}{a_{2}}=\frac{m_{2}}{m_{1}}\)

Question 25.
If a force \(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\) N produces an acceleration of 10 ms^-2 on a body, then the mass of a body is –
(a) 10 kg
(b) 9 kg
(c) 0.9 kg
(d) 0.5 kg
Answer:
\(\overline{\mathrm{F}}\) = 3\(\hat{i}\) – 4\(\hat{j}\)
Magnitude:
|\(\overline{\mathrm{F}}\)| = \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5N
F = ma
⇒ m = \(\frac{|\mathrm{F}|}{a}\) = \(\frac{5}{10}\) = \(\frac{1}{2}\) = 0.5 kg

Question 26.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
(a) 0.75 s
(b) 1.33 s
(c) 6 s
(d) 35 s
Answer:
Acceleration a = \(\frac{-F}{m}\) = \(\frac{50}{20}\) = – 2.5 ms^-2
u = l5 ms^-1
v = 0
t = ?
v = u + at
0 = 15 – 2.5t
t = \(\frac{15}{2.5}\) = 6s

Question 27.
Rain drops come down with –
(a) zero acceleration and non zero velocity
(b) zero velocity with non zero acceleration
(c) zero acceleration and non zero net force
(d) none
Answer:
(a) zero acceleration and non zero velocity

Question 28.
If force is the cause then the effect is –
(a) mass
(b) potential energy
(c) acceleration
(d) Inertia
Answer:
(c) acceleration

Question 29.
In free body diagram, the object is represented by a –
(a) line
(b) arrow
(c) circle
(d) point
Answer:
(d) point

Question 30.
When an object of mass m slides on a friction less surface inclined at an angle 0, then normal force exerted by the surface is –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg tan θ
Answer:
(b) mg cos θ

Question 31.
The acceleration of the sliding object in an inclined plane –
(a) g cos θ
(b) mg cos θ
(c) g sin θ
(d) mg sin θ
Answer:
(c) g sin θ

Question 32.
The speed of an object sliding in an inclined plane at the bottom is –
(a) mg cos θ
(b) \(\sqrt{2 s g sin θ}\)
(c) \(\sqrt{2 s g cos θ}\)
(d) \(\sqrt{2 s g tan θ}\)
Answer:
(b) \(\sqrt{2 s g sin θ}\)

Question 33.
The acceleration of two bodies of mass m₁ and m₂ in contact on a horizontal surface is –
(a) \(a=\frac{\mathbf{F}}{m_{1}}\)
(b) \(a=\frac{F}{m_{2}}\)
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(d) \(a=\frac{\mathrm{F}}{m_{1} m_{2}}\)
Answer:
(c)  \(a=\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 34.
Two blocks of masses m₁ and m₂ (m₁ > m₂) in contact with each other on frictionless, horizontal surface. If a horizontal force F is given on m₁, set into motion with acceleration a, then reaction force on mass m₁ by m₂, is –
(a) \(\frac{\mathrm{F} m_{1}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{1}}\)
(c) \(\frac{m_{1} m_{2}}{\mathrm{F} m_{2}}\)
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)
Answer:
(d) \(\frac{\mathrm{F} m_{2}}{m_{1}+m_{2}}\)

Question 35.
If two masses m₁ and m₂ (m₁ > m₂) tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 36.
if two masses m₁ and m₂ (m₁ > m₂)tied to string moving over a frictionless pulley, then acceleration of masses –
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g
(b) \(\frac{m_{1}+m_{2}}{\left(m_{1}-m_{2}\right)}\) g
(c) \(\frac{2 m_{1} m_{2}}{m_{1}+m_{2}}\) g
(d) \(\frac{m_{1} m_{2}}{2 m_{1} m_{2}}\) g
Answer:
(a) \(\frac{\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}}\) g

Question 37.
Three massses is in contact as shown. If force F is applied to mass m₁, the acceleration of three masses is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 38.
Three masses in contact is as shown above. If force F is applied to mass m₁ then the contact force acting on mass m₂ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)

Question 39.
Three masses is contact as shown. It force F is applied to mass m₁, then the contact force acting on mass m₃ is –
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
(b) \(\frac{m_{1} F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(c) \(\frac{\left(m_{2}+m_{3}\right) F}{\left(m_{1}+m_{2}+m_{3}\right)}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)

Question 40.
Two masses connected with a string. When a force F is applied on mass m₂. The acceleration produced is –

(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{\mathbf{F}}{m_{1}-m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{\mathrm{F}}\)
(d) \(\frac{m_{3} \mathrm{F}}{m_{1}+m_{2}+m_{3}}\)
Answer:
(a) \(\frac{\mathrm{F}}{m_{1}+m_{2}}\)

Question 41.
Two masses connected with a string. When a force F is applied on mass m₂. The force acting on m₁ is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}+m_{2}}\)
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)
(c) \(\frac{m_{1}+m_{2}}{m_{1}} \mathbf{F}\)
(d) \(\frac{m_{1}+m_{2}}{m_{2}} \mathbf{F}\)
Answer:
(b) \(\frac{m_{2} \mathrm{F}}{m_{1}+m_{2}}\)

Question 42.
If a block of mass m lying on a frictionless inclined plane of length L height h and angle of inclination θ, then the velocity at its bottom is –
(a) g sin θ
(b) g cos θ
(c) \(\sqrt{2 g h}\)
(d) \(\sqrt{2 a sin θ}\)
Answer:
(c) \(\sqrt{2 g h}\)

Question 43.
If a block of mass m lying on a frictionless inclined plane of length L, height h and angle of inclination θ, then the time take taken to reach the bottom is –
(a) g sing θ
(b) sin θ \(\sqrt{\frac{2 h}{g}}\)
(c) sin θ \(\sqrt{\frac{g}{h}}\)
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
Answer:
(d) \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)

Question 44.
A rocket works on the principle of conservation of –
(a) energy
(b) mass
(c) angular momentum
(d) linear momentum
Answer:
(b) mass

Question 45.
A bomb at rest explodes. The total momentum of all its fragments is –
(a) zero
(b) infinity
(c) always 1
(d) always greater then 1
Answer:
(a) zero

Question 46.
A block of mass m₁ is pulled along a horizontal friction-less surface by a rope of mass m₂ If a force F is given at its free end. The net force acting on the block is –
(a) \(\frac{m_{1} \mathrm{F}}{m_{1}-m_{2}}\)
(b) F
(c) \(\frac{m_{2} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
(d) \(\frac{m_{1} \mathrm{F}}{\left(m_{1}+m_{2}\right)}\)
Answer:
(b) F

Question 47.
A block of mass m is pulled along a horizontal surface by a rope. The tension in the rope will be same at all the points –
(a) if the rope is accelerated
(b) if the rope is mass less
(c) always
(d) none of the above
Answer:
(b) if the rope is mass less

Question 48.
The lines of forces act at a common point is called as –
(a) concurrent forces
(b) co-planar forces
(c) equilibrium
(d) resultant
Answer:
(a) concurrent forces

Question 49.
If the lines of forces act in the same plane, they can be –
(a) concurrent forces
(b) coplanar forces
(c) either concurrent force or coplanar forces
(d) Lami’s force
Answer:
(d) concurrent forces

Question 50.
Lami’s theorem is applicable only when the system of forces are is –
(a) same plane
(b) different plane
(c) equilibrium
(d) none of the above
Answer:
(c) equilibrium

Question 51.
Due to the action of internal forces of the system, the total linear momentum of the system is –
(a) a variable
(b) a constant
(c) always zero
(d) always infinity
Answer:
(c) always zero

Question 52.
The velocity with which a gun suddenly moves backward after firing is –
(a) linear velocity
(b) positive velocity
(c) recoil velocity
(d) v₁ + v₂
Answer:
(c) recoil velocity

Question 53.
If a very large force acts on an object for a very short duration, then the force is called as –
(a) Newtonian force
(b) impulsive force
(c) concurrent force
(d) coplanar force
Answer:
(A) impulsive force

Question 54.
The unit of impulse is –
(a) Nm
(b) Ns
(c) Nm²
(d) Ns^-2
Answer:
(b) Ns

Question 55.
The force which always opposes the relative motion between an object and the surface where it is placed is –
(a) concurrent force
(b) frictional force
(c) impulsive force
(d) coplanar force
Answer:
(b) frictional force

Question 56.
The force which opposes the initiation of motion of an object on the surface is –
(a) static friction
(b) kinetic friction
(c) friction
(d) zero
Answer:
(d) static friction

Question 57.
When the object is at rest, the resultant of gravitational force and upward normal force is –
(a) Static force
(b) zero
(c) one
(d) infinity
Answer:
(b) zero

Question 58.
The magnitude of static frictional force d lies between –
(a) 0 ≤ f ≤ µ_sN
(b) 0 ≥f ≥ µ_sN
(c) 0 and 1
(d) 0 and minimal static frictional force.
Answer:
(a) 0 ≤ f ≤ µ_sN

Question 59.
The unit of co-efficient of static friction is –
(a) N
(b) N m
(c) N s
(d) no unit
Answer:
(d) no unit

Question 60.
If the object is at rest and no external force is applied on the object, the static friction acting on the object is –
(a) µ_sN
(b) zero
(c) one
(d) infinity
Answer:
(d) no unit

Question 61.
When object begins to slide, the static friction acting on the object attains –
(a) zero
(b) minimum
(c) maximum
(d) infinity
Answer:
(c) maximum

Question 62.
The static friction does not depend upon –
(a) the area of contact
(b) normal force
(c) the magnitude of applied force
(d) none of the above
Answer:
(a) the area of contact

Question 63.
Which of the following pairs of materials has minimum amount of coefficient of static friction is –
(a) Glass and glass
(b) wood and wood
(c) ice and ice
(d) steel and steel
Answer:
(c) ice and ice

Question 64.
Kinetic friction is also called as –
(a) sliding friction
(b) dynamic friction
(c) both (a) and (b)
(d) static friction
Answer:
(c) both (a) and (b)

Question 65.
The unit of coefficient of kinetic friction is/has –
(a) Nm
(b) Ns
(c) Nm²
(d) no unit
Answer:
(d) no unit

Question 66.
The nature of materials in mutual contact decides –
(a) µs
(b) µk
(c) µs or µk
(d) none
Answer:
(c) µs or µk

Question 67.
Coefficient of kinetic friction is less than –
(a) O
(b) one
(c) µs
(d) µsN
Answer:
(c) µs

Question 68.
The static friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(a) increases linearly

Question 69.
The kinetic friction –
(a) increases linearly
(b) is constant
(c) zero
(d) varies parabolically
Answer:
(b) is constant

Question 70.
Kinetic friction is independent of –
(a) nature of materials
(b) temperature of the surface
(c) applied force
(d) none of the above
Answer:
(c) applied force

Question 71.
The angle between the normal force and the resultant force of normal force and maximum frictional force is –
(a) angle of friction
(b) angle of repose
(c) angle of inclination
(d) none of the above
Answer:
(a) angle of friction

Question 72.
The angle friction θ is given by –
(a) tan µ_s
(b) tan^-1 µ_s
(c) \(\frac{f S^{\mathrm{max}}}{N}\)
(d) sin^-1 µ_s
Answer:
(b) tan^-1 µ_s

Question 73.
The angle of inclined plane with the horizontal such that an object placed on it begins to slide is –
(a) angle of friction
(b) angle of repose
(c) angle of response
(d) angle of retardation
Answer:
(b) angle of repose

