NCERT Exemplar Class 11 Physics Chapter 14 Waves are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 14 Waves. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-14-waves/

NCERT Exemplar Class 11 Physics Chapter 14 Waves

Multiple Choice Questions
Single Correct Answer Type

Q1. Water waves produced by a motorboat sailing in water are
(a) neither longitudinal nor transverse
(b) both longitudinal and transverse
(c) only longitudinal
(d) only transverse
Sol: (b) Water waves produced by a motorboat sailing on the surface of deep water are both longitudinal and transverse because the waves, produce transverse as well as lateral vibrations in the particles of the medium. The water molecules at the surface move up and down; and back and forth simultaneously describing nearly circular paths as shown in Figure.

As the wave passes, water molecules at the crests move in the direction of the wave while those at the troughs move in the opposite direction.

Q3. Speed of sound wave in air
(a) is independent of temperature
(b) increases with pressure
(c) increases with increase in humidity
(d) decreases with increase in humidity

Q4. Change in temperature of the medium changes
(a) frequency of sound waves
(b) amplitude of sound waves
(c) wavelength of sound waves
(d) loudness of sound waves

Q5. With propagation of longitudinal waves through a medium, the quantity transmitted is •
(a) matter
(b) energy
(c) energy and matter
(d) energy, matter and momentum
Sol: (b) A wave is a disturbance which propagates energy and momentum from one place to the other without the transport of matter. In propagation of longitudinal waves through a medium leads to transmission of energy through the medium without matter being transmitted. There is no movement of matter (mass) and hence momentum.

Important point:
Characteristics of wave motion: ;
• It is a sort of disturbance which travels through a medium. ,
• Material medium is essential for the propagation of mechanical waves. .
• When a wave motion passes through a medium, particles of the medium only vibrate simple harmonically about their mean position. They do leave their position and move with the disturbance.
• There is a continuous phase difference amongst successive particles of the medium, i.e. particle 2 starts vibrating slightly later than particle 1 and so on.

•The velocity of the particle during there vibration is differnet at different positions different positions.
• The velocity of wave motion througha particular medium is constant.
It depends only on die nature of medium not op the frequency, wavelength or intensity,
•Energy is, propagated along with thewave motion without any net transport of themedium.

Q6. Which of the following statements are true for wave motion?
(a) Mechanical transverse waves can propagate through all mediums.
(b) Longitudinal waves can propagate through solids only.
(c) Mechanical transverse waves can propagate through solids only.
(d) Longitudinal waves can propagate through vacuum.
Sol:(c) In case of mechanical transverse wave propagates through a medium, the medium particles oscillate right angles to the direction of wave motion or energy propagation. It travels in the form of crests and troughs.
When mechanical transverse wave propagates through a medium element of the medium is subjected to shearing stress. Solids and strings have shear modulus, that is why, sustain shearing stress. Fluids have no shape of their own, they yield to shearing stress. Transverse waves can be transmitted through solids, they can be setup on the surface of liquids. But they cannot be transmitted into liquids and gases.

Q7. A sound wave is passing through air column in the form of compression and rarefaction. In consecutive compressions and rarefactions,
(a) density remains constant
(b) Boyle’s law is obeyed
(c) bulk modulus of air oscillates
(d) there is no transfer of heat
Sol: (d)
Key concept:
• When sound wave travels through a medium, say air, the particles of medium disturb in the same fashion, i.e. compression and rarefaction (depression). When air particles come closer it is called compression. On the other hand, when particles go farther than their normal position it is called rarefaction. In the condition of compression, molecules of medium come closer to each other and in the condition of rarefaction, molecules of the medium go farther from each other; compared to their normal positions.
• When compression takes place in the medium, the density and pressure of the medium increase. When rarefaction takes place in the medium, density and pressure of the medium decrease. This increase and decrease in density and pressure are temporary. Thus, compression is called the region of high density and pressure. Rarefaction is called the region of low density and pressure.

When sound wave travels through a medium:
• Due to compression and rarefaction, density of the medium (air) changes. As density is changing, so Boyle’s law is not obeyed.
• The Bulk modulus of medium remains unchanged.
• The time of compression and rarefaction is too small, i.e. we can assume adiabatic process and hence no transfer of heat.

Sol: (b)
Key concept:
Reflection of Mechanical Waves:

Medium	Longi­ tudinal wave	Trans­ verse wave	Change in direc­tion	Phase change	Time change	Path change
Reflec­tion from rigid end/ denser medium	Compres­sion as rarefac­tion and vice- versa	Crest as crest and Trough as trough	Reversed		T/2	λ/2
Reflec­tion from free end/ rarer medium	Compres­sion as compres­sion and rarefac­tion as rarefac­tion	Crest as trough and Trough as crest	No change	Zero	Zero	Zero

Q9. A string of mass 2.5 kg is under tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in
(a) 1 s
(b) 0.5 s
(c) 2 s
(d) data given is insufficient

Q10. A train whistling at constant frequency is moving towards a station at a constant speed v. The train goes past a stationary observer on the station. The frequency n of the sound as heard by the observer is plotted as a function of time t (figure). Identify the expected curve.

Key concept: General expression for Apparent Frequency. Suppose observer (O) and source (S) are moving in the same direction along a line with velocities v₀ and v_s respectively. Velocity of sound is v and velocity of medium is v_m, then apparent frequency observed by observer is given

Sign convention for different situations:

• The direction of v is always taken from source to observer.
• All the velocities in the direction of v are taken positive.
• All the velocities in the opposite direction of v are taken negative.

More Than One Correct Answer Type
Q11. A transverse harmonic wave on a string is described by

where x and y are in cm and t is in sec. The positive direction of x is from left to right.
(a) the wave is travelling from right to left
(b) the speed of the wave is 20 m/s
(c) frequency of the wave is 5.7 Hz
(d) the least distance between two successive crests in the wave is 2.5 cm
Sol: (a, b, c)
Key concept: The general equation of a plane progressive wave with initial phase is

(a) It represents a progressive wave of frequency 60 Hz.
(b) It represents a stationary wave of frequency 60 Hz.
(c) It is the result of superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m/s in opposite direction.
(d) Amplitude of this wave is constant.
Sol: (b, c)
Key concept: Standing Waves or Stationary Waves:
When two sets of progressive wave trains of same type (both longitudinal or both transverse) having the same amplitude and same time period/ frequency/wavelength travelling with same speed along the same straight line in opposite directions superimpose, a new set of waves are formed. These are called stationary waves or standing waves.

Q13. Speed of sound waves in a fluid depends upon

(a) directly on density of the medium
(b) square of Bulk modulus of the medium
(c) inversely on the square root of density
(d) directly on the square root of bulk modulus of the medium

Q14. During propagation of a plane progressive mechanical wave,
(a) all the particles are vibrating in the same phase
(b) amplitude of all the particles is equal
(c) particles of the medium executes SHM
(d) wave velocity depends upon the nature of the medium
Sol: (b, c, d)
Key concept:
Characteristics of wave motion:
• When a wave motion passes through a medium, particles of the medium only vibrate simple harmonically about their mean position. They do not move with the disturbance.
• Medium particles oscillate with same frequency and also the amplitude of oscillation of all the particles is equal. All the particles marked as 1, 2, 3,4 and 5 oscillate with the same frequency.
• There is a continuous phase difference amongst successive particles of the medium, i.e. particle 2 starts vibrating slightly later than particle 1 and so on.
• The velocity of the particle during their vibration is different at different positions.
• The velocity of wave motion through a particular medium is constant. It depends only on the nature of medium not on the frequency, wavelength or intensity.

Option (a): Clearly, the particles 1, 2 and 3 are having different phase.
Option (b) and (c): Particles of the wave shown in the figure executes SHM with same amplitude.
Option (d): The wave velocity of mechanical wave depends only on elastic and inertia property of medium for a progressive wave propagating in a fluid. Hence wave velocity depends upon the nature of the medium.

Q15. The transverse displacement of a string (clamped at its both ends) is given by

All the points on the string between two consecutive nodes vibrate with
(a) same frequency
(b) same phase
(c) same energy
(d) different amplitude

Sol: (a, b, d)
Key concept:
• The points for which amplitude is minimum are called nodes in a stationary wave, nodes are equally spaced at a distance λ/2.
• The points for which amplitude is maximum are called antinodes. Like nodes, antinodes are also equally spaced with spacing (λ/2) Furthermore, nodes and antinodes are alternate with spacing (λ/4).
• The nodes divide the medium into segments (or loops). All the particles in a segment vibrate in same phase, but in opposite phase with the particles in the adjacent segment. Twice in one period all the particles pass through their mean position simultaneously with maximum velocity (A at), the direction of motion being reversed after each half cycle.