Question 74.
Comparatively, which of the following has lesser value than others?
(a) static friction
(b) kinetic friction
(c) Rolling friction
(d) skiping friction
Answer:
(c) Rolling friction

Question 75.
The origin of friction is –
(a) electrostatic interaction
(b) electromagnetic interaction magnetic
(c) photon interaction
(d) interaction
Answer:
(b) electromagnetic interaction

Question 76.
Friction can be reduced by –
(a) polishing
(b) lubricating
(c) using ball bearings
(d) all the above
Answer:
(c) using ball bearings

Question 77.
For a particle revolving in a circular path, the acceleration of the particle is –
(a) along the tangent
(b) along the radius
(c) along the circumference of the circle
(d) zero
Answer:
(b) along the radius

Question 78.
A particle moves along a circular path under the action of a force. The work done by the force is –
(a) Positive and non zero
(b) zero
(c) Negative and non zero
(d) none of the above
Answer:
(b) zero

Question 79.
A bullet hits and gets embedded in a solid block resting on a horizontal frictionless table. Which of the following is conserved?
(a) Momentum and kinetic energy
(b) kinetic energy alone
(c) Momentum alone
(d) potential energy alone
Answer:
(c) Momentum alone

Question 80.
The origin of the centripetal force can be –
(a) gravitational force
(b) frictional force
(c) coulomb force
(d) all the above
Answer:
(d) all the above

Question 81.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) \(\frac{v^{2}}{r}\)
(c) r v²
(d) rω
Answer:
(b) \(\frac{v^{2}}{r}\)

Question 82.
Centripetal acceleration is –
(a) \(\frac{m v^{2}}{r}\)
(b) r ω²
(c) rv²
(d) rω
Answer:
(c) rω²

Question 83.
The centripetal force is –
(a) \(\frac{m v^{2}}{r}\)
(b) rω²
(c) both (a) and (b)
(d) none
Answer:
(c) both (a) and (b)

Question 84.
When a car is moving on a circular track the centripetal force is due to –
(a) gravitational force
(b) frictional force
(c) magnetic force
(d) elastic force
Answer:
(b) frictional force

Question 85.
If the road is horizontal then the normal force and gravitational force are –
(a) equal and along the same direction
(b) equal and opposite
(c) unequal and along the same direction
(d) unequal and opposite
Answer:
(b) equal and opposite

Question 86.
The velocity of a car for safe turn on leveled circular road –
(a) \(v \leq \sqrt{\mu_{s} r g}\)
(b) \(v \geq \sqrt{\mu_{s} r g}\)
(c) \(v=\sqrt{\mu_{s} rg}\)
(d) \(v \leq \mu_{s} rg\)
Answer:
(a) \(v \leq \sqrt{\mu_{s} r g}\)

Question 87.
In a leveled circular road, skidding mainly depends on –
(a) µ_s
(b) µ_k
(c) acceleration
(d) none
Answer:
(a) µ_s

Question 88.
The speed of a car to move on the banked road so that it will have safe turn is –
(a) µ_srg
(b) \(\sqrt{r g \tan \theta}\)
(c) rg tan θ
(d) r²g tan θ
Answer:
(b) \(\sqrt{r g \tan \theta}\)

Question 89.
Centrifugal force is a –
(a) pseudo force
(b) real force
(c) forced acting towards center
(d) none of the above
Answer:
(a) pseudo force

Question 90.
Origin of centrifugal force is due to –
(a) interaction between two
(b) inertia
(c) electromagnetic interaction
(d) inertial frame
Answer:
(b) inertia

Question 91.
Centripetal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (h)
(d) linear motion
Answer:
(c) both (a) and (b)

Question 92.
Centrifugal force acts in –
(a) inertial frame
(b) non inertial frame
(c) both (a) and (b)
(d) linear motion
Answer:
(b) non inertial frame

Question 93.
A cricket ball of mass loo g moving with a velocity of 20 ms-¹ is brought to rest by a player in 0.05s the impulse of the ball is –
(a) 5 Ns
(b) – 2 Ns
(c) – 2.5 Ns
(d) zero
Answer:
(b) – 2 Ns
mass = 0.1 kg
Initial velocity t = 20 ms^-1
Final velocity y = 0
Change in momentum in impulse = m(v – u) = 0.1(0 – 20) = – 2 Ns

Question 94.
If a stone tied at the one end of a string of length 0.5 m is whirled in a horizontal circle with a constant speed 6 ms^-1  then the acceleration of the shone is –
(a) 12 ms^-2
(b) 36 ms^-2
(c) 2π² ms^-2
(d) 72 ms^-2
Answer:
(d) Centripetal acceleration = \(\frac{v^{2}}{r}\) = \(\frac{6^{2}}{0.5}\) = \(\frac{36}{0.5}\) = 72 ms^-2

Question 95.
A block of mass 3 kg is at rest on a rough inclined plane with angle of inclination 30° with horizontal. If .is 0.7, then the frictional force is –
(a) 17.82 N
(b) 1.81 N
(c) 3.63 N
(d) 2.1 N
Answer:
(a) Frictional force = µmg cos θ = 0.7 x 3 x 9.8 cos 30° = 17.82 N

Question 96.
Two masses 2 kg and 4 kg are tied at the ends of a mass less string and which is passing over a friction-less pulley. The tension in the string is –
(a) 3.68 N
(b) 78.4 N
(c) 26 N
(d) 13.26 N
Answer:
(c) Tension in the string T = \(\frac{2 m_{1} m_{2}}{\left(m_{1}+m_{2}\right)}\)g
T = \(\frac{2 x 2 x 4}{2 + 4}\) x 9.8 = \(\frac{16}{6}\) x 9.8 = 26.13 N

Question 97.
A bomb of 10 kg at rest explodes into two pieces of mass 4 kg and 6 kg. if the velocity of 4 kg mass is 6 ms-¹ then the velocity of 6 kg is –
(a) – 4 ms^-1
(b) – 6 ms^-1
(c) – 24 ms^-1
(d) – 2.2 ms^-1
Answer:
(a) According to law of conservation of momentum
m₁v₁ + m₂v₂ = 0
v₂ = –\(\frac{m_{1} v_{1}}{m_{2}}\) = \(\frac{4 x 6}{6}\) = -4ms^-1

Question 98.
A body is subjected under three concurrent forces and it is in equilibrium. The resultant of any two forces is –
(a) coplanar with the third force
(b) is equal and opposite to third force
(c) both (a) and (b)
(d) none of the above
Answer:
(c) both (a) and (b)

Question 99.
An impulse is applied to a moving object with the force at an angle of 20° with respect to velocity vector. The angle between the impulse vector and the change in momentum vector is –
(a) 0°
(b) 30°
(c) 60°
(d) 120°
Answer:
(a) Impulse and change in momentum are in same direction. So the angle is zero.

Question 100.
A bullet of mass m and velocity v₁ is fired into a large block of wood of mass M. The final velocity of the system is-
(a) \(\frac{v_{1}}{m+\mathrm{M}}\)
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)
(c) \(\frac{m+m}{m} v_{1}\)
(d) \(\frac{m+m}{m-M} v_{1}\)
Answer:
(b) \(\frac{m v_{1}}{m+\mathrm{M}}\)

Question 101.
A block of mass 2 kg is placed on the floor. The co – efficient of static friction is 0.4. The force of friction between the block and floor is –
(a) 2.8 N
(b) 7.8 N
(c) 2 N
(d) zero
Answer:
(b) The force required to move = = µR = µmg = 0.4 x 2 x 9.8 = 7.84 N

Question 102.
A truck weighing 1000 kg is moving with velocity of 50 km/h on smooth horizontal roads. A mass of 250 kg is dropped into it. The velocity with which it moves now is –
(a) 12.5 km/h
(b) 20 km/h
(c) 40 km/h
(d) 50 km/h
Answer:
(c) According to law of conservation of linear momentum
m₂ v₂ = (m₁ + m₂)v₂
v₂ = \(\frac{m_{1} v_{1}}{m_{1}+m_{2}}\) = \(\frac{1000 \times 50}{1250}\) = 40 km/h

Question 103.
A body of mass loo g is sliding from an inclined plane of inclination 30°. if u = 1.7, then the frictional force experienced is –
(a) \(\frac{3.4}{\sqrt{3}}\)N
(b) 1.47 N
(c) \(\frac{\sqrt{3}}{3.4}\)N
(d) 1.38 N
Answer:
(b) Frictional force F = µ mg cos θ = 1.7 x 0.1 x 10 cos 30°= \(\frac{1.7}{2}\) x \(\sqrt{3}\) = 1.47 N

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (1 Mark)

Question 1.
A passenger sitting in a car at rest, pushes the car from within. The car doesn’t move, why?
Answer:
For motion, there should be external force.

Question 2.
Give the magnitude and directions of the net force acting on a rain drop falling with a constant speed.
Answer:
as \(\overline{\mathrm{a}}\) = 0 so \(\overline{\mathrm{F}}\) = 0.

Question 3.
Why the passengers in a moving car are thrown outwards when it suddenly takes a turn?
Answer:
Due to inertia of direction.

Question 4.
You accelerate your car forward. What is the direction of the frictional force on a package resting on the floor of the car?
Answer:
The package in the accelerated car (a non inertial frame) experiences a Pseudo force in a direction opposite to that of the motion of the car. The frictional force on the package which acts opposite to this pseudo force is thus in the same direction (forward) as that of the car.

Question 5.
What is the purpose of using shockers in a car?
Answer:
To decrease the impact of force by increasing the time for which force acts.

Question 6.
Why are types made of rubber not of steel?
Answer:
Since coefficient of friction between rubber and road is less than the coefficient of friction between steel and road.

Question 7.
Wheels are made circular. Why?
Answer:
Rolling friction is less than sliding friction.

Question 8.
If a ball is thrown up in a moving train, it comes back to the thrower’s hands. Why?
Answer:
Both during its upward and downward motion, the ball continues to move inertia of motion with the same horizontal velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it comes back to the thrower’s hand.

Question 9.
Calculate the force acting on a body which changes the momentum of the body at the rate of 1 kg-m/s² .
Answer:
As F = rate change of momentum
F = 1 kg-m/s² = 1N

Question 10.
On a rainy day skidding takes place along a curved path. Why?
Answer:
As the friction between the types and road reduces on a rainy day.

Question 11.
Why does a gun recoils when a bullet is being fired?
Answer:
To conserve momentum.

Question 12.
Why is it difficult to catch a cricket ball than a tennis ball even when both are moving with the same velocity?
Answer:
Being heavier, cricket ball has higher rate of change of momentum during motion so more force sumed.

Question 13.
The distance travelled by a moving body is directly proportional to time. Is any external force acting on it?
Answer:
As s ∝ t, so acceleration a = 0, therefore, no external force is acting on the body.

Question 14.
Calculate the impulse necessary to stop a 1500 kg car moving at a speed of 25 ms^-1.
Answer:
Use formula I = change in momentum = m(v – u) (Impulse – 37500 Ns)

Question 15.
Lubricants are used between the two parts of a machine. Why?
Answer:
To reduce friction and so to reduce wear and tear.

Question 16.
What provides the centripetal force to a car taking a turn on a level road?
Answer:
Force of friction between the type and road provides centripetal force.

Question 17.
A body is acted upon by a number of external forces. Can it remain at rest?
Answer:
Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Question 18.
Bodies of larger mass need greater initial effort to put them in motion. Why?
Answer:
As F = ma so for given a, more force will be required to put a large mass in motion.

Question 19.
An athlete runs a certain distance before taking a long jump Why?
Answer:
So that inertia of motion may help him in his muscular efforts to take a longer jump.

Question 20.
Action and reaction forces do not balance each other. Why?
Answer:
As they acts on different bodies.

Question 21.
The wheels of vehicles are provided with mudguards. Why?
Answer:
When the wheel rotates at a high speed, the mud sticking to the wheel flies off tangentially, this is due to inertia of direction. If order that the flying mud does not spoil the clothes of passer by the wheels are provided with mudguards.

Question 22.
China wares are wrapped in straw paper before packing. Why?
Answer:
The straw paper between the China ware increases the Time of experiencing the jerk during transportation. Hence impact of force reduces on China wares.

Question 23.
Why is it difficult to walk on a sand?
Answer:
Less reaction force.

Question 24.
The outer edge of a curved road is generally raised over the inner edge Why?
Answer:
In addition to the frictional force, a component of reaction force also provides centripetal force.

Question 25.
Explain why the water doesn’t fall even at the top of the circle when the bucket full of water is upside down rotating in a vertical circle?
Answer:
Weight of the water and bucket is used up in providing the necessary centripetal force at the top of the circle.

Question 26.
Why does a speedy motor cyclist bends towards the center of a circular path while taking a turn on it?
Answer:
So that in addition of the frictional force, the horizontal component of the normal reaction also provides the necessary centripetal forces.

Question 27.
An impulse is applied to a moving object with a force at an angle of 20° wr.t. velocity vector, what is the angle between the impulse vector and change in momentum vector ?
Answer:
Impulse and change in momentum are along the same direction. Therefore angle between these two vectors is zero.

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (2 Marks)

Question 28.
A man getting out of a moving bus runs in the same direction for a certain distance. Comment.
Answer:
Due to inertia of motion.

Question 29.
If the net force acting upon the particle is zero, show that its linear momentum remains constant.
Answer:
As F x \(\frac {dp}{dt}\)
when F = 0, \(\frac {dp}{dt}\) = 0 so P = constant

Question 30.
A force of 36 dynes is inclined to the horizontal at an angle of 60°. Find the acceleration in a mass of 18 g that moves in a horizontal direction.
Answer:
F = 36 dyne at an angle of 60°
F_x = F cos 60° = 18 dyne
F_x = ma_x
So a_x = \(\frac{F_{x}}{m}\) = 1 cm /s²

Question 31.
The motion of a particle of mass m is described by h = ut + \(\frac {1}{2}\) gt². Find the force acting on particle.
Answer:
a = ut + \(\frac {1}{2}\) gt²
find a by differentiating h twice w.r.t.
a = g
As F = ma so F = mg (answer)

Question 32.
A particle of mass 0.3 kg is subjected to a force of F = -kx with k= 15 Nmr^-1. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
Answer:
As F = ma so F = -kx = ma
a = \(\frac {-kx}{m}\)
for x = 20 cm, ⇒ a = -10 m/s².

Question 33.
A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms-¹. What is the speed of the recoil of the gun?
Answer:
Initial momentum = 0
Using conservation of linear momentum mv + MV = 0
V = \(\frac {-mv}{M}\) ⇒ V = 2.5 m/s

Question 34.
Smooth block is released at rest on a 45° incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction, µ = \(\left(1-\frac{1}{n^{2}}\right)\)
Answer:
When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ
when there is friction, the downward acceleration of the block is a’ = g (sin θ – µ cos θ)
As the block Slides a distance d in each case so
d = \(\frac {1}{2}\) at² = \(\frac {1}{2}\) a’t’²
\(\frac{a}{a^{\prime}}=\frac{t^{\prime 2}}{t^{2}}=\frac{(n t)^{2}}{t^{2}}\) = n²
or \(\frac {g sin θ}{g(sin θ – µ cos θ)}\) = n²
Solving, we get (Using θ = 45°)
µ = 1 – \(\frac{1}{n^{2}}\)

Question 35.
A spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads 49 N of a body hang on it. If the lift moves:

1. Downward
2. upward, with an acceleration of 5 ms²
3. with a constant velocity.

What will be the reading of the balance in each case?
Answer:
1.  R = m(g – a) = 49 N
so = m = \(\frac {49}{9.8}\) = 5 kg
R = 5 (9.8 – 5)
R = 24 N

2. R = m(g + a)
R = 5 (9.8 + 5)
R = 74 N

3.  as a = 0 so R = mg = 49 N

Question 36.
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms^-1. What is the trajectory of the ‘oscillating bob if the string is cut when the bob is –

1. At the mean position
2. At its extreme position.

Answer:

1. Parabolic
2. vertically downwards

Question 37.
A block placed on a rough horizontal surface is pulled by a horizontal force F. Let f be the force applied by the rough surface on the block. Plot a graph of f versus F.
Answer:

Unto point A, f = F (50 Long as block is stationary) beyond A, when F increases, block starts moving f remains constant.

Question 38.
A mass of 2 kg is suspended with thread AB. Thread CD of the same type is attached to the other end of 2 kg mass.

• Lower end of the lower thread is pulled gradually, hander and hander is the downward direction so as to apply force on AB Which of the thread will break & why?
• If the lower thread is pulled with a jerk, what happens?

Answer:

• Thread AB breaks down
• CD will break.

Question 39.
A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is p and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to held the block against the wall?
Answer:

For the block not to fall f = Mg
But f = µR = µF so
µF = Mg
F = \(\frac {Mg}{µ}\)

Samacheer Kalvi 11th Physics Laws of Motion Short Answer Questions (3 Marks) & Numericals

Question 40.
A block of mass 500 g is at rest on a horizontal table. What steady force is required to give the block a velocity of 200 cm s^-2 in 4 s?
Answer:
Use F – ma
a = \(\frac {v – u}{ t }\) = \(\frac {200- 0}{ 4 }\) = 50 cm/s²
F = 500 x 50 = 25,000 dyne.

Question 41.
A force of 98 N is just required to move a mass of 45 kg on a rough horizontal surface. Find the coefficient of friction and angle of friction?
Answer:
F = 48 N,R = 45 x 9.8 = 441 N
µ = \(\frac {F’}{ R}\) = 0.22
Angle of friction θ = tan^-1 0.22 = 12°24′

Question 42.
Calculate the force required to move a train of 2000 quintal up on an incline plane of 1 in 50 with an acceleration of 2 ms^-2. The force of friction per quintal is 0.5 N.
Answer:
Force of friction = 0.5 N per quintal
f = 0.5 x 2000 = 1000 N
m = 2000 quintals = 2000 x 100 kg
sin θ = \(\frac {1}{50}\), a – 2 m/s²
In moving up an inclined plane, force required against gravity
mg sin θ = 39200 N
And force required to produce acceleration = ma
= 2000 x 100 x 2 = 40,0000 N
Total force required = 1000 + 39,200 + 40,0000 = 440200 N.

Question 43.
A force of 100 N gives a mass m₁, an acceleration of 10 ms^-2 and of 20 ms^-2 to a mass m₂.
What acceleration must be given to it if both the masses are tied together?
Answer:
Suppose, a = acceleration produced if m₁ and m₂ are tied together,
F = 100 N
Let a₁ and a₂ be the acceleration produced in m₁ and m₂ respectively.
∴ a₁ and a₂ = 20ms^-2 (given)
Again m₁ = \(\frac{\mathrm{F}}{a_{1}}\) and m₂ = \(\frac{\mathrm{F}}{a_{2}}\)
⇒ m₁ = \(\frac {100}{10}\) = 10kg
and m₂ = \(\frac {100}{20}\) = 5kg
∴ m₁ + m₂ = 10 + 5 = 15
so, a = \(\frac{\mathbf{F}}{m_{1}+m_{2}}\) = \(\frac {100}{15}\) = \(\frac {20}{3}\) = 6.67 ms²

Question 44.
The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. In which case, the acceleration of m is more?

Answer:
Case (a):
a = \(\frac {2m – m}{2m + m}\) g = a = \(\frac {g}{3}\)
Case (b):
FBD of mass m
ma’ = T – mg
ma’ = 2 mg – mg
⇒ ma’ = mg
a’ = g
So in case (b) acceleration of m is more.

Question 45.
Figure shows the position-time graph of a particle of mass 4 kg. What is the

(a) Force on the particle for t < 0, t > 4s, 0 < t < 4s?
(b) Impulse at t = 0 and t = 4s?
(Consider one dimensional motion only)
Answer:
(a) For t < 0. No force as Particles is at rest. For t > 4s, No force again particle comes at rest.
For 0 < t < 4s, as slope of OA is constant so velocity constant i.e., a = 0, so force must be zero.

(b) Impulse at t = 0
Impulse = change in momentum
I = m(v – w) = 4(0 – 0.75) = 3 kg ms^-1
Impulse at t = 4s
1 = m(v – u) = 4 (0 – 0.75) = -3 kg ms^-1

Question 46.
What is the acceleration of the block and trolley system as the figure, if the coefficient of kinetic friction between the trolley and the surface is 0.04? Also Calculate friction in the string: Take g = 10 m/s², mass of the string is negligible.

Answer:
Free body diagram of the block
30 – T = 3a
Free body diagram of the trolley

T – f_k = 20 a ………….(2)
where f_k = µ_k= 0.04 x 20 x 10 = 8 N
Solving (i) & (ii), a = 0.96 m/s² and T = 27.2 N

Question 47.
Three blocks of masses ml = 10 kg, m² = 20 kg are connected by strings on smooth horizontal surface and pulled by a force of 60 N. Find the acceleration of the system and frictions in the string.

Solution:
All the blocks more with common acceleration a under the force F = 60 N.
F = (m₁ + m₂ + m₃)a
a = \(\frac{\mathrm{F}}{\left(m_{1}+m_{2}+m_{3}\right)}\) = 1 m/s²
to determine, T₁ →Free body diagram of m₁.

T₁ = m₁a = 10 x 1 = 10 N
to determine, T₂ →Free body diagram of m₃

F – T₂ = m₃a
Solving, we get T₂ = 30 N

Question 48.
The rear side of a truck is open and a box of 40 kg mass is placed 5m away from the open end. The coefficient of friction between the box and the surface below it is 0.15 on a straight road, the truck starts from rest and accelerates with 2 m/s². At what distance from the starting point does the box fall off the truck ? (ignore the size of the box)
Answer:
Force on the box due to accelerated motion of the truck
F = ma = 40 x 2 = 80 N (in forward direction)
Reaction on the box, F’ = F = 80 N (in backward direction)
Force of limiting friction, f = µR = 0 .15 x 40 x 10 = 60 N
Net force on the box in backward direction is P = F’ f = 80 – 60 = 20 N
Backward acceleration in the box = a= \(\frac {p}{m}\) = \(\frac {20}{40}\) = 0.5 ms^-2
t = time taken by the box to travel s = 5 m and falls off the truck, then from
s = ut + \(\frac {1}{2}\) at²
5 = 0 x t + \(\frac {1}{2}\) x 0.5 x t²
t = 4.47
If the truck travels a distance x during this time
then x = 0 x 4.34 +\(\frac {1}{2}\) x 2 x (4.471)²
x = 19.98 m

Question 49.
A block slides down as incline of 30° with the horizontal. Starting from rest, it covers 8 m in the first 2 seconds. Find the coefficient of static friction.
Use s = ut + \(\frac {1}{2}\) at²
a = \(\frac{2 s}{t^{2}}\) at² as u = 0
µ = \(\frac{g sin θ – a}{g Cos θ }\)
Putting the value and solving, µ = 0.11

Question 50.
A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m/s² . The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the:
(a) Force on the floor of the helicopter by the crew and passenger.
(b) Action of the rotor of the helicopter on the surrounding air
(c) Force on the helicopter due to the surrounding air (g = 10 m/s² )
Answer:
(a) Force on the floor of the helicopter by the crew and passengers
= apparent weight of crew and passengers
= 500(10+ 15)
=12500 N

(b) Action of rotor of helicopter on surrounding air is Obviously vertically downwards, because helicopter rises on account of reaction of this force. Thus force of action
= (2000 + 500) (10 + 15)
= 2500 x 25
= 62,500 N

(c) Force on the helicopter due to surrounding air is obviously a reaction. As action and reaction are equal and opposite, therefore
Force of reaction F’ = 62,500 vertically upwards.