Q16. A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m/s. Given that the speed of sound in still air is 340 m/s. Then.
(a) the frequency of sound as heard by an observer standing on the platform is 400 Hz
(b) the speed of sound for the observer standing on the platform is 350 m/s
(c) the frequency of sound as heard by the observer standing on the platform will increase
(d) the frequency of sound as heard by the observer standing on the platform will decrease
Sol: (a, b) When the wind is blowing in the same direction as that of sound wave, then net speed of the wave is sum of speed of sound wave and speed of the wind.
Given, f₀ = 400 Hz, v = 340 m/s Speed of wind v_w = 10 m/s
Option (a): As there is no relative motion between the source and observer, hence frequency observed will be the same as natural frequency f₀ = 400 Hz
Option (b): When the wind is blowing in the same direction as that of sound wave, then net speed of the wave is sum of speed of sound wave and speed of the wind. The speed of sound v = v + v_w
= 340+ 10 = 350 m/s

Option (b) and (c): There will be no change in frequency because there is no relative motion between source and observer. Hence (c) and (d) are incorrect.

Q17. Which of the following statement are true for a stationary waves?
(a) Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
(b) All the particles cross their mean position at the same time.
(c) All the particles are oscillating with same amplitude.
(d) There is no net transfer of energy across any plane.
(e) There are some particles which are always at rest.
Sol: (a, b, d, e) Consider the equation of a stationary wave
y(x, t) = 2a sin (kx) cos (wt)
Option (a): In stationary wave any particle at a given position have amplitude 2a sin kx.
Option (b): The time period of oscillation of all the particles is same, hence all the particles cross their mean position at the same time.
Option (c): Amplitude of all the particles are 2a sin kx which is different for different particles at different values of x.
Options (d) and (e): Nodes are the points which is always at rest hence no transfer of energy across the nodes. It means the energy is a stationary wave is confined between two nodes.

Q18. A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?
Sol: Wire of sonometer is twice the length which it vibrates in its second harmonic. Thus, if the tuning fork resonates at L, it will resonate at 2L. This can be explained as below:
The frequency of sonometer is given by

Hence, when the wire is doubled the number of loops also get doubled to produce the resonance. That is it resonates in second harmonic.

Q19. An organ pipe of length L open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480 Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?

Q20. A tuning fork A, marked 512 Hz, produces 5 beats per second, when sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?
Sol: When the prong of B is loaded with wax, its frequency becomes less than the original frequency.
If we assume that the original frequency of B is 507, then on loading its frequency will be less than 507. The beats between A and B will be more than 5.
If we assume that the original frequency of B is 517, then on loading its frequency will be less than 517. The beats between A and B may be equal to 5. Hence the frequency of the tuning fork B when not loaded should be 517.

Q22. The displacement of an elastic wave is given by the function, y = 3 sin ωx + 4 cos ωt, where y is in cm and t is in second. Calculate the resultant amplitude.

Sol: Given, displacement of the wave
y = 3 sin ωx + 4 cos ωx
Let us assume, 3 = A cos θ                                   .                                                .. .(i)
3=Acos θ                                                                                …(ii)
On dividing Eq. (ii) by Eq. (i)
tan θ = 4/3 => ϕ= tan^-1(4/3)
Squaring and adding equations (i) and (ii),
A² cos² θ + A² sin² θ = 3² + 4²
=>              A² (cos² θ + sin² θ) = 25
A² = 25 => A = 5. Hence, amplitude = 5 cm

Q23. At what temperatures (in °C) will the speed of sound in air be 3 times its value at 0°C?

Q24. When two waves of almost equal frequencies n_l and n₂ reach at a point simultaneously, what is the time interval between successive maxima?

Short Answer Type Questions

Q25. A steel wire has a length of 12 m and a mass of 2.10 kg. What will be the speed of a transverse wave on this wire when a tension of 2.06 x 10⁴N is applied?

Q26. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz? (Sound velocity in air = 330 ms^-1)

Q27. A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. The train begins tomove with a speed of 10 ms’¹ towards the platform. What is the frequency of the sound for an observer standing on the  platform? (Sound velocity in air = 330 ms^-1)

Q28. The wave pattern on a stretched string is shown in figure. Interpret what kind of wave this is and find its
wavelength.

Sol. If we observe the graph there are some points on the graph which are always at rest. The points on positions x = 10,20,30,40 never move, always at mean position with respect to time. These are forming nodes which characterize a stationary wave.
We know the distance between two successive nodes is equal to λ/2
=>      λ = 2 x (node to node distance)
= 2 x (20 – 10) = 20 cm

Q29. The pattern of standing waves formed on a stretched string at two instants of time are shown in figure. The velocity of two waves superimposing to form stationary waves is 360 ms^-1 and their frequencies are 256 Hz.

(a) In second plot, the displacement of each particle is zero. It means all the points are crossing mean position. At first plot point at A¹ is at amplitude position. Time taken to move from amplitude position to mean position to is equal to one-fourth of the time period.

Q30. A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water (figure). The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, maximum intensity of sound is heard.
If the room temperature is 20°C, calculate
(a) speed of sound in air at room temperature.
(b) speed of sound in air at 0°C.
(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Sol: If a pipe partially filled with water whose upper surface of the water acts as a reflecting surface of a closed organ pipe. If the length of the air column is varied until its natural frequency equals the frequency of the fork, then the column resonates and emits a loud note.
The frequency of tuning fork,f= 512 Hz.
For observation of first maxima of intensity,

(c) The resonance will still be observed for 17 cm length of air column above mercury. However, due to more complete
reflection of sound waves at mercury surface, the intensity of reflected sound increases.

Q31. Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1 : 2 : 3 : 4.

Sol: Key concept: For a string fixed at its two ends vibrate in a normal mode if it vibrates according to the equation

 NCERT Exemplar Problems Class 11 Physics

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NCERT Exemplar Class 11 Physics Chapter 13 Oscillations are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 13 Oscillations. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-13-oscillations/

NCERT Exemplar Class 11 Physics Chapter 13 Oscillations

NCERT Exemplar Class 11 Physics Chapter 13 Oscillations

Multiple Choice Questions
Single Correct Answer Type

Q1. The displacement of a particle is represented by the equation

. The motion of the particle is
(a) simple harmonic with period 2KUO
(b) simple hannonic with period nia)
(c) periodic but not simple harmonic
(d) non-periodic
Sol: (b)
Key concept:

Q2. The displacement of a particle is represented by the equation y= sin³ ωtThe motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period 2π/ω
(d) simple harmonic with period π/ω
Sol: (b)
Key concept: There are certain motions that are repeated at equal intervals of time. Let the the interval of time in which motion is repeated. Then x(t) =x(t + T), where T is the minimum change in time. The function that repeats itself is known as a periodic function.

Q3. The relation between acceleration and displacement of four particles are given below:
(a) a_x = +2x            (b) a_x = +2x²            (c) a_x = -2x²               (d) a_x = -2x
Which one of the particle is exempting simple harmonic motion?

Sol: (d)
Key Concept: In case of simple harmonic motion, the acceleration is always directed towards the mean position and so is always opposite to displacement, i.e. a α = -x or a = -ω²x
In option (d) a_x = -2x or a α -x, the acceleration of the particle is proportional to negative of displacement. Hence it represents S.H.M.

Q4. Motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the density of the  liquid
(d) simple harmonic and time period is directly proportional to the density of the liquid

Sol: (c)
Key Concept: If the liquid in U-tube is filled to a height h and cross¬section of the tube is uniform and the liquid is incompressible and non- viscous. Initially the level of liquid in the two limbs will be at the same height equal to h. If the liquid is-pressed by y in one limb, it will rise by y along the length of the tube in the other limb, so the restoring force will be developed by hydrostatic pressure difference

Q5. A particle is acted simultaneously by mutually perpendicular simple harmonic motion x = a cos ωt and y = a sin ωt. The trajectory of motion of the particle will be
(a) an ellipse (b) a parabola (c) a circle (d) a straight line
Sol: (c)
Key concept: If two S.H.M’s act in perpendicular directions, then their resultant motion is in the form of a straight line or a circle or a parabola etc. depending on the frequency ratio of the two S.H.Ms and initial phase difference. These figures are called Lissajous figures.
Let the equations of two mutually perpendicular S.H.M’s of same frequency be

Q6. The displacement of a particle varies with time according to the relation
y = a sint + b cost
(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude a² + b²
(d) The motion is SHM with amplitude √a² + b<sup>2


Q7. Four pendulums A, B, C and D are suspended from the same elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,
(a) D will vibrate with maximum amplitude
(b) C will vibrate with maximum amplitude
(c) B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude

Sol: (b) Here A is given a transverse displacement. Through the elastic support the disturbance is transferred to all the pendulums.
A and C are having same length, hence they will be in resonance, because of their time period of oscillation. Since length of pendulums A and C is same  and T =2 π√L/g , hence their time period is same and they will have
frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.</sup>

Q8. Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is


Sol: (a)
Key concept: Suppose a particle P is moving uniformly on a circle of radius A with angular speed. Q and R are the two feets of the perpendicular drawn from P on two diameters one along .Y-axis and the other along Y-axis.