Question 51.
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is (µ). Let the mass of the box be m.

1.  At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane ?
2. What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a > θ.
3. What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ?
4. What is the force needed to be applied upwards along the plane to pk kg f make the box move up the plane with acceleration a ?

Answer:
1. When the box just starts sliding
µ = tanθ
or 0 = tan^-1 µ

2. Force acting on the box down the plane
= mg (sin a – µ cos a)

3. Force needed mg (sin a + µ cos a)

4. Force needed = mg (sin a + µ cos a) + ma.

Question 52.
Two masses of 5 kg and 3 kg are suspended with help of mass less in extensible string as shown. Calculate T₁ and T₂ when system is going upwards with acceleration m/s². (Use g 9.8 m/s²)
Answer:
According Newton’s second law of motion
(1) T₁ – (m₁ + m₂)g = (m₁ + m₂)a
T₁ = (m₁ + m₂)(a + g) = (5 + 3) (2 + 9.8)
T₁ = 94.4 N

(2) T₂ – m₂g = m₂a
T₂ = m₂ (a + g)
T₂ = 3(2 + 9.8)
T₂ = 35.4 N

Question 53.
There are few forces acting at a Point P produced by strings as shown, which is at rest. Find the forces F₁ & F₁

Answer:
Using Resolution of forces IN and 2N and then applying laws of vector addition. Calculate for F₁ & F₁.
F₁ = \(\frac{1}{\sqrt{2}}\) N, F₂ = \(\frac{3}{\sqrt{2}}\)N

Question 54.
A hunter has a machine gun that can fire 50g bullets with a velocity of 150 ms A 60 kg tiger springs at him with a velocity of 10 ms^-1. How many bullets must the hunter fire into the target so as to stop him in his track?
Answer:
Given m = mass of bullet = 50 gm = 0.50 kg
M = mass of tiger = 60 kg
v = Velocity of bullet – 150 m/s
V = Velocity of tiger = – 10 m/s
(v It is coming from opposite direction n = no. of bullets fired per second at the tiger so as to stop it.)
P_i = 0, before firing ……..(i)
P_f = n (mv) + MV …………(ii)
∴ From the law of conservation of momentum,
P_i = P_f
⇒ 0 = n (mv) + MV
n = \(\frac{MV}{mv}\) = \(\frac{-60 \times(-10)}{0.05 \times 150}\) = 80

Question 55.
Two blocks of mass 2 kg and 5 kg are connected by an ideal string passing over a pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° with the horizontal whereas 5 kg block hangs freely. Find the acceleration of the system and the tension in the string.

Let a be the acceleration of the system and T be the Tension in the string. Equations of motions for 5 kg and 2 kg blocks are
5g – T = 5a
T – 2g sin θ – f = 2a
where f = force of limiting friction
= µR = µ mg cos θ = 0.3 x 2 g x cos 30°
Solving (1) & (2)
a = 4.87 m/s²

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Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 5.3 நாட்டுப்புறக் கைவினைக் கலைகள் Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 5.3 நாட்டுப்புறக் கைவினைக் கலைகள்

கற்பவை கற்றபின்

Question 1.
உங்கள் பகுதியில் கிடைக்கும் களிமண், பனையோலை போன்ற பொருள்களைப் பயன்படுத்திக் கைவினைப் பொருள்களைச் செய்து காட்சிப்படுத்துக.
Answer:

Question 2.
பனையோலையால் செய்யப்படும் பல்வேறு கைவினைப்பொருள்களின் படங்களைச் சேகரித்து படத்தொகுப்பு உருவாக்குக.
Answer:

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
பழந்தமிழ் இலக்கியங்களைப் பாதுகாத்து வைத்தவை ……………………
அ) கல்வெட்டுகள்
ஆ) செப்பேடுகள்
இ) பனையோலைகள்
ஈ) மண்பாண்டங்கள்
Answer:
இ) பனையோலைகள்

Question 2.
பானை ………………….. ஒரு சிறந்த கலையாகும்.
அ) செய்தல்
ஆ) வனைதல்
இ) முடைதல்
ஈ) சுடுதல்
Answer:
ஆ) வனைதல்

Question 3.
‘மட்டுமல்ல’ எனும் சொல்லைப் பிரித்து எழுதக் கிடைப்பது ……………………
அ) மட்டு + மல்ல
ஆ) மட்டம் + அல்ல
இ) மட்டு + அல்ல
ஈ) மட்டும் + அல்ல
Answer:
ஈ) மட்டும்+அல்ல

Question 4.
கயிறு + கட்டில் என்பதனைச் சேர்த்தெழுதக் கிடைக்கும் சொல் …………………….
அ) கயிற்றுக்கட்டில்
ஆ) கயிர்க்கட்டில்
இ) கயிறுக்கட்டில்
ஈ) கயிற்றுகட்டில்
Answer:
அ) கயிற்றுக்கட்டில்

பின்வரும் சொற்களைச் சொற்றொடரில் அமைத்து எழுதுக.

1. முழுவதும் – பாடநூல் முழுவதும் வாசித்தால்தான் தெளிவு கிடைக்கும்.
2. மட்டுமல்லாமல் – ஏட்டுக் கல்வி மட்டுமல்லாமல் தொழில் கல்வியும் கற்க வேண்டும்.
3. அழகுக்காக – பல அரங்குகளில் சுடுமண் சிற்பங்களை அழகுக்காக வைத்திருப்பார்கள்.
4. முன்பெல்லாம் – முன்பெல்லாம் மண்பாண்டங்களை அதிகம் பயன்படுத்தினார்கள்.

குறுவினா

Question 1.
எவற்றையெல்லாம் கைவினைக்கலைகள் எனக் கூறுகிறோம்?
Answer:
மண் பொம்மைகள் செய்தல், மரப்பொம்மைகள் செய்தல், காகிதப்பொம்மைகள் செய்தல், தஞ்சாவூர்த்தட்டு செய்தல், சந்தன மாலையும் ஏலக்காய் மாலையும் செய்தல், மாட்டுக்கொம்பினால் கலைப் பொருட்கள் செய்தல் ஆகியவற்றை எல்லாம் கைவினைக் கலைகள் எனக் கூறுகிறோம்.

Question 2.
மண்பாண்டம், சுடுமண் சிற்பம் – ஒப்பிடுக.
Answer:

Question 3.
பனையோலையால் உருவாக்கப்படும் பொருள்கள் யாவை?
Answer:
குழந்தைகளுக்கான கிளுகிளுப்பை பொம்மைகள், பொருள்களை வைத்துக்கொள்ள உதவும் சிறிய கொட்டான், பெரிய கூடை, சுளகு, விசிறி, தொப்பி, ஓலைப்பாய், பனை மட்டை நாரிலிருந்து கயிறு, கட்டில், கூடை போன்றவை செய்யப்படுகின்றன.

சிறுவினா

Question 1.
பிரம்பினால் பொருள்கள் செய்யும் முறையைக் கூறுக.
Answer:
(i) பிரம்பு என்பது ஒரு தாவரம். முதலில் பிரம்புகளை நெருப்பில் காட்டிச் சூடுபடுத்த வேண்டும்.

(ii) சூடான பிரம்பை நட்டு வைத்திருக்கும் இரண்டு பாறைகளுக்கு இடையே செலுத்தி வளைக்க வேண்டும். அது வேண்டிய வடிவத்தில் கம்பி போல வளையும்.

(iii) பின்னர் அதனை தண்ணீரில் நனைத்து வைத்து விட்டால், அப்படியே நிலைத்து விடும். பிறகு அவற்றை இணைத்துச் சிறு ஆணிகளை அறைந்தும், சிறு பிரம்பு இழைகளைக் கொண்டும் தேவையான பொருட்களாக மாற்ற வேண்டும்.

Question 2.
மூங்கிலால் செய்யப்படும் பொருள்கள் குறித்து எழுதுக.
Answer:
மட்டக்கூடை, தட்டுக்கூடை, கொட்டுக்கூடை, முறம், ஏணி, சதுரத்தட்டி, கூரைத்தட்டி, தெருக்கூட்டும் துடைப்பம், மாடுகளுக்கான மூஞ்சிப்பெட்டி, பழக்கூடை, பூக்கூடை, பூத்தட்டு, கட்டில், புல்லாங்குழல், புட்டுக்குழாய், கால்நடைகளுக்கு மருந்து புகட்டும் குழாய், தொட்டில், பாடை ஆகியவை அனைத்தும் மூங்கிலால் செய்யப்படும் பொருள்கள் ஆகும்.

நெடுவினா

Question 1.
தமிழகக் கைவினைக் கலைகளைப் பற்றிய செய்திகளைத் தொகுத்து எழுதுக.
Answer:
மண்பாண்டக் கலை :

• குடம், தோண்டி, கலயம், கடம், மூடி, உழக்கு, அகல், உண்டியல், அடுப்பு, தொட்டி ஆகிய அனைத்துப் பொருட்களும் சுத்தமான களிமண்ணால் செய்யப்பட்டவை.
• பக்குவப்படுத்தப்பட்ட களிமண், மெல்லிய மணல் சாம்பல் ஆகியவற்றைக் கலந்து எடுத்துக் கொள்ள வேண்டும்.
• சக்கரத்தின் நடுவே வைத்து உரிய வடிவத்தால் அதைக் கொண்டு வர வேண்டும். பிறகு அடிப்பகுதியில் நூல் அல்லது ஊசியால் அறுத்து எடுத்து காயவைக்க வேண்டும். பிறகு உரிய மண்பாண்டம் தயாராகிவிடும்.
• மண்பாண்டங்களில் சமைத்த உணவு உடலுக்கு நல்லது.
• திருவிழாக் காலங்களிலும் சமயச் சடங்குகளிலும் மண்பானைகள் இன்றுவரை பயன்படுத்தப்பட்டு வருகின்றன.

மூங்கில் கலை :
(i) மூங்கில் கொண்டு பல கைவினைப் பொருட்கள் செய்யப்படுகின்றன.