Suppose the particle P is on the X-axis at t = 0. Radius OP makes an angle with the X-axis at time t, then
x = A cosωt and y = A sinωt
Here, x and v are the displacements of Q and R from the origin at time t, which are the displacement equations of SHM. It implies that although P is under uniform circular motion, Q and R are performing SHM about O with the same angular speed as that of P.

Q9. The equation of motion of a particle is x = a cos(∝t)² . The motion
(a) periodic but not oscillatory
(b) periodic and oscillatory
(c) oscillatory but not periodic
(d) neither periodic nor oscillatory
Sol: (c) The equation of motion of a particle is
x = a cos(∝t)²
is a cosine function and x varies between -a and +a, the motion is oscillatory. Now checking for periodic motion, putting t+T in place of t. T is supposed as period of the function ω(t).

Q10. A particle executing SHM has a maximum speed of 30 cm/s and a maximum acceleration of 60 cm/s². The period of oscillation is

(a) sec
(b) /2 sec
(c) 2 sec
(d) /t

=

Q11. When a mass in is connected individually to two springs S₁ and S₂, the oscillation frequencies are V₁ and V₂. If the same mass is attached to the two springs as shown in figure, the oscillation frequency would be

More Than One Correct Answer Type
Q12. The rotation of earth about its axis is
(a) periodic motion
(b) simple harmonic motion
(c) periodic but not simple harmonic motion
(d) non-periodic motion
Sol: (a, c) Rotation of earth about its axis repeats its motion after a fixed interval of lime, so its motion is periodic.
The rotation of earth is obviously not a to and fro type of motion about a fixed point, hence its motion is not an oscillation. Also this motion does not follow S.H.M equation, a ∝ -x.
Hence, this motion is not a S.H.M.

Q13. Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
(a) simple harmonic motion
(b) non-periodic motion
(c) periodic motion
(d) periodic but not SHM
Sol: (a, c) For small angular displacement, the situation is shown in the figure. Only one restoring force creates .
motion in ball inside bowl.

Q14. Displacement versus time curve for a particle executing SHM is shown in figure. Choose the correct statements.
(a) Phase of the oscillator is same at t = 0s and t = 2 s
(b) Phase of the oscillator is same at t = 2s and t = 6 s
(c) Phase of the oscillator is same at t = 1s and t = 7 s
(d) Phase of the oscillator is same at t = 1s and t = 5 s

Sol: (b,d)
Key concept:
Phase: The physical quantity which represents the state of motion of particle (e.g. its position and direction of motion at any instant).
The time varying quantity (t + ) is called the phase of the motion, and the constant is called the phase constant (or phase angle). Phase determines the status of the particle at t = 0.
Suppose we choose t= 0, at an instant when the particle is passing through its mean position and is going towards the positive direction. The phase (t + )becomes zero.
=> = 0 and x= Asint and v = Acost
If we choose t = 0, at an instant when the particle is at its position extreme position, then is π/2 at that instant.
Thus t + = π/2 at t = 0 ⇒ = π/2, or x = A sin (t + π/2) – A cost

In option (a) at t =0 s and t = 2 s, the displacements are in opposite directions, hence phase of the oscillator is not same which makes option incorrect.

In option (b) it is clear from the curve that points corresponding to t = 2 s and t = 6 s are separated by a distance belonging to one time period. Hence, these points must be in same phase.
In option (c) t = 1 s and t – 7 s though the displacement is zero but the particle moves in opposite directions, downwards at t = 1 s and upwards at t — 7 s. Hence phase of the oscillator is not same which makes option incorrect.
In option (d) points belong to t = 1 s and t = 5 s are at separation of one time period, hence must be in phase.

Q15. Which of the following statements is/are true for a simple hannonic oscillator?
(a) Force acting is directly proportional to displacement from the mean position and opposite to it
(b) Motion is periodic
(c) Acceleration of the oscillator is constant
(d) The velocity is periodic
Sol: (a, b, d)
Key concept: The simple harmonic motion is a type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. When the system is displaced from its equilibrium position, a restoring force that obeys Hooke’s law tends to restore the system to equilibrium. As a result, it accelerates and starts going back to the equilibrium position. An oscillation follows simple harmonic motion if it fulfils the following two rules:
1. Acceleration is always in the opposite direction to the displacement from the equilibrium position.
2. Acceleration is proportional to the displacement from the equilibrium position.

Q16. The displacement-time graph of a particle executing SHM is shown in figure. Which of the following statement is/are true? 

Displacement is maximum, i.e corresponds to extreme position, it means PE is maximum and KE is zero.

Q17. A body is performing SHM, then its
(a) average total energy per cycle is equal to its maximum kinetic energy
(b) average kinetic energy per cycle is equal to half of its maximum kinetic energy                                                                         
(c) mean velocity over a complete cycle is equal to 2/π times of its maximumvelocity                                          
(d) root mean square velocity is 1/√2 times of its maximum velocity

Q18. A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (figure). Take the direction from A to B as the positive direction and choose the correct statements. __ _
AO = OB = 5 cm
BC= 8 cm

(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive
(b) The sign of velocity of the particle at C going towards B is negative
(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative
(d) The sign of acceleration and force on the particle when it is at point B is negative

In option (a): When the particle is 3 cm away from A going towards B. So, velocity is towards AB, i.e. positive. In SHM, acceleration is always towards mean position (O) it means both force and acceleration act towards O, have positive sign.
Hence option (a) is correct.
In option (b): When the particle is at C, velocity is towards B hence positive. Hence option (b) is not correct.
In option (c): When the particle is 4 cm away from B going towards A velocity is negative and acceleration is towards mean position (O), hence negative. Hence option (c) is correct.
In option (d): Acceleration is always towards mean position (O). When the particle is at B, acceleration and force are towards BA that is negative. Hence option (d) is correct.

Very Short Answer Type Questions
Q19. Displacement versus time curve for a particle executing SHM is shown in figure. Identify the points marked at which (i) velocity of the oscillator is zero, (ii) speed of the oscillator is maximum.
Sol: Key concept: In displacement-time graph of SHM, zero displacement values correspond to mean position; where velocity of the oscillator is maximum. Whereas the crest and troughs represent amplitude positions, where displacement is maximum and velocity of the oscillator is zero.
(i) The points A, C, E, G lie at extreme positions (maximum displacement, y = A). Hence the velocity of the oscillator is zero.
(ii) The points B, D, F, H lie at mean position (zero displacement, y = 0). We know the speed is maximum at mean position.

Q20. Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in figure. When the mass is displaced from equilibrium position by a distance x towards right, find the restoring force.

When mass is displaced from equilibrium position by a distance x towards right, the right spring gets compressed by x developing a restoring force kx towards left on the block. The left spring is stretched by an amount x developing a restoring force kx left on the block.

Q21. What are the two basic characteristics of a simple harmonic motion?
Sol: The two basic characteristics of a simple harmonic motion
(i) Acceleration is directly proportional to displacement.
(ii) The direction of acceleration is always towards the mean position, that is opposite to displacement.

Q22. When will the motion of a simple pendulum be simplehannonic?
Sol: Simple pendulum perform angular S.H.M. Consider the bob of simple pendulum is displaced through an angle θ shown. Q
The restoring torque about the fixed point O is

τ = mgl sinθ
If θ is small angle in radians, then sin θ =  0
=> mglθ
In vector form τ ∝ θ
Hence, motion of a simple pendulum is SHM for small angle of oscillations.

Q23. What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?

Q24. What is the ratio between the distance travelled by the oscillator in one time period and amplitude?
Sol: In the diagram shown a particle is executing SHM between P and Q. The particle starts from mean position ‘O’ moves to amplitude position ‘P’, then particle turn back and moves from ‘P’ to iQ\ Finally the particle turns back again and return to mean position ‘O’. In this way the particle completes one oscillation in one time period.

Total distance travelled while it goes from O → P  →  O  →  Q  →  O
= OP + PO + OQ+QO = A+A+A+A=4A
Amplitude = OP = A
Hence, ratio of distance and amplitude = 4A/A = 4

Q25. In figure, what will be the sign of the velocity of the point P’, which is the projection of the velocity of the reference particle P. P is moving in a circle of radius R in anti-clockwise direction.

Sol. As the particle on reference circle moves in anti-clockwise direction. The projection will move from P’ to O towards left, i.e. from right to left, hence sign is negative.

Q26. Show that for a particle executing SHM, velocity and displacement have a phase difference of π/2 .

Q27. Draw a graph to show the variation of PE, KE and total energy of a simple harmonic oscillator with displacement.

Important point: From the graph we note that potential energy or kinetic energy completes two vibrations in a time during which S.H.M. completes one vibration. Thus the frequency of potential energy or kinetic energy is double than that of S.H.M.

Q28. The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the moon?

Short Answer Type Questions
Q29. Find the time period of mass M when displaced from its equilibrium position and then released for the system as shown in figure.

Sol: Key concept: For observing oscillation, we have to displace the block slightly beyond equilibrium position and find the acceleration due to the restoring force.
Let in the equilibrium position, the spring has  extended by an amount x₀.
Tension in the spring = kx₀
For equilibrium of the mass M, Mg = 2kx₀

Let the mass be pulled through a distance y and then released. But, string is inextensible, hence the spring alone will contribute the total extension y + y = 2y, to lower the mass down by y from initial equilibrium mean position x₀. So, net extension in the spring (x₀ + 2y). From F.B.D of the block,

Q30.Show that the motion of a particle represented by y = sin ax – cos cot is simple harmonic with a period of 2π/ω
Sol: The given equation is in the form of combination of two harmonic functions. We can write this equation in the form of a single harmonic (sine or cosine) function.

We have displacement function: y = sin ωt – cos ωt

Hence the function represents SHM with a period T = 2π/ω

Q31. Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.
Sol: Let us assume that the required displacement where PE is half of the maximum energy of the oscillator be x.
The potential energy of the oscillator at this position,

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NCERT Exemplar Class 11 Physics Chapter 12 Kinetic Theory

Q1. A cubic vessel (with face horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of500 ms’¹ in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground
(a) remains the same because 500 ms’¹ is very much smaller than v_rms of the gas.
(b) remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls.
(c) will increase by a factor equal to (v²_rms + (500)²)/v²_rms where v_rms was the original mean square velocity of the gas.
(d) will be different on the top wall and bottom wall of the vessel.

Sol: (b) According to the ideal gas law,

P=nRT/V, here temperature of the vessel remain unchanged hence, the
pressure remains same from that point of view.
Now, let us discuss the phenomenon inside the vessel. The gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic.
The number of collisions per unit volume in a gas remains constant. So, the pressure of the gas inside the vessel remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules with respect to the walls.

Q2. 1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time,
(a) the pressure on EFGH would be zero
(b) the pressure on all the faces will be equal
(c) the pressure of EFGH would be double the pressure on ABCD
(d) the pressure on EFGH would be half that on ABCD

Sol: (d) In an ideal gas, the gas molecules keep on colliding among themselves as well as with the walls of containing vessel. These collisions are perfectly elastic. So, their kinetic energy and momentum remains conserved.
So, the momentum transferred to the face ABCD = 2mv And the gas molecule is absorbed by the face EFGH. Hence it does not rebound. So, momentum transferred to the face EFGH = mv.
And the pressure on the faces is due to the total momentum to the faces. So, pressure on EFGH would be half that on ABCD.

Q3. Boyle’s law is applicable for an
(a) adiabatic process
(b) isothermal process
(c) isobaric process
(d) isochoric process
Sol: (b)
Key concept: Boyle’s law: For a given mass of an ideal gas at constant temperature, the volume of a gas is inversely proportional to its pressure.

So we can say that when temperature is constant, Boyle’s law is applicable.

i.e.,           PV= nRT= constant
=>PV = constant (at constant temperature)
i.e.. p ∝ 1/V— [where, P = pressure. V= volume]

So, this law is applicable for an isothermal process, in which temperature remain constant.

Q4. A cylinder containing an ideal gas is in vertical position and has a piston of mass M that is able to move up or down without friction ( figure). If the temperature is increased
(a) both P and V of the gas will change
(b) only P will increase according to Charles’ law
(c) V will change but not P
(d) P will change but not V

Q5. Volume versus temperature graphs for a given mass of an ideal gas are shown in figure. At two different values of constant pressure. What can be inferred about relation between P_l and P₂ ?

(a)P₁ > P₂
(b) P_{1 =} P₂
(c) P₁ < P₂
(d) Data is insufficient

Q6. 1 mole of H₂ gas is contained in a box of volume V = 1.00 m³ at T = 300 K. The gas is heated to a temperature of T= 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

(a) same as the pressure initially
(b) 2 times the pressure initially
(c) 10 times the pressure initially
(d) 20 times the pressure initially

Sol: (d) The situation is shown in the diagram, H₂ gas is contained in a box is heated and gets converted to a gas of hydrogen atoms. Then the number of moles would become twice.
According to gas equation,
PV= nRT

P = Pressure of gas, n = Number of moles
R = Gas constant, T = Temperature PV=nRT
As volume (V) of the container is constant. Hence, when temperature (T) becomes 10 times, (from 300 K to 3000 K) pressure (P) also becomes 10 times, asP∝ T.
Pressure is due to the bombardment of particles and as gases break, the number of moles becomes twice of initial, so n₂ = 2n₁

Q7. A vessel of volume V contains a mixture of 1 mole of hydrogen and 1 mole of oxygen (both considered as ideal). Let f₁(v)dv denotes the fraction of molecules with speed between v and (v + dv) with f₂(v)dv, similarly for oxygen. Then,
(a) f₁(v) + f₂(v) = f (v) obeys the Maxwell’s distribution law
(b) f₁(v), f₂(v) will obey the Maxwell’s distribution law separately
(c) neither f₁(v)nor f₂(v)will obey the Maxwell’s distribution law
(d) f₂(v) and f₁(v)will be the same

Q8. An inflated rubber balloon contains one mole of an ideal gas, has a pressure P. volume V and temperature T. If the temperature rises to 1.1 T, and the volume is increased to 1.05 V, the final pressure will be
(a) 1.1 P
(b) P
(c) less than P
(d) between P and 1.1
Sol:(d) According to the equation of ideal gas, PV= nRT
P = pressure
V = volume
n = number of moles of gases
R = gas constant
T = temperature
Thus we have to rewrite this equation in such a way that no. of moles is given by,

More Than One Correct Answer Type

Q9. ABCDEFGH is a hollow cube made of an insulator (figure) face A BCD has positive charge on it. Inside the cube, we have ionised hydrogen.
The usual kinetic theory expression for pressure
(a) will be valid
(b) will not be valid, since the ions would experience forces other than due to collisions with the walls
(c) will not be valid, since collisions with walls would not be elastic
(d) will not be valid because isotropy is lost

Sol: (b, d) According to the problem, ionized hydrogen is present inside the cube, they are having charge. Now, due to the presence of positive charge on the surface A BCD hydrogen ions would experience forces other than the forces due to collision with the walls of container. So, these forces must be of electrostatic nature. Hence, Isotropy of system is lost at only one face ABCD because of the presence of external positive charge. The usual expression for pressure on the basis of kinetic theory will be valid.

Q10. Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. From the equation in kinetic theory PV = 2/3 E,E is
(a) the total energy per unit volume
(b) only the translational part of energy because rotational energy is very small compared to the translational energy
(c) only the translational part of the energy because during collisions with the wall pressure relates to change in linear momentum
(d) the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero

Sol: (c) According to kinetic theory equation, PV = 2/3 E [where P= Pressure V = volume]
E is representing only translational part of energy. Internal energy contains all types of energies like translational, rotational, vibrational etc. But the molecules of an ideal gas is treated as point masses in kinetic theory, so its kinetic energy is only due to translational motion. Point mass does not have rotational or vibrational motion. Here, we assumed that the walls only exert perpendicular forces on molecules. They do not exert any parallel force, hence there will not be any type of rotation present. The wall produces only change in translational motion.