(ii) மட்டக்கூடை, தட்டுக்கூடை, கொட்டுக்கூடை, முறம், ஏணி, சதுரத்தட்டி, கூரைத்தட்டி, தெருக்கூட்டும் துடைப்பம், மாடுகளுக்கான மூஞ்சிப்பெட்டி, பழக்கூடை, பூக்கூடை, பூத்தட்டு, கட்டில், புல்லாங்குழல், புட்டுக்குழாய், கால்நடைகளுக்கு மருந்து புகட்டும் குழாய், தொட்டில், பாடை ஆகிய அனைத்தும் மூங்கிலால் செய்யப்படும் பொருள்கள் ஆகும்.

(iii) முன்பு எல்லாம் திருமணத்தின் போது சீர்த்தட்டுகளாகப் பயன்படுத்தினர்

பனையோலைக் கலை :
(i) பனையோலையில் பல கைவினைப் பொருட்கள் உருவாக்கப்படுகின்றன.

(ii) குழந்தைகளுக்கான கிளுகிளுப்பை பொம்மைகள், பொருள்களை வைத்துக் கொள்ள உதவும் சிறிய கொட்டான், பெரிய கூடை, சுளகு, விசிறி, தொப்பி, ஓலைப்பாய், பனை மட்டை நாரிலிருந்து கயிறு, கட்டில், கூடை போன்றவை செய்யப்படுகின்றன.

பிரம்புக் கலை :
(i) பிரம்பு என்பது ஒரு தாவரம்.

(ii) முதலில் பிரம்புகளை நெருப்பில் காட்டி சூடுபடுத்த வேண்டும். சூடான பிரம்பை நட்டு வைத்திருக்கும் இரண்டு பாறைகளுக்கு இடையே செலுத்தி வளைக்க வேண்டும்.

(iii) அது வேண்டிய வடிவத்தில் கம்பி போல வளையும். பின்னர் அதனைத் தண்ணீரில் நனைத்து வைத்து விட்டால், அப்படியே நிலைத்து விடும். பிறகு அவற்றை இணைத்துச் சிறு ஆணிகளை அறைந்தும், சிறு பிரம்பு இழைகளைக் கொண்டு கூட்டியும் தேவையான பொருட்களாக மாற்ற வேண்டும்.

(iv) பிரம்பு மிகவும் குளிர்ச்சியானது. எனவே அதில் அமர்வது உடல்நலத்துக்கு நல்லது.
(v) மேலும் பிரம்புப்பொருள் வீட்டுக்கு அழகையும் கொடுக்கும்.

சிந்தனை வினா

Question 1.
கைவினைக் கலைகளுக்கும் சுற்றுச்சூழல் பாதுகாப்பிற்கும் இடையேயுள்ள தொடர்பு குறித்து எழுதுக.
Answer:
(i) கைவினைப் பொருட்கள் அனைத்தும் இயற்கையான பொருளால் தயாரிக்கப் படுபவை.

(ii) செயற்கையான பொருளோ தீங்கு விளைவிக்கும் இரசாயனமோ இதில் பயன்படுத்தப்படுவது இல்லை.

(iii) இயற்கையாகக் கிடைக்கும் களிமண், பனை ஓலை, மூங்கில், பிரம்பு ஆகியவற்றை முதன்மைப் பொருளாகக் கொண்டு கைவினைப் பொருட்கள் செய்யப்படுகின்றன.

(iv) கைவினைக் கலைகளுக்குப் பயன்படுத்தப்படக்கூடிய பொருட்களின் மீதத்தைப் பூமியில் புதைத்தாலும், அவை மக்கி விடும். இதனால் சுற்றுப்புறத்திற்கு எந்தத் தீங்கும் ஏற்படாது.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
பானை ஓடுகள் கிடைத்துள்ள இடம் …………………..
அ) சிந்துசமவெளி
ஆ) ஆதிச்சநல்லூர்
இ) செம்பியன் கண்டியூர்
ஈ) கீழடி
Answer:
அ) சிந்துசமவெளி

Question 2.
முதுமக்கள் தாழிகள் கிடைத்துள்ள தமிழக இடம் ……………………..
அ) சிந்துசமவெளி
ஆ) ஆதிச்சநல்லூர்
இ) செம்பியன் கண்டியூர்
ஈ) கீழடி
Answer:
ஆ) ஆதிச்சநல்லூர்

Question 3.
மிகவும் பழமையான கைவினைக் கலைகளில் ஒன்று ………………….
அ) பிரம்புக் கலை
ஆ) மண்பாண்டக் கலை
இ) பனை ஓலைக்கலை
ஈ) மூங்கில் கலை
Answer:
ஆ) மண்பாண்டக் கலை

Question 4.
குளங்கள், ஆற்றங்கரைகள், வயல்வெளிகள் ஆகிய இடங்களில் கிடைக்கும் ……………………
அ) செம்மண்
ஆ) களிமண்
இ) வண்டல் மண்
ஈ) உவர்மன்
Answer:
ஆ) களிமண்

Question 5.
பானை செய்யும் சக்கரத்தின் வேறு பெயர் ………………..
அ) ஊசி
ஆ) நூல்
இ) திருவை
ஈ) சுழல்
Answer:
இ) திருவை

Question 6.
குழந்தைகளைப் படுக்க வைக்க உதவும் பாய் …………………..
அ) பந்திப்பாய்
ஆ) திண்ணைப் பாய்
இ) பட்டுப் பாய்
ஈ) தடுக்குப் பாய்
Answer:
ஈ) தடுக்குப் பாய்

Question 7.
உட்காரவும் படுக்கவும் உதவும் ………………… பாய்.
அ) பந்திப்பாய்
ஆ) திண்ணைப் பாய்
இ) பட்டுப் பாய்
ஈ) தடுக்குப் பாய்
Answer:
ஆ) திண்ணைப் பாய்

Question 8.
“முற்காலத்தில் பாய்மரக் கப்பல்களில் பயன்படுத்தியது கூட பாய் தான்” என்பதைக் கூறும் நூல் ………………….
அ) அகநானூறு
ஆ) பரிபாடல்
இ) புறநானூறு
இ) பதிற்றுப்பத்து
Answer:
இ) புறநானூறு

Question 9.
தமிழ்நாட்டின் மாநில மரம் ……………………
அ) தென்னை மரம்
ஆ) மாமரம்
இ) பனை மரம்
ஈ) மூங்கில் மரம்
Answer:
இ) பனை மரம்

Question 10.
பிரம்பு என்பது …………………. வகையைச் சேர்ந்த தாவரம்.
அ) செடி
ஆ) கொடி
இ) மரம்
ஈ) நீர்நிலை
Answer:
ஆ) கொடி

Question 11.
கலாமஸ் ரொடாங் என்னும் தாவரவியல் பெயர் கொண்டது …………………
அ) பிரம்பு
ஆ) மூங்கில்
இ) மண்பாண்டம்
ஈ) பனையோலை
Answer:
அ) பிரம்பு

குறுவினா

Question 1.
கைவினைக் கலைகள் என்றால் என்ன?
Answer:
அன்றாடப் பயன்பாட்டுக்காக அழகிய பொருள்களைத் தொழில் முறையில் உருவாக்கும் கலையைக் கைவினைக் கலை என்பர்.

Question 2.
மூங்கில் வகைகள் யாவை?
Answer:

• கல் மூங்கில்
• மலை மூங்கில்
• கூட்டு மூங்கில்

என மூங்கில் மூன்று வகைப்படும்.

Question 3.
கைவினைப் பொருட்கள் செய்யப் பயன்படும் மூங்கில் எவை?
Answer:
கூட்டு மூங்கிலே கைவினைப் பொருட்கள் செய்யப் பயன்படும் மூங்கில் ஆகும்.

Question 4.
மண்பாண்டப் பொருட்கள் யாவை?
Answer:
பானை, சட்டி, குடம், தோண்டி, கலயம், கடம், மூடி, உழக்கு, அகல், உண்டியல், அடுப்பு, தொட்டி ஆகியன மண்பாண்டப் பொருட்கள் ஆகும்.

Question 5.
களிமண் கிடைக்கும் இடங்கள் யாவை?
Answer:
குளங்கள், ஆற்றங்கரைகள் மற்றும் வயல்வெளிகள் ஆகியவற்றில் களிமண் கிடைக்கும்.

Question 6.
திருவை என்றால் என்ன?
Answer:
மண் பானை செய்யும் சக்கரத்தையே திருவை என்பர்.

Question 7.
மண்பாண்டங்களால் கிடைக்கும் நன்மைகள் யாவை?
Answer:

• மண்பாண்டங்களில் சமைத்த உணவு நல்ல சுவையுடன் இருக்கும்.
• மண்பாண்டத்தில் செய்யப்படும் உணவே உடல் நலத்திற்கும் நல்லது.
• மண் பானையில் வைத்த தண்ணீர் குளிர்ச்சியாக இருக்கும்.

Question 8.
சுடுமண் சிற்பங்கள் என்றால் என்ன?
Answer:
மண்பாண்டங்களைப் போன்றே களிமண்ணால் செய்யப்பட்டுச் சூளையில் சுட்டு எடுக்கப்படுபவை சுடுமண் சிற்பங்கள் ஆகும்.

Question 9.
பாயின் வகைகள் யாவை?
Answer:

• படுக்கும் பாய்
• பந்திப் பாய்
• திண்ணைப் பாய்
• பட்டுப்பாய்
• தொழுகைப் பாய்

Question 10.
பிரம்பு தருவிக்கப்படும் இடங்கள் யாவை?
Answer:
அஸ்ஸாம், அந்தமான், மலேசியா ஆகிய இடங்களில் இருந்து பிரம்பு தருவிக்கப் படுகிறது.

Question 11.
பிரம்புப் பொருட்களால் கிடைக்கும் நன்மைகள் யாவை?
Answer:

• பிரம்பு மிகவும் குளிர்ச்சியானது. எனவே, அதில் அமர்வது உடல்நலத்துக்கு நல்லது.
• மேலும், பிரம்புப் பொருள் வீட்டுக்கு அழகையும் கொடுக்கும்.

சிறுவினா

Question 1.
தமிழருக்கும் மண்பாண்டக் கலைக்கும் உள்ள தொடர்பைக் காட்டும் சான்றுகள் யாவை?
Answer:

• சிந்து சமவெளி அகழாய்வில் பானை ஓடுகள் கிடைத்துள்ளன.
• தமிழ்நாட்டில் ஆதிச்சநல்லூரில் முதுமக்கள் தாழிகள் கிடைத்துள்ளன.
• நாகை மாவட்டம் செம்பியன் கண்டியூரில் கலையழகு மிகுந்த மண்கலங்கள் கண்டுபிடிக்கப்பட்டுள்ளன.
• மதுரைக்கு அருகில் உள்ள கீழடியில் ஏராளமான சுடுமண் பொருட்கள் கிடைத்துள்ளன.
• இவையெல்லாம் தமிழருக்கும் மண்பாண்டக் கலைக்கும் உள்ள தொடர்பைக் காட்டும் சான்றுகளாகும்.

Question 2.
பாயின் வகைகளை விளக்குக.
Answer:

• குழந்தைகளைப் படுக்க வைப்பது தடுக்குப் பாய்.
• உட்கார்ந்து உண்ண உதவுவது பந்திப் பாய்.
• உட்காரவும் படிக்கவும் உதவுவது திண்ணைப் பாய்.
• திருமணத்துக்குப் பயன்படுவது பட்டுப் பாய்.
• இஸ்லாமியர் தொழுகைக்குப் பயன்படுத்துவது தொழுகைப் பாய்.