Q11. In a diatomic molecule, the rotational energy at a given temperature
(a) obeys Maxwell’s distribution
(b) have the same value for all molecules
(c) equals the translational kinetic energy for each molecule
(d) is (2/3)rd the translational kinetic energy for each molecule
Sol: (a, d)
Key concept: Kinetic Energy of Ideal Gas:
Molecules of ideal gases possess only translational motion. So they possess only translational kinetic energy.

According to the problem we have to find the rotational energy of a diatomic molecule in the terms of translation kinetic energy.
First let us check the options by picking them one by one.
(a) Translational kinetic energy and rotational kinetic energy both obey Maxwell’s distribution independent of each other.
(b) Rotational kinetic energy is different for different molecule.
(c) Molecules of ideal gases possess only translational motion. So they possess only translational kinetic energy. But in case of non-ideal gas there is a smaller rotational energy.
(d) Here, 2 rotational and 3 translational energies are associated with each molecule. Translation kinetic energy of each molecule,

Important points: Kinetic energy per molecule of a gas does not depend upon the mass of the molecule but only depends upon the temperature of the gas.

Kinetic energy per mole of gas depends only upon the temperature of gas.
Kinetic energy per gram of gas depend upon the temperature as well molecular weight (or mass of one molecule) of the gas.

From the above expressions it is clear that higher the temperature of the gas, more will be the average kinetic energy possessed by the gas molecules at T= 0, E = 0, i.e. at absolute zero the molecular motion stops.

Q12. Which of the following diagrams (figure) depicts ideal gas behaviour ?

So, graph of PV versus T will be a straight line parallel to the temperature axis (x-axis).
i.e., slope of this graph will be zero.
So, (d) is not correct.

Q13. When an ideal gas is compressed adiabatically, its temperature rises the molecules on the average have more kinetic energy than before. The kinetic energy increases,
(a) because of collisions with moving parts of the wall only
(b) because of collisions with the entire wall
(c) because the molecules gets accelerated in their motion inside the volume
(d) because of redistribution of energy amongst the molecules
Sol:(a) Since the gas is ideal and the collisions of the molecules are elastic. When the molecules collides with the moving parts of the wall, its kinetic energy increases. But the total kinetic energy of the system will remain conserved. When the gas is compressed adiabatically, the total work done on the gas increases, its internal energy which in turn increases the KE of gas molecules and hence, the collisions between molecules also increases.

Very Short Answer Type Questions
Q14. Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197 g mole^-1

Q15. The volume of a given mass of a gas at 27°C, 1 atm is 100 cc. What will be its volume at 327°C?
Sol:
Key concept: Here the temperatures are given in Celsius. To apply ideal gas equation, we must convert the given temperature in kelvin. So, to convert them in kelvin we use the relation

Q16. The molecules of a given mass a gas have root mean square speeds of 100 ms^-1 at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?
Sol: Key concept:Root Mean Square Speed: It is defined as the square root of mean of squares of the speed of different molecules

Q17. Two molecules of a gas have speeds of 9 x 10¹⁶ ms^-1 and 1 x I0⁶ ms^-1 respectively. What is the root mean square speed of these molecules? 

Q18. A gas mixture consists of 2.0 moIes of oxygen and 4.0 moles of neon at temperature T. Neglecting all vibrational modes, calculate the total internal energy of the system. (Oxygen has two rotational modes.)

Sol: Key concept: Degree of Freedom:
The term degree of freedom of a system refers to the possible independent motions, systems can have or
The total number of independent modes (ways) in which a system can possess energy is called the degree of freedom (f).
The independent motions can be translational, rotational or vibrational or any combination of these.
So the degree of freedom are of three types:
(i) Translational degree of freedom
(ii) Rotational degree of freedom
(iii) Vibrational degree of freedom
General expression for degree of freedom
f=3A- B; where A = Number of independent particles.
B = Number of independent restriction

Diatomic gas: Molecules of diatomic gas are made up of two atoms joined rigidly to one another through a bond. This cannot only move bodily, but also rotate about one of the three co-ordinate axes. However its moment of inertia about the axis joining the two atoms is negligible compared to that about the other two axes.
Hence it can have only two rotational motion. Thus a diatomic molecule has 5 degree of freedom: 3 translational and 2 rotational.
Monoatomic gas: Molecules of
monoatomic gas can move in any direction in space so it can have three independent motions and hence 3 degrees of freedom (all translational).

Neon (Ne) is a monoatomic gas having 3 degrees of freedom.
Energy per mole = 3/2RT
Hence, Energy = 4 x3/2 RT = 6RT ….(ii)
[Using Eqs. (i) and (ii)]
Total energy = 5RT = 6RT= 11RT

Q19. Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 A and 2 A. The gases may be considered under identical conditions of temperature, pressure and volume.

Short Answer Type Questions

When the partition is removed, the gases get mixed without any loss of energy. The mixture now attains a common equilibrium pressure and the total volume of the system is sum of the volume of individual chambers V1 and V2. Let P be the pressure after the partition is removed.

Q21. A gas mixture consists of molecules of A, B and C with masses m_A > m_B > m_c. Rank the three types of molecules in decreasing order of (a) average KE, (b) rms speeds.

Q22. We have 0.5 g of hydrogen gas in a cubic chamber of size 3 cm kept at NTP. The gas in the chamber is compressed keeping the temperature constant till a final pressure of 100 atm. Is one justified in assuming the ideal gas law, in the final state? (Hydrogen molecules can be consider as spheres of radius 1 A).

Q23. When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle’s law in this case?
Sol: Here, according to the question, when air is pumped, more molecules are pumped and Boyle’s law is stated for situation where, mass of molecules remains constant.

In this case, when air is pumped into a cycle tyre, mass of air in it increases as the number of air molecules keep increasing. Hence, this is a case of variable mass, Boyle’s law (and even Charle’s law) is only applicable in situations, where mass of gas molecules remains fixed. Hence, Boyle’s law is not applicable in this case.

Q24. A balloon has 5.0 mole of helium at 7°C. Calculate
(a) the number of atoms of helium in the balloon.
(b) the total internal energy of the system.

Important point: The above degrees of freedom are shown at room temperature. Further at high temperature, in case of diatomic or polyatomic molecules, the atoms with in the molecule may also vibrate with respect to each other. In such cases, the molecule will have an additional degrees of freedom, due to vibrational motion.
An object which vibrates in one dimension has two additional degrees of freedom. One for the potential energy and one for the kinetic energy of vibration. Helium is a mono atomic gas and It has only 3 degrees of freedom. But after addition its degree of freedom will be 5.

Q25. Calculate the number of degrees of freedom of molecules of hydrogen in 1 cc of hydrogen gas at NTP.

Q26. An insulated container containing monoatomic gas of molar mass m is moving with a velocity v₀. If the container is suddenly stopped, find the change in temperature.
Sol: Since, the container is suddenly stopped which is initially moving with velocity v₀, there is no time for exchange of heat in the process. Then total KE of the container is transferred to gas molecules in the form of translational KE, thereby increasing the absolute temperature.
Let n be the no. of moles of the monoatomic gas in the container. Since molar mass of the gas is m.
Total mass of the container, M = mn
KE of molecules due to velocity v₀,
KE = 1/2(mn) v₀²

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We hope the NCERT Exemplar Class 11 Physics Chapter 12 Kinetic Theory help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 12 Kinetic Theory, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics.

NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics

Multiple Choice Questions
Single Correct Answer Type

Q1. An ideal gas undergoes four different processes from the same initial state (figure). Four processes are adiabatic, isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which one is adiabatic?
(a) 4
(b) 3
(c) 2
(d) 1

Solution:

Q2. If an average person jogs, he produces 14.5 x 10³ cal/min. This is removed by the evaporation of sweat. The amount of sweat evaporated per minute (assuming 1 kg requires 580 x 10³ cal for evaporation) is
(a) 0.25 kg (b) 2.25 kg (c) 0.05 kg (d) 0.20 kg
Sol:(a) Rate of bum calories is equivalent to sweat produced. Then, Amount of sweat evaporated/minute

Q3. Consider P-Vdiagram for an ideal gas is shown in figure. Out of the following diagrams, which figure represents the T-P diagram?

i.e., V 1/p or PV =  Constant

Hence, we can say that the gas is going through an isothermal process. Clearly, from the graph that between process 1 and 2 temperature is constant and the gas expands and pressure decreases, i.e., P₂<P_l. So, we have to keep in mind while drawing the T-P graph, that temperature (T) is constant and pressure at point 2 is greater than the pressure at 1, which corresponds to diagram (iii).

 

Q4. An ideal gas undergoes cyclic process ABCDA as shown in given P-V diagram. The amount of work done by the gas is
(a) 6P_gV₀
(b) -2P₀V₀
(c) +2 P₀V_o
(d) +4Po V₀

Sol:

Important point: In a cyclic process work done is
1. positive if the cycle is clockwise.
2. negative if the cycle is anticlockwise.

Q5. Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is 

Sol: (a) According to the P-V diagram shown for the container A (which is going through isothermal process) and for container B (which is going through adiabatic process).

Q6. Three copper blocks of masses M₁ M₂ and M₃ kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at T₁, T₂, T₃ (T₁ > T₂> T₃). Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)

Sol: (b) According to question, since there is no net loss to the surroundings and the equilibrium temperature of the system is T.
Let us assume that T_l,T₂< T< T₃.
Heat lost by M₃ = Heat gained by M₁+ Heat gained by M₂


More Than One Correct Answer Type
Q7. Which of the processes described below are irreversible?
(a) The increase in temperature of an iron rod by hammering it.
(b) A gas in a small container at a temperature T₁, is brought in contact with a big reservoir at a higher temperature T₂ which increases the temperature of the gas.
(c) A quasi-static isothermal expansion of an ideal gas in cylinder fitted with a frictionless piston.
(d) An ideal gas is enclosed in a piston cylinder arrangement with adiabatic walls. A weight w is added to the piston, resulting in compression of gas.
Sol: (a, b, d)
Key concept: Reversible process: A reversible process is one which can be reversed in such a way that all changes occurring in the direct process are exactly repeated in the opposite order and inverse sense and no change is left in any of the bodies taking part in the process or in the surroundings.
The conditions for reversibility are:
• There must be complete absence of dissipative forces such as friction, viscosity, electric resistance etc. ~
• The direct and reverse processes must take place infinitely slowly.
• The temperature of the system must not differ appreciably from its surroundings.
Irreversible process: Any process which is not reversible exactly is an irreversible process. All natural processes such as conduction, radiation, radioactive decay etc. are irreversible. All practical processes such as free expansion, Joule-Thomson expansion, electrical heating of a wire are also irreversible.
(a) In this case internal energy of the rod is increased from external work done by hammer which in turn increases its temperature. So, the process cannot be retraced itself.
(b) In this process energy in the form of heat is transferred to the gas in the small container by big reservoir at temperature T₂.
(c) In a quasi-static isothermal expansion, the gas is ideal, this process is reversible because the cylinder is fitted with frictionless piston.
(d) As the weight is added to the cylinder arrangement in the form of external pressure hence, it cannot be reversed back itself.

Q8. An ideal gas undergoes isothermal process from some initial state i to final state f Choose the correct alternatives

Sol: (a, d)
Key concept: First Law of Thermodynamics:
It is a statement of conservation of energy in thermodynamical process.
According to it heat given to a system (∆Q) is equal to the sum of increase in its internal energy (AIT) and the work done (AW) by the system against the surroundings.
∆Q=∆U+∆W
According to the first law of thermodynamics. ∆AQ = ∆U + ∆Wbut
∆U ∝∆T
∆U=0 [As ∆T= 0]
∆Q = ∆W, i.e., heat supplied in an isothermal change is used to do work against external surrounding.
or if the work is done on the system then equal amount of heat energy will be liberated by the system

Q9. Figure shows the P-V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.
(a) Change in internal energy is same in IV and III cases, but not in I and II.
(b) Change in internal energy is same in all the four cases.
(c) Work done is maximum in case I.
(d) Work done is minimum in case II.

Sol: (b, c)

Key concept: Internal energy (U): Internal energy of a system is the energy possessed by the system due to molecular motion and molecular configuration.
The energy due to molecular motion is called internal kinetic energy U_K and that due to molecular configuration is called internal potential energyUp.
i.e., Total internal energy U= U_K+ U_P
(i) For an ideal gas, as there is no molecular attraction U_P = 0
i.e., internal energy of an ideal gas is totally kinetic and is given by
U = U_K = 3/2 RT

Change in internal energy does not depend on the path of the process. So it is called a point function, i.e. it depends only on the initial and final states (A and B) of the system, i.e. ∆U = Uf – U_i
Hence internal energy is same for all four paths I, II, III and IV.
The work done by an ideal gas is equal to the area bounded between P-V curve.
Work done from A to B, ∆W_A→B = Area under the P-V curve which is maximum for the path I.

Q10. Consider a cycle followed by an engine (figure).
1 to 2 is isothermal
2 to 3 is adiabatic
3 to 1 is adiabatic
Such a process does not exist, because
(a) heat is completely converted to mechanical energy in such a process, which is not possible
(b) mechanical energy is completely converted to heat in this process, which is not possible
(c) curves representing two adiabatic processes don’t intersect
(d) curves representing an adiabatic process and an isothermal process don’t intersect

Sol. (a, c)
(a) The given process is a cyclic process, i.e. it returns to the original state 1. And change in internal energy in a cyclic process is always zero as for cyclic process U_f = U_i So, ∆U = U_f – U_i = 0
Hence, total heat is completely converted to mechanical energy. Such a process is not possible by second law of thermodynamics.
(c) Here, two curves are intersecting, when the gas expands adiabatically from 2 to 3. It is not possible to return to the same state without being heat supplied, hence the process 3 to 1 cannot be adiabatic. So, we conclude that such a process does not exist because curves representing two adiabatic processes do not intersect.

Q11. Consider a heat engine as shown in figure. Q₁ and Q₂ are heat added both to T₁ and heat taken from T₂ in one cycle of engine. W is the mechanical work done on the engine.If W > 0, then possibilities are:

Sol: (a, c)
Key concept: Refrigerator or Heat Pump:
A refrigerator or heat pump is basically a heat engine run in reverse direction. It essentially consists of three parts:
Source: At higher temperature T₁
Working substance: It is called refrigerant liquid ammonia and freon works as a working substance.

Sink: At lower temperature T₂.
The working substance takes heat Q₂ from a sink (contents of refrigerator) at lower temperature, has a net amount of work done W on it by an external agent (usually compressor of refrigerator) and gives out a larger amount of heat Q₁, to a hot body at temperature T₁ (usually atmosphere). Thus, it transfers heat from a cold body to a hot body at the expense of mechanical energy supplied to it by an external agent. The cold body is thus cooled more and more.

We know that the diagram represents the working of a refrigerator. So, we can write

Very Short Answer Type Questions

Q12. Can a system be heated and its temperature remains constant?
Sol: Yes, this is possible when the entire heat supplied to the system is utilised in expansion.
As ∆Q = ∆U + ∆W and ∆U = nC_v∆T
∆Q = nC_v∆T+ ∆W
If temperature remains constant, then ∆T = 0, this implies ∆Q = ∆W. This implies that heat supplied should perform work against the surroundings.

Q13. A system goes from P to Q by two different paths in the P-V diagram as shown in figure.
Heat given to the system in path 1 is 1000 J.
The work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2?

Sol: According to the first law of thermodynamics,
∆Q = AU + ∆W. Let us apply this for each path.
For path 1: Heat given Q₁ = +1000 J
Let work done for path 1 = W₁                                          .
For path 2:
Work done (W₂) = (W_I– 100) J
Heat given Q₂ – ?
As change in internal energy between two states for different path is same.
∆ U=Q_i-W₁ = Q₂-W₂
1000– W_!=Q₂-(W₁ – 100)
=> Q₂= 1000- 100 = 900 J

Q14. If a refrigerator’s door is kept open, will the room become cool or hot? Explain.
Sol: A refrigerator is a heat engine it extracts heat from low temperature reservoir and transfer it to high temperature. If a refrigerator’s door is kept open, then room will become hot, because then refrigerator exhaust more heat into the room than earlier. In this way, temperature of the room increases and room becomes hot. No refrigerator is efficient. Thus it exhaust more heat into the room than it extract from it. Thus, a room cannot be cooled by keeping the door of a refrigerator open.

Q15. Is it possible to increase the temperature of a gas without adding heat to it? Explain.
Sol: Yes, it is possible to increase the temperature of a gas without adding heat to it, during adiabatic compression the temperature of a gas increases while no heat is given to it.
For an adiabatic compression, no heat is given or taken out in adiabatic process.
Therefore, ∆Q = 0
According to the first law of thermodynamics,
∆Q=∆U+∆W
∆U = -∆W ( ∆Q =0)
In compression work is done on the gas, i.e. work done is negative. Therefore, ∆U = Positive
Hence, internal energy of the gas increases due to which its temperature increases.