Question 3.
பாய்களில் எவையெவை இடம்பெற்றிருந்தன?
Answer:
(i) திருமணத்திற்குப் பயன்படுத்தும் பாய்களில் மணமக்கள் பெயர்கள் இடம் -5 பெற்றிருந்தன.

(ii) குத்துவிளக்கு, மயில், பூக்கள், வழிபாட்டுச் சின்னங்கள் ஆகியவையும் பாய்களில் இடம்பெற்றிருந்தன.

Question 4.
பிரம்பினால் செய்யப்பட்ட பலவகைப் பொருட்கள் யாவை?
Answer:

• கட்டில்
• ஊஞ்சல்
• நாற்காலி
• மேசை
•  பூக்கூடை
• பழக்கூடை
• இடியாப்பத் தட்டு
• அர்ச்சனைத் தட்டு
• வெற்றிலைப் பெட்டி

Students can Download Maths Chapter 1 Rational Numbers Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Ex 1.1

8th Maths Exercise 1.1 Samacheer Kalvi Question 1.
Fill in the blanks:
(i) \(\frac{-19}{5}\) lies between the integers _____ and _____
(ii) The rational number that is represented by 0.44 is ______.
(iii) The standard form of \(\frac{+58}{-78}\) is _____.
(iv) The value of \(\frac{-5}{12}+\frac{7}{15}\) = ______
(v) The value of \(\left(\frac{-15}{23}\right) \div\left(\frac{+30}{-46}\right)\) is ______
Solution:
(i) -4 and -3
(ii) \(\frac{11}{25}\)
(iii) \(\frac{-29}{39}\)
(iv) \(\frac{1}{20}\)
(v) 1

8th Maths Exercise 1.1 In Tamil Question 2.
Say True or False.
(i) 0 is the smallest rational number.
(ii) There are an unlimited rationals between 0 and 1.
(iii) The rational number that does not have a reciprocal is 0.
(iv) The only rational number which is its own reciprocal is -1.
(v) The rational numbers that are equal to their additive inverses are 0 and -1.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) False

8th Maths Exercise 1.1 Question 3.
List five rational numbers between 2 and 0
(i) -2 and 0
(ii) \(\frac{-1}{2}\) and \(\frac{3}{5}\)
(iii) 0.25 and 0.35
(iv) -1.2 and -2.3
Solution:
(i) -2 and 0

Samacheer Kalvi 8th Maths Solutions Term 1 Pdf Question 4.
Write four rational numbers equivalent to -3 7
(i) \(\frac{-3}{5}\)
(ii) \(\frac{7}{-6}\)
(iii) \(\frac{8}{9}\)
Solution:

8th Standard Maths Exercise 1.1 Answers Question 5.
Draw the number line and represent the following rational numbers on it.

Solution:

8th Maths Book Example Sums Question 6.
Find the rational numbers for the points marked on the number line.

Solution:
(i) The number lies between -3 and -4. The unit part between -3 and -4 is divided into 3 equal parts and the second part is asked.
∴ The required number is \(-3 \frac{2}{3}=-\frac{11}{3}\).
(ii) The required number lies between 0 and -1. The unit part between 0 and -1 is divided into 5 equal parts, and the second part is taken.
∴The required number is \(-\frac{2}{5}\)
(iii) The required number lies between 1 and 2. The unit part between 1 and 2 is divided into 4 equal parts and the third part is taken.
∴ The required number is \(1 \frac{3}{4}=\frac{7}{4}\)

Samacheer Kalvi.Guru 8th Maths Question 7.
Using average, write 3 rational numbers between \(\frac{14}{5}\) and \(\frac{16}{3}\)
Solution:

Maths 8th Guide Samacheer Kalvi Question 8.
Verify that -(-x) is the same x for:
(i) x = \(\frac{11}{15}\)
(ii) x = \(\frac{-31}{45}\)
Solution:

Samacheer Kalvi 8th Maths Solutions Question 9.
Re-arrange suitable and add :
\(\frac{-3}{7}+\frac{5}{6}+\frac{4}{7}+\frac{1}{3}+\frac{13}{-6}\)
Solution:

8th Maths 1.1 Question 10.
What should be added to \(\frac{-8}{9}\) to get \(\frac{2}{5}\).
Solution:
Let the number to be added = x

8th Std Maths Exercise 1.1 Question 11.
Subtract \(\frac{-8}{44}\) from \(\frac{-17}{11}\)
Solution:

Maths Term 1 Question 12.
Evaluate:
(i) \(\frac{9}{2} \times \frac{-11}{3}\)
(ii) \(\frac{-7}{27} \times \frac{24}{-35}\)
Solution:

Maths 8th Class Chapter 1 Exercise 1.1 Question 13.
Divide
(i) \(\frac{-21}{5}\) by \(\frac{-7}{-10}\)
(ii) \(\frac{-3}{13}\) by -3
(iii) -2 by \(\frac{-6}{15}\)
Solution:

8th Maths In Tamil Question 14.
Simplify \(\left(\frac{2}{5}+\frac{3}{2}\right)+\frac{3}{10}\) as a rational number and show that it is between 6 and 7.
Solution:

8th Standard Maths Book Exercise 1.1 Question 15.
Write the five rational numbers which are less than -2.
Solution:
All the integers are rational numbers
∴ Rational numbers less than -2 are -10, -15, -20, -25, -30

8th Maths Guide Question 16.
Compare the following pairs of rational numbers

Solution:


\(\frac{10}{15}<\frac{12}{15}\)
∴ \(\frac{2}{3}<\frac{4}{5}\)

8th Maths Book Answer Question 17.
Arrange the following rational numbers is ascending and descending order.

Solution:

Objective Type Questions

8th Std Maths Guide Question 18.
The number which is subtracted from \(\frac{-6}{11}\) to get \(\frac{8}{9}\) is

Solution:
(B) \(\frac{-142}{99}\)
Hint:
Let x be the number be subtracted

Samacheer Kalvi Guru 8th Maths Book Solutions Question 19.
Which of the following rational numbers is the greatest?

Solution:
(A) \(\frac{-17}{24}\)
Hint:
LCM of 24, 16, 8, 32 = 8 × 2 × 3 × 2 = 96


∴ \(\frac{-17}{24}\) is the greatest number.

8th Std Maths Book Answers Question 20.
\(\frac{-5}{4}\) is a rational number which lies between
(A) 0 and \(\frac{-5}{4}\)
(B) -1 and 0
(C) -1 and -2
(D) -4 and -5
Solution:
(C) -1 and -2
Hint:
\(\frac{-5}{4}\) = \(-1 \frac{1}{4}\)
∴ \(\frac{-5}{4}\) lies between -1 and -2.

Samacheer Kalvi 8th Maths Book Question 21.
The standard form of \(\frac{3}{4}+\frac{5}{6}+\left(\frac{-7}{12}\right)\) is

Solution:
(D) 1
Hint:

8th Maths Book Samacheer Kalvi Question 22.
The sum of the digits of the denominator in the simplest form of \(\frac{112}{528}\)
(A) 4
(B) 5
(C) 6
(D) 7
Solution:
(C) 6
Hint:

Sum of digits in the denominator = 3 + 3 = 6

Question 23.
The rational number (numbers) which has (have) additive inverses is (are)
(A) 7
(B) \(\frac{-5}{7}\)
(C) 0
(D) all of these
Solution:
(D) all of these
Hint:
Additive inverse of 7 is -7
Additive inverse of \(\frac{-5}{7}\) is \(\frac{-5}{7}\)
Additive inverse of 0 is 0.

Question 24.
Which of the following pairs is equivalent?

Solution:
(B) \(\frac{16}{-30}, \frac{-8}{15}\)
Hint:

∴ \(\frac{16}{-30}\) and \(\frac{8}{15}\) are equivalent fraction.

Question 25.
\(\frac{3}{4} \div\left(\frac{5}{8}+\frac{1}{2}\right)\) =

Solution:
(C) \(\frac{2}{3}\)
Hint:

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All these concepts of Chapter 1 Laws of Motion are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 1 Laws of Motion State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 1 Laws of Motion.

Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 2 Optics

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Samacheer Kalvi 10th Science Optics Textual Solved Problems

10th Science Optics Book Back Answers Question 1.
Light rays travel from vacuum into a glass whose refractive index is 1.5. If the angle of incidence is 30°, calculate the angle of refraction inside the glass.
Solution:
According to Snell’s law,

Optics Lesson Book Back Answers Question 2.
A beam of light passing through a diverging lens of focal length 0.3 m appears to be focused at a distance 0.2 m behind the lens. Find the position of the object.
Solution:
f = -0.3 m, v = -0.2 m

10th Science Optics Lesson Question 3.
A person with myopia can see objects placed at a distance of 4 m. If he wants to see objects at a distance of 20 m, what should be the focal length and power of the concave lens he must wear?
Solution:
Given that x = 4 m and y = 20 m.
Focal length of the correction lens is, \(f=\frac{x y}{x-y}\)
\(\begin{aligned} f &=\frac{4 \times 20}{4-20} \\ &=\frac{80}{-16}=-5 \mathrm{m} \end{aligned}\)
Power of the correction lens \(=\frac{1}{f}=-\frac{1}{5}=-0.2 \mathrm{D}\).

Optics 10th Class Question 4.
For a person with hypermetropia, the near point has moved to 1.5 m. Calculate the focal length of the correction lens in order to make his eyes normal.
Solution:
Given that, d = 1.5 m; D = 25 cm = 0.25 m (For a normal eye).
The focal length of the correction lens is,

Samacheer Kalvi 10th Science Optics Textual Evaluation

I. Choose the correct answer.

10th Standard Optics Lesson Question 1.
The refractive index of four substances A, B, C and D are 1.31, 1.43, 1.33, 2.4 respectively. The speed of light is maximum in ____.
(a) A
(b) B
(c) C
(d) D.
Answer:
(a) A

10th Optics Lesson Question 2.
A small bulb is placed at the principal focus of a convex lens. When the bulb is switched on, the lens will produce:
(a) a convergent beam of light
(b) a divergent beam of light
(c) a parallel beam of light
(d) a coloured beam of light
Answer:
(a) a convergent beam of light

Class 10 Science Optics Question 3.
A small bulb is placed at the principal focus of a convex lens. When the bulb is switched on, the lens will produce ____.
(a) a convergent beam of light
(b) a divergent beam of light
(c) a parallel beam of light
(d) a coloured beam of light.
Answer:
(c) a parallel beam of light

Samacheer Kalvi Guru 10th Science Question 4.
Magnification of a convex lens is:
(a) Positive
(b) negative
(c) either positive or negative
(d) zero
Answer:
(a) Positive

Samacheer Kalvi 10th Science Solutions Question 5.
A convex lens forms a real, diminished point sized image at the focus. Then the position of the object is at _____.
(a) focus
(b) infinity
(c) at 2f
(d) between f and 2f.
Answer:
(b) infinity

Samacheer Kalvi 10th Science Solution Question 6.
Power of a lens is -4D, then its focal length is _____.
(a) 4m
(b) -40m
(c) -0.25 m
(d) -2.5 m.
Answer:
(c) -0.25m

10th Science Samacheer Kalvi Question 7.
In a myopic eye, the image of the object is formed:
(a) behind the retina
(b) on the retina
(c) in front of the retina
(d) on the blind spot
Answer:
(c) in front of the retina

Science Solution Class 10 Samacheer Kalvi Question 8.
The eye defect ‘presbyopia’ can be corrected by ______.
(a) convex lens
(b) concave lens
(c) convex mirror
(d) bifocal lenses.
Answer:
(d) bifocal lenses.