Q16 Air pressure in a car tyre increases during driving. Explain.
Sol: Volume of a car tyre is fixed. During driving, temperature of the gas increases while its volume remains constant. So, according to Charle’s law, at constant volume (V),
Pressure (P) ∝Temperature (T)
Therefore, pressure of gas increases

Short Answer Type Questions
Q17. Consider a Carnot’s cycle operating between T₁ = 500 K and T₂ = 300 K producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
Sol: Key concept: Carnot theorem: The efficiency of Carnot’s heat engine depends only on the temperature of source (T₁) and temperature of sink(T₂), and heat supplied (Q₁) i.e., η= W/ Q₁  = 1 – T₂/ T₁
(The efficiency of engine is defined as the ratio of work done to the heat supplied.)
Carnot stated that no heat engine working between two given temperatures of source and sink can be more efficient than a perfectly reversible engine (Carnot engine) working between the same two temperatures. Carnot’s reversible engine working between two given temperatures is considered to be the most efficient engine.

Q18. A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he bums twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kcal, how many times must he go up and down to reduce his weight by 5 kg?

Q19. Consider a cycle tyre being filled with air by a pump. Let Vbe the volume of the tyre (fixed) and at each stroke of the pump ∆V(<< V) of air is transferred to the tube adiabatically. What is the work done when the pressure in the tube is increased from P_l to P₂
Sol: Since the process is adiabatic, there is no exchange of heat in the process, Let, pressure is increased by AP and volume is increased by AV at each stroke.
For just before and after an stroke, we can write

Q20. In a refrigerator one removes heat, from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power and heat transferred from -3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.
Sol: Carnot designed a theoretical engine which is free from all the defects of a practical engine. The Carnot engine is the most efficient heat engine operating between two given temperatures. The efficiency of Carnot engine is

Q21. If the coefficient of performance of a refrigerator is 5 and operates at the room temperature (27°C), find the temperature inside the refrigerator.
Sol:
Key concept: The performance of a refrigerator is expressed by means of “coefficient of performance” β which is defined as the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body.

Q22. The initial state of a certain gas is (P_i,V_i T_i). It undergoes expansion till its volume becomes V_f Consider the following two cases.
a)the expansion takes place at constant temperature.
b)the expansion takes place at constant pressure.
Plot the P-V diagram for each case. In which of the two cases, is work done by the gas more?
Sol:

The situation is shown in the given P-V graph, where variation is shown for each process.
It is clear from the graph that Process 1 is isobaric and Process 2 is isothermal.
Since, work done is equal to the area under the P-V curve. Here, area under the P-V curve 1 is more. So, work done is more when the gas expands in isobaric process as in comparison of gas expands in isothermal.

Long Answer Type Questions

Q23. Consider a P-V diagram in which the path followed by one mole of perfect gas in a cylindrical container is shown in figure.
(a) Find the work done when the gas is taken from state 1 to state 2.
(b) What is the ratio of temperature T₁/T₂, if V₂ = 2V₁?




This is the amount of heat supplied.

Q24. A cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in figure.
A to B: volume constant
B to C: adiabatic
C to D: volume constant
D to A: adiabatic

V_c  =V_D  =2V_A = 2V_B

(a) In which part of the cycle heat is supplied to the engine from outside?
(b) In which part of the cycle heat is being given to the surrounding by the engine?
(c) What is the work done by the engine in one cycle? Write your answer in term of P_A, P_B, V_A?
(d) What is the efficiency of the engine?

Sol: (a) For the process AB (which is isochoric process), volume is constant. So,
dV= 0 => dW= 0
dQ = dU + dW = dU
=> dQ = dU = Change in internal energy
Hence, in this process heat supplied is utilised to increase, internal  energy of the system.

(b) For the process CD (which is also isochoric process), volume is constant but pressure decreases.
Hence, temperature also decreases (because Pα T) so heat is given to the surroundings.
(c) To calculate work done by the engine in one cycle, we calculate work done in each part separately.

Q25. A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle. [Cv = (3/2)/?]
(a) AB: constant volume
(b) BC: constant pressure
(c) CD: adiabatic
(d) DA : constant pressure

Sol: (a)

 By using first law of thermodynamics, we can find amount of heat associated with each process

For process AB

Volume is constant, hence work done dW = 0

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We hope the NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 11 Thermodynamics, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar  Class 11 Physics Chapter 10 Thermal Properties of Matter are part of NCERT Exemplar Class 11 Physics. Here we have given NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter. https://www.cbselabs.com/ncert-exemplar-problems-class-11-physics-chapter-10-thermal-properties-matter/

NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter

Q1. A bimetallic strip is made of aluminium and steel (aAI > astee|) On heating, the strip will
(a) remain straight (bj get twisted
(c) will bend with aluminium on concave side.
(d) will bend with steel on concave side
Sol: (d)
Key concept: Bi-metallic strip-. Two strips of equal lengths but of different materials (different coefficient of linear expansion) when join together, it is called “bi-metallic strip”, and can be used in thermostat to break or make electrical contact. This strip has the characteristic property of bending on heating due to unequal linear expansion of the two metal. The strip will bend with metal of greater a on outer side, i.e. convex side.

On heating, the metallic strip with higher coefficient of linear expansion (∝_Al) will expand more.
According to the question, ∝_AI > ∝_steel, so aluminum will expand more. So, it should have larger radius of curvature. Hence, aluminium will be on convex side. The metal of smaller ∝ (i.e., steel) bends on inner side, i.e., concave side.

Q2. A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly
(a) its speed of rotation increases
(b) its speed of rotation decreases
(c) its speed of rotation remains same
(d) its speed increases because its moment of inertia increases
Sol: (b) When the rod is heated uniformly to raise its temperature slightly, it expands. So, moment of inertia of the rod will increase.
Moment of inertia of a uniform rod about its perpendicular bisector

If the temperature increases, moment of inertia will increase.
No external torque is acting on the system, so angular momentum should be conserved.

Q3. The graph between two temperature scales A and B is shown in figure between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by

Sol: Key concept: Temperature on one scale can be converted into other scale by using the following identity.
Reading on any scale – LFP /UFP – LFP = Constant for all scales
where, LFP —> Lower fixed point
UFP —>Upper fixed point
From the graph it is clear that the lowest point for scale A is 30° and highest point for the scale A is 180°.
Lowest point for scale B is 0° and highest point for scale B is 100°. Hence, the relation between the two scales A and B is given by

Q4. An aluminium sphere is dipped into water.
Which of the following is true?
(a) Buoyancy will be less in water at 0°C than that in water at 4°C.
(b) Buoyancy will be more in water at 0°C than that in water at 4°C.
(c) Buoyancy in water at 0°C will be same as that in water at 4°C.
(d) Buoyancy may be more or less in water at 4°C depending on the radius of the sphere.
Sol: (a)
Key concept: Liquids generally increase in volume with increasing temperature but in case of water, it expands on heating if its temperature is greater than 4°C. The density of water reaches a maximum value of 1.000 g/cm3 at 4°C.
This behaviour of water in the range from 0°C to 4°C is called anomalous expansion.

Q5. As the temperature is increased, the period of a pendulum
(a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob
(b) decreases as its effective length increases even though its centre of mass ‘ still remains at the centre of the bob
(c) increases as its effective length increases due to shifting to centre of mass below the centre of the bob
(d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

Sol: (a) A pendulum clock keeps proper time at temperature θ₀. If temperature is increased to θ (>θ ₀), then due to linear expansion, length of pendulum increases and hence its time period will increase

So, as the temperature increases, length of pendulum increases and hence time period of pendulum increases. Due to increment in its time period, a pendulum clock becomes slow in summer and will lose time.

Q6. Heat is associated with
(a) kinetic energy of random motion of molecules
(b) kinetic energy of orderly motion of molecules
(c) total kinetic energy of random and orderly motion of molecules
(d) kinetic energy of random motion in some cases and kinetic energy of orderly motion in other
Sol: (a) When a body is heated its temperature rises and in liquids and gases vibration of molecules about their mean position increases, hence kinetic energy associated with random motion of molecules increases.
So, thermal energy or heat associated with the random and translatory motions of molecules.

Q7. The radius of a metal sphere at room temperature Tis Rand the coefficient of linear expansion of the metal is .The sphere heated a little by a temperature ∆Tso that its new temperature is T + ∆T.The increase in the volume of the sphere is approximately.