Samacheer Kalvi 10th Science Question 9.
Which of the following lens would you prefer to use while reading small letters found in a dictionary?
(a) A convex lens of focal length 5 cm
(b) A concave lens of focal length 5 cm
(c) A convex lens of focal length 10 cm
(d) A concave lens of focal length 10 cm.
Answer:
(c) A convex lens of focal length 10 cm

10th Science Solution Samacheer Kalvi Question 10.
If V_B, V_G, V_R be the velocity of blue, green and red light respectively in a glass prism, then which of the following statement gives the correct relation?
(a) V_B = V_G = V_R
(b) V_B > V_G > V_R
(c) V_B < V_G < V_R
(d) V_B < V_G > V_R
Answer:
(c) V_B < V_G < V_R

II. Fill in the Blanks.

Samacheer Kalvi Guru 10th Science Book Back Answers Question 1.
The path of the light is called as ______.
Answer:
ray of light.

Samacheer Kalvi 10 Science Question 2.
The refractive index of a transparent medium is always greater than _____.
Answer:
one.

10th Samacheer Kalvi Science Question 3.
If the energy of the incident beam and the scattered beam are the same, then the scattering of light is called _____ scattering.
Answer:
elastic.

Samacheer Kalvi 10th Science Solution Book Question 4.
According to Rayleigh’s scattering law, the amount of scattering of light is inversely proportional to the fourth power of its ____.
Answer:
wavelength.

Question 5.
Amount of light entering into the eye is controlled by ______.
Answer:
Iris.

III. True or False. If False Correct it.

Question 1.
The velocity of light is greater in a denser medium than in a rarer medium?
Answer:
False.
Correct Statement: Velocity of light is lesser in a denser medium than in rarer medium

Question 2.
The power of the lens depends on the focal length of the lens?
Answer:
True.

Question 3.
Increase in the converging power of eye lens cause ‘hypermetropia’.
Answer:
True.

Question 4.
The convex lens always gives a small virtual image.
Answer:
False.
Correct Statement: The convex lens give enlarged. Virtual image when it is placed between pole and principal focus.

IV. Match the Following.

Question 1.

Answer:
1. (d) Screen of the eye
2. (a) Pathway of light
3. (e) Power of accommodation
4. (b) Far point comes closer
5. (c) Near point moves away

V. Assertion and Reasoning Type Questions

Mark the correct choice as?
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but the reason is true.

Question 1.
Assertion: If the refractive index of the medium is high (denser medium) the velocity of the light in that medium will be small
Reason: Refractive index of the medium is inversely proportional to the velocity of the light
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.

Question 2.
Assertion: Myopia is due to the increase in the converging power of the eye lens.
Reason: Myopia can be corrected with the help of the concave lens.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.

VI. Answer Briefly.

Question 1.
What is refractive index?
Answer:
The ratio of the speed of light in a vacuum to the speed of light in a medium is defined as the refractive index ‘µ’ of that medium.

Question 2.
State Snell’s law.
Answer:
The ratio of the sine of the angle of incidence and sine of the angle of refraction is equal to the ratio of refractive indices of the two media. This law is also known as Snell’s law.
\(\frac{\sin i}{\sin r}=\frac{\mu_{2}}{\mu_{1}}\)

Question 3.
Draw a ray diagram to show the image formed by a convex lens when the object is placed between F and 2F.
Answer:
Ray diagram for an object placed between F and 2F

Question 4.
Define dispersion of light.
Answer:
When a beam of white light or composite light is refracted through any transparent media such as glass or water, it is split into its component colours. This phenomenon is called as ‘dispersion of light’.

Question 5.
State Rayleigh’s law of scattering.
Answer:
Rayleigh’s scattering law states that “The amount of scattering of light is inversely proportional to the fourth power of its wavelength”.
Amount of scattering \(\mathrm{s} \propto \frac{1}{\lambda^{4}}\)

Question 6.
Differentiate convex lens and concave lens.
Answer:

Question 7.
What is power of accommodation of eye?
Answer:
The ability of the eye lens to focus nearby as well as the distant objects is called power of accommodation of the eye.

Question 8.
What are the causes of ‘Myopia’?
Answer:
Myopia, also known as short-sightedness, occurs due to the lengthening of the eyeball. With this defect, nearby objects can be seen clearly but distant objects cannot be seen clearly.

Question 9.
Why does the sky appear in blue colour?
Answer:
When sunlight passes through the atmosphere, the blue colour (shorter wavelength) is scattered to a greater extent than the red colour (longer wavelength). This scattering causes the sky to appear in blue colour.

Question 10.
Why are traffic signals red in colour?
Answer:

1. The wavelength of red colour is more than other colours.
2. Red colour will travel longer distance without scattering.
3. Red colour gets least scattered and reaches people.

VII. Give the Answer in Detail.

Question 1.
List any five properties of light.
Answer:

1. Light is a form of energy.
2. Light always travels along a straight line.
3. Light does not need any medium for its propagation. It can even travel through vacuum.
4. The speed of light in vacuum or air is, c = 3 × 10⁸ ms^-1.
5. Since, light is in the form of waves, it is characterized by a wavelength (k) and a frequency (v), which are related by the following equation: c = v λ (c – velocity of light).
6. Different coloured light has different wavelength and frequency.
7. Among the visible light, violet light has the lowest wavelength and red light has the highest wavelength.
8. When light is incident on the interface between two media, it is partly reflected and partly refracted.

Question 2.
Explain the rules for obtaining images formed In a convex lens with the help of a ray diagram.
Answer:
Rule 1: When a ray of light strikes the convex or concave lens obliquely at its optical centre, it continues to follow its path without any deviation.

Rule 2: When rays parallel to the principal axis strikes a convex or concave lens, the refracted rays are converged to (convex lens) or appear to diverge from (concave lens) the principal focus.

Rule 3: When a ray passing through (convex lens) or directed towards (concave lens) the principal focus strikes a convex or concave lens, the refracted ray will be parallel to the principal axis.

Question 3.
Differentiate the eye defects: Myopia and Hypermetropia?
Answer:

Question 4.
Explain the construction and working of a ‘Compound Microscope’.
Answer:
Construction:

• A compound microscope consists of two convex lenses. The lens with the shorter focal length is placed near the object, and is called an ‘objective lens’ or ‘objective piece’.
• The lens with larger focal length and larger aperture placed near the ‘observer’s eye is called as ‘eye lens’ or ‘eyepiece’.
• Both the lenses are fixed in a narrow tube with adjustable provision.

Working:

• The object (AB) is placed at a distance slightly greater than the focal length of the objective lens (u > f₀).
• A real, inverted and magnified image (A’B’) is formed at the other side of the objective lens.
• This image behaves like the object for the eye lens. The position of the eye lens is adjusted in such a way, that the image (A’B’) falls within the principal focus of the eyepiece.
• This eyepiece forms a virtual, enlarged and erect image (A” B”) on the same side of the object.
• A compound microscope has 50 to 200 times more magnification power than the simple microscope.

VIII. Numerical Problems:

Question 1.
An object is placed at a distance 20 cm from a convex lens of focal length 10 cm. Find the image distance and nature of the image.
Solution:


Nature of the image real enlarged and inverted image.

Question 2.
An object of height 3 cm is placed at 10 cm from a concave lens of focal length 15 cm. Find the size of the image.
Solution:
u = -10 cm [left side of lens]
f = -15 cm [left side of lens]
h = 3 cm


Nature of image: When an object is placed at 10 cm from the left side of the lens, a virtual image is formed between the optical centre and focus of a concave lens. The size of the image is smaller than that of the object.

IX. Higher-Order Thinking (HOT) Questions:

Question 1.
While doing an experiment for the determination of the focal length of a convex lens, Raja Suddenly dropped the lens. It got broken into two halves along the axis. If he continues his experiment with the same lens,

1. Can he get the image?
2. Is there any change in the focal length?

Answer:

1. Yes, he can get the image, because the lens is broken into two halves along the axis.
2. No, focal length remains the same for the lens if it’s broken (or) not broken.

Question 2.
The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them?
Answer:

1. Owl has large pupil so that it provides a larger pathway for light to flow towards the retina.
2. In addition it has large comer so that a bigger image of the insects would be formed on the retina.
3. In this way the eyes of the nocturnal birds help them to identify the elements in the surroundings.

Samacheer Kalvi 10th Science Optics Additional Questions

I. Choose the correct answer.

Question 1.
The speed of light in vacuum or air is _____.
(a) 3 × 10^-8 ms^-1
(b) 3 × 10⁸ ms^-1
(c) 3 × 10⁸ m^-1s
(d) 3 × 10^-18 ms^-1.
Answer:
(b) 3 × 10⁸ ms^-1

Question 2.
If the energy of the incident beam of light and the scattered beam of light are same, then it is known as:
(a) Inelastic scattering
(b) Elastic scattering
(c) Raman scattering
(d) Mie scattering
Answer:
(b) Elastic scattering

Question 3.
Which is the following is not an example of colloid _____.
(a) milk
(b) ice – cream
(c) pure water
(d) smoke.
Answer:
(c) pure water

Question 4.
The band of colours is termed as _____.
(a) monochromatic source
(b) composite light
(c) spectrum
(d) dispersion of light.
Answer:
(c) spectrum

Question 5.
Water droplets, pollen and dust cause scattering.
(a) Elastic
(b) Inelastic
(c) Mie
(d) Rayleigh
Answer:
(c) Mie

Question 6.
_____ scattering takes place when the diameter of the Scatterer is similar to or larger than the wavelength of the incident light.
(a) Rayleigh
(b) Tyndall
(c) Mie
(d) Raman.
Answer:
(c) Mie

Question 7.
The colour of the sun is red at sunrise and sunset. This occurs due to _____ scattering.
(a) Rayleigh
(b) Mie
(c) Tyndall
(d) Raman.
Answer:
(a) Rayleigh

Question 8.
A lens may be considered to be made up of:
(a) lenses
(b) prisms
(c) mirrors
(d) transparent medium
Answer:
(b) prisms

Question 9.
If one of the faces of a bi-convex lens is plane, it is known as a ______.
(a) convex lens
(b) Plano-convex lens
(c) concave lens
(d) Plano concave lens.
Answer:
(b) Plano-convex lens

Question 10.
The spectral lines having a frequency equal to the incident ray frequency is called ______.
(a) Spectral lines
(b) Raman lines
(c) Colour lines
(d) Rayleigh lines.
Answer:
(d) Rayleigh lines.

Question 11.
The lens that is thinner in the middle than at the edges is called:
(a) concave lens
(b) bifocal lens
(c) cylindrical lens
(d) convex lens
Answer:
(a) concave lens

Question 12.
Magnification of a concave lens is ______.
(a) positive
(b) negative
(c) either positive or negative
(d) zero.
Answer:
(b) negative

Question 13.
Where should an object be placed so that a real image is formed at infinity is obtained by a convex lens _____.
(a) 2F
(b) 0
(c) F
(d) >2F.
Answer:
(c) F

Question 14.
The focal length of a lens is the distance between:
(a) optic centre and principal focus
(b) optic centre and centre of curvature
(c) principal focus and centre of curvature
(d) none
Answer:
(a) optic centre and principal focus

Question 15.
Lens Maker’s formula is _____.