ρ = mass / volume

So, the volume of all object will also be same.
Here the cooling will be done in the form of radiations that is according to Stefan’s law. Since, emissive power is directly proportional to the surface. Here, for given volume, sphere has least surface area and circular plate of greatest surface area.

As thickness of the plate is least, hence surface area of the plate is maximum. According to Stefan’s law, heat loss (cooling) is directly proportional to the surface area.

H_sphere : H_cube: H_plate  = A_sphere : A_cube: A_plate
As A_plate is maximum, hence the plate will cool fastest.
As the sphere is having minimum surface area, hence the sphere cools slowest.

More Than One Correct Answer Type
Q9. Mark the correct options.
(a) A system X is in thermal equilibrium with Y but not with Z. The systems Y and Z may be in thermal equilibrium with each other.
(b) A system X is in thermal equilibrium with Y but not with Z. The systems Y and Z are not in thermal equilibrium with each other.
(c) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other.
(d) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y-and Z may be in thermal equilibrium with each other.
Sol: (b, d)
Key concept:Two bodies are said to be in thermal equilibrium with each other, when no heat flows from one body to the other. That is when both the bodies are at the same temperature.

Q10. Gulab jamuns (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice bigger (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following.
(a) Both size gulab jamuns will get heated in the same time
(b) Smaller gulab jamuns are heated before bigger ones
(c) Smaller pizzas are heated before bigger ones
(d) Bigger pizzas are heated before smaller
Sol: (b, c) Between these four which has the least surface area will be heated first because of less heat radiation. So, smaller gulab jamuns are having least surface area, hence they will be heated first.
Similarly, smaller pizzas are heated before bigger ones because they are of small surface areas.

Q11. Refer to the plot of temperature versus time (figure) showing the changes in the state if ice on heating (not to scale). Which of the following is correct?
(a) The region AB represents ice and water in thermal equilibrium
(b) At B water starts boiling
(c) At C all the water gets converted into steam
(d) C to D represents water and steam in equilibrium at boiling point

Sol: (a, d) During phase change process, temperature of the system remains constant.
(a) In region AB, a phase change takes place, heat is supplied and ice melts but temperature of the system is 0°C. it remains constant during process. The heat supplied is used to break bonding between molecules.
(b) In region CD, again a phase change takes place from a liquid to a vapour state during which temperature remains constant. It shows water and steam are in equilibrium at boiling point.

Q12. A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?
(a) The rate of cooling is constant till milk attains the temperature of the surrounding.
(b) The temperature of milk falls off exponentially with time.
(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools.
(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.
Sol: (b, c, d) When hot milk spread on the table heat is transferred to the surroundings by conduction, convection and radiation. Because the surface area of poured milk on a table is more than the surface area of milk filled in a glass. Hence, its temperature falls off exponentially according to Newton’s law of cooling. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to the surroundings. So, option (b), (c) and (d) are correct.

Very Short Answer Type Questions
Q13. Is the bulb of a thermometer made of diathermic or adiabatic wall?
Sol: The bulb of a thermometer is made up of diathermic wall because diathermic walls allow exchange of heat energy between two systems but adiabatic walls do not. So it receives heat from the body to measure the temperature of body.

Q14. A student records the initial length l , change in temperature ∆ T and change in length ∆ l of a rod as follows:

S. No	l(m)	∆T(°C)	∆l(m)
1.	2	10	4 x 10-4
2.	1	10	4 x 10-4
3.	2	20	2 x 10-4
4.	3	10	6x 10-4

If the first observation is correct, what can you say about observations 2, 3 and 4.

Sol: If the first observation is correct, hence from the 1st observation we get the coefficient of linear expansion

For 4th observation, ∆l = ∝l∆T

= 2 x 10^-5 x 3 x 10 = 6 x 10^-4 m [i.e., observed value is correct]

Q15. Why does a metal bar appear hotter than a wooden bar at the same temperature? Equivalently it also appears cooler than wooden bar if they are both colder than room temperature.

Q16. Calculate the temperature which has numeral value on Celsius and Fahrenheit scale.
Sol: To construct a scale of temperature, two fixed points are taken. First fixed point is the freezing point of water, it is called lower fixed point. The second fixed point is the boiling point of water, it is called upper fixed point. Temperature on one scale can be converted into other scale by using the following identity.

Q17. These days people use steel utensils with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor. Junction

Sol: The copper bottom of the steel utensil gets heated quickly.
Because of the reason that copper is a good conductor of heat as compared to steel. But steel does not conduct as quickly, thereby allowing food inside to get heated uniformly.

Short Answer Type Questions
Q18. Find out the increase in moment of inertia I of a uniform rod (coefficient of linear expansion a) about its perpendicular bisector when its temperature is slightly increased by ∆T.
Sol: Moment of inertia of a uniform rod of mass M and length l about its perpendicular bisector

Q19. During summers in India, one Of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavored sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter in those areas where it snows, people make snow balls and throw around. Explain the formation of ball out of crushed ice or snow in the light of p-T diagram of water.
Sol : Given diagram shows the variation of pressure with temperature for water. When the pressure is increased in solid state (at 0°, 1 atm), ice changes into liquid state while decreasing pressure in liquid state (at 0°, 1 atm), water changes to ice.

When crushed ice is squeezed, some of it melts, filling up the gap between ice flakes upon releasing pressure. This water freezes, binding all ice flakes and making the ball more stable.

Q20. 100 g of water is super cooled to -10°C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?

 

Q21. One day in the morning Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 min before he could take bath. Now, he had two options (i) fill the remaining bucket completely by cold water and then attend to the work, (ii) first attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain

Sol:According to the Newton’s law o‘f cooling, the rate of loss of heat is directly proportional to the difference of temperature. Or we can say which gives a consequence about rate of fall of temperature of a body with respect to the difference of temperature of body and surroundings.
The first option would have kept water warmer because by adding hot water to cold water, the temperature of the mixture decreases. Due to this temperature difference between the mixed water in the bucket and the surrounding decreases, thereby the decrease in the rate of loss of the heat by the water.
In second option, the hot water in the bucket will lose heat quickly. So if he first attend to the work and fill the remaining bucket with cold water which already lose much heat in 5-10 minutes then the water become more colder as comparison with first case.

Long Answer Type Questions
Q22. We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that difference between their length B remain constant. If α_iron = 1.2 x 10 ^-5/K and α_brass = 1.8xl0^-5/K, what should we take as length of each strip?

Sol: According to the problem, L₁-L_b = 10 cm where,
L₁ = length of iron scale
L_b = Length of brass scale
This condition is possible if change in length both the rods is remain same at all temperatures.

Q23. We would like to make a vessel whose volume does not change with temperature (take a hint from the problem above). We can use brass and iron (β_vbrass ⁼ 6 x 10 ⁵ / K and β_viron = 3.55 x 10^-5/K) to create a volume of 100 cc. How do you think you can achieve this?
Sol:Here we are making a vessel whose Brass volume does not change with temperature.
To make the desired vessel, we should have an iron vessel with a brass rod inside as shown in the diagram.

Therefore, an iron vessel with a volume of 249.9 cm³ fitted with a brass rod of volume 144.9 cm³ will serve as a vessel of volume 100 cm³, which will not change with temperature.
Important points:

• Solids can expand in one dimension (linear expansion), two dimensions (superficial expansion) and three dimensions (volume expansion) while liquids and gases usually suffers change in volume only.
• Thermal expansion is minimum in case of solids but maximum in case of gases because intermoleeular force is maximum in solids but minimum in gases.

Q24. Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57°C is drunk. You can take body (tooth) temperature to be 37°C and a = 1.7x 10^-5/°C bulk modulus for copper = 140x 10⁹N/m².

This is about 10³ times of atmospheric pressure

Q25. A rail track made of steel having length 10 m is clamped on a railway line at its two ends (figure). On a summer day due to rise in temperature by 20°C. It is deformed as shown in figure. Find x (displacement of the centre) if27

Q26. A thin rod. having length L₀ at 0°C and coefficient of linear expansion α has its two ends maintained at temperatures θ₁, and θ₂, Find its new length.
Sol. When temperature of a rod varies linearly, then average temperature of the middle point of the rod can be taken as mean of temperatures at the two ends. According to the diagram,

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We hope the NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter help you. If you have any query regarding NCERT Exemplar Class 11 Physics Chapter 10 Thermal Properties of Matter, drop a comment below and we will get back to you at the earliest.