Answer:

Question 16.
The power of a lens is numerically defined as the reciprocal of its _____.
(a) wavelength
(b) frequency
(c) focal length
(d) refractive index.
Answer:
(c) focal length

Question 17.
When the object is placed at the principal focus of concave lens, the image will be formed at:
(a) 2F
(b) F
(c) infinity
(d) between F and 2F
Answer:
(c) infinity

Question 18.
Astigmatism can be corrected by using _____ lens.
(a) convex
(b) concave
(c) cylindrical
(d) bifocal.
Answer:
(c) cylindrical

Question 19.
The least distance of distinct vision for the normal human eye is ______.
(a) 30 cm
(b) 25 cm
(c) 35 cm
(d) infinity.
Answer:
(b) 25 cm

Question 20
To get a real and diminished image using convex lens, the object must be placed:
(a) at F
(b) beyond 2F
(c) at 2F
(d) beyond F
Answer:
(b) beyond 2F

II. Fill in the Blanks.

Question 1.
The interacting particle of the medium is called ______.
Answer:
Scatterer.

Question 2.
When a beam of light is refracted through any transparent media, it split into its component colours. This is called as _____.
Answer:
dispersion of light.

Question 3.
The speed of light in a medium is high if the refractive index of the medium is _____?
Answer:
low.

Question 4.
The spectral lines which are having frequencies other than the incident ray frequency are called _____?
Answer:
Raman lines.

Question 5.
_____ lenses are used to correct the defect of vision called hypermetropia.
Answer:
Convex.

Question 6.
______ lenses are used as the eye lens of the Galilean Telescope.
Answer:
Concave.

Question 7.
By convention, the power of a convex lens is taken as _____ whereas the power of a concave lens is taken as _____
Answer:
positive, negative.

Question 8.
The diameter of a human eye is _____?
Answer:
2.3 cm.

Question 9.
_____ is achieved by changing the focal length of the eye lens with the help of ciliary muscles.
Answer:
Power of accommodation.

Question 10.
_____ is the coloured part of an eye.
Answer:
Iris.

Question 11.
______ is used to observe parts of flower, insects and fingerprints in the field of forensic science.
Answer:
Simple microscope.

Question 12.
A _____ works based on the principle of the vernier.
Answer:
travelling microscope.

Question 13.
A _____ telescope is used to view heavenly bodies like stars, planets galaxies and satellites.
Answer:
astronomical.

Question 14.
_______ is an optical instrument used to see distant objects clearly.
Answer:
Telescope.

Question 15.
_______ is the most sensitive part of the human eye.
Answer:
Retina.

Question 16.
A convex lens is ______ in the middle than at edges.
Answer:
Thicker.

Question 17.
_____ lens produces mostly real images.
Answer:
Convex.

Question 18.
The SI unit of power of a lens is _____.
Answer:
dioptre.

Question 19.
The amount of scattering of light is _____ proportional to the fourth power of its wavelength.
Answer:
inversely.

Question 20.
Refractive index of the medium is dependent on the ______ of the light.
Answer:
wavelength.

III. True or False. If False Correct it.

Question 1.
Light always travels along a straight line.
Answer:
True.

Question 2.
The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the different plane.
Answer:
False.
Correct Statement: The incident ray, the refracted ray of light and the normal to the refracting surface all lie in the same plane.

Question 3.
The angle of refraction is the same for different colours.
Answer:
False.
Correct Statement: The angle of refraction is different for different colours.

Question 4.
The scattering of light rays by the colloidal particles in the colloidal solution is called the Tyndall effect.
Answer:
True.

Question 5.
The convex lens is also called a diverging lens.
Answer:
False.
Correct Statement: Concave lens is also called the diverging lens.

Question 6.
Convex lenses are used in making microscope, telescope and slide projectors.
Answer:
True.

Question 7.
The distances measured against the direction of incident light are taken as negative.
Answer:
True.

Question 8.
Converging lenses are used in wide-angle spy hole indoors.
Answer:
False.
Correct Statement: Diverging lenses or concave lenses are used in wide-angle spy hole indoors.

Question 9.
If the magnification is less than 1, then we get a diminished image.
Answer:
True.

Question 10.
A compound microscope has 50 to 200 times more magnification power than the simple microscope.
Answer:
True.

IV. Match the following.

Question 1.

Answer:
1. (h) \(\frac{\sin i}{\sin r}=\frac{\mu_{2}}{\mu_{1}}\)
2. (g) The sky to appear in blue colour
3. (i) White appearance of the clouds
4. (c) converging lens
5. (b) Diverging lens
6. (d) Front surface of an eyeball
7. (a) 25 cm
8. (f) Infinity
9. (k) Bifocal lenses
10. (j) Torrid lenses

V. Assertion and Reasoning Type Questions

Mark the correct choice as?
(a) If both assertion and reason are true and the reason is the correct explanation of assertion.
(b) If both assertion and reason are true but the reason is not the correct explanation of assertion.
(c) Assertion is true but the reason is false.
(d) Assertion is false but the reason is true.

Question 1.
Assertion: The refractive index of the medium is different for different coloured lights.
Reason: The refractive index of a medium is dependent on the wavelength of the light.
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.

Question 2.
Assertion: A parallel beam of light passing through the concave lens, is diverged or spread out.
Reason: A convex lens is also called a diverging lens.
Answer:
(c) Assertion is true but the reason is false.

Question 3.
Assertion: Magnification power of microscopes can be decreased by increasing the focal length of the lens used.
Reason: Due to constructional limitations, the focal length of the lens cannot be decreased beyond certain unit.
Answer:
(d) The assertion is false but the reason is true.

VI. Answer Briefly.

Question 1.
Why does the cloud appear in white colour?
Answer:

• Mie scattering is responsible for the white appearance of the clouds.
• When white light falls on the water drop, all the colours are equally scattered which together form the white light.

Question 2.
Define Raman Scattering.
Answer:
Raman Scattering is defined as “The interaction of light ray with the particles of pure liquids or transparent solids, which leads to a change in wavelength or frequency.”

Question 3.
Differentiate stokes line and anti-stokes lines.
Answer:

1. Stokes lines: The lines having frequencies lower than the incident frequency is called stokes lines.
2. Antistokes lines: The lines having frequencies higher than the incident frequency are called Antistokes lines.

Question 4.
What is meant by colloid?
Answer:
A colloid is a microscopically small substance that is equally dispersed throughout another material.
Example: Milk, Ice cream, muddy water, smoke.

Question 5.
What are the applications of the convex lens?
Answer:

1. Convex lenses are used as camera lenses.
2. They are used as magnifying lenses.
3. They are used in making microscope, telescope and slide projectors.
4. They are used to correct the defect of vision called hypermetropia.

Question 6.
Give the lens formula.
Answer:
The lens formula gives the relationship among distance of the object (u), the distance of the image (v) and the focal length (f) of the lens. It is expressed as
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
It is applicable to both convex and concave lenses.

Question 7.
What are the applications of a concave lens?
Answer:

1. Concave lenses are used as eye lens of ‘Galilean Telescope’
2. They are used in wide-angle spy hole indoors.
3. They are being used to correct the defect of vision called ‘myopia’

Question 8.
What are the uses of a simple microscope?
Answer:
Simple microscopes are used

1. By watch repairers and jewellers.
2. To read small letters clearly.
3. To observe parts of flower, insects etc.
4. To observe Anger prints in the field of forensic science.

Question 9.
Define power of a lens.
Answer:

1. The ability of a lens to converge (convex lens) or diverge (concave lens) is called its power.
2. The power of a lens can be defined as the deep of convergence or divergence of light rays.
3. Power of a lens is numerically defined as the reciprocal of its focal length.
   \(\mathrm{P}=\frac{1}{f}\)
   The SI unit of power of a lens is dioptre.

Question 10.
Define magnification of a lens.
Answer:
It is defined as the ratio of the height of the image to the height of an object. Magnification is denoted by the letter ‘m’.If the height of the object is ‘h’ and the height of the image is h’, the magnification produced by a lens is,
\(m=\frac{\text { height of the image }}{\text { height of the object }}=\frac{h^{\prime}}{h}\)
Also, it is related to the distance of the object (w) and the distance of the image (v) as follows:
\(m=\frac{\text { distance of the image }}{\text { distance of the object }}=\frac{v}{u}\)

Question 11.
Give the lens maker’s formula.
Answer:
The lens maker’s formula is one such equation. It is given as
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)
where µ is the refractive index of the material of the lens; R1 and R2 are the radii of curvature of the two faces of the lens; f is the focal length of the lens.

Question 12.
Based on the initial and final energy of the light beam, classify the scattering.
Answer:
Based on the initial and final energy of the light beam, scattering can be classified as

• Elastic scattering
• Inelastic scattering

Question 13.
Based on the nature and size of the scatter. Classify the scattering.
Answer:
The nature and size of the scatterer result in different types of scattering. They are

• Rayleigh scattering
• Mie scattering
• Tyndall scattering
• Raman scattering

Question 14.
Why does the sky appear in red colour at sunrise and sunset?
Answer:

1. At sunrise and sunset, the light rays from the Sun have to travel a larger distance in the atmosphere than at noon.
2. Most of the blue lights are scattered away and only the red light which gets least scattered reaches us. Therefore, the colour of the Sun is red at sunrise and sunset.

Question 15.
Discuss the advantages and disadvantages of the telescope.
Answer:
Advantages:

• Elaborate view of the Galaxies, Planets, stars and other heavenly bodies is possible.
• The camera can be attached for taking photographs for the celestial objects.
• The telescope can be viewed even with the low intensity of light.

Disadvantages:

• Frequent maintenances needed.
• It is not easily portable one.

VII. Give the Answer in Detail.

Question 1.
What is scattering? Explain two types of scattering.
Answer:
When a beam of light, interacts with a constituent particle of the medium, it undergoes many kinds of scattering. Based on initial and final energy of the light beam,scattering can be classified as,
(i) Elastic scattering: If the energy of the incident beam of light and the scattered beam of light are same, then it is called as ‘elastic scattering’.
(ii) Inelastic scattering: If the energy of the incident beam of light and the scattered beam of light are not same, then it is called as ‘inelastic scattering’.

Question 2.
Explain the types of Scattering.
Answer:
Types of Scattering.
When a beam of light, interacts with a constituent particle of the medium, it undergoes many kinds of scattering.
Based on the initial and final energy of the light beam, scattering can be classified as

1. Elastic scattering: If the energy of the incident beam of light and the scattered beam of light are same, then it is called as ‘elastic scattering’.
2. Inelastic scattering: If the energy of the incident beam of light and the scattered beam of light are not same, then it is called as ‘inelastic scattering’.

The nature and size of the Scatterer result in different types of scattering. They are

• Rayleigh scattering
• Mie scattering
• Tyndall scattering
• Raman scattering

Question 3.
Explain the Cartesian sign convention.
Answer:
According to Cartesian sign convention,

1. The object is always placed on the left side of the lens.
2. All the distances are measured from the optical centre of the lens.
3. The distances measured in the same direction as that of incident light are taken as positive.
4. The distances measured against the direction of incident light are taken as negative.
5. The distances measured upward and perpendicular to the principal axis is taken as positive.
6. The distances measured downward and perpendicular to the principal axis is taken as negative.

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