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NCERT Exercises

Question 1.
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Answer:
Here, M = 2.50 kg, T = 200 N, length of the string, 1 = 20 m
Therefore, mass per unit length of the string,

Question 2.
A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top? Given that the speed of sound in air is 340 m s^_1?
(g = 9.8 m s^-2)
Answer:
Here, h = 300 m, g = 9.8 m s ^-2, v = 340 m s^-1 if h = time taken by stone to strike the surface of water in the pond, then from

Question 3.
A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals to the speed of sound in dry air at 20 °C = 343 m s^_1.
Answer:
Here, l = 12.0 m, M = 2.10 kg, T = ?, ν = 343 m s^-1
Mass per unit length ,μ

Question 4.
Use the formula   to explain why the speed of sound in air
(a) is independent of pressure,
(b) increases with temperature,
(c) increases with humidity.
Answer:
(a) According to Boyle’s law, we know that
PV = constant at a constant temperature … (i)
Now volume of gas, 
∴ From (i) and (ii), we get
 = constant.
Since mass of a gas remains constant.
 = constant.
g is always constant for a given gas or air
∴  from  equation ,we conclude that velocity of sound in air always remaun constant if its temperature is constant.

where M = ρV = molecular weight of the  gas
As ϒ,R and M are Constants, So α√T
i.e. velocity of sound in a gas is directly proportional to the square root of its I      temperature, hence we conclude that the velocity of sound in air increases with increase in temperature.
(c) Effect of humidity : The presence of water vapours in  air changes  the  density of air, thus the velocity of sound changes  with  humidity of air.
Let ρ_m = density of moist air.
ρ_d = density of dry air.
_υm = velocity of sound in moist air.
_υd = velocity of sound in dry air.

Also we know that density of water vapours is less than the density of dry air i.e. dry air is heavier than water vapours as the molecular mass of water is less than that of N₂ (28) and O₂ (32), so p_m <  ρ_d.

i.e. velocity of sound in air increases with humidity, i.e. velocity of sound in moist air is greater than velocity of sound in dry air. That is why sound travels faster on rainy day than on a dry day.

Question 5.
You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination (x – vt) to (x + vt), i.e. y = f (x ± vt). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:
(a) (x-vt)²
(b) log [(x + vt)/x₀]
(c) 1/(x + vt)
Answer:
No, the converse is not true.
The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, the wave function must have a finite value.
Out of the given functions for y, none of these satisfies this condition, so these functions do not represent a travelling wave.

Question 6.
A bat emits ultrasonic sound of frequency 1000 kHz in air. If this sound meets a water surface, what is the wavelength of
(a) the reflected sound,
(b) the transmitted sound? Speed of sound in air is 340 m s-¹ and in water 1486 ms-¹.
Answer:
Here, υ = 1000 kHz = 1000 x 10³ Hz = 10⁶ Hz;
speed of sound in air,υ_a = 340 m s^_1;
speed of sound in water, v_w = 1,486 m s^_1
(a) The reflected sound: After reflection, the ultrasonic sound continues to travel in air. If λ_a is wavelength in air, then

(b) The transmitted sound : The transmitted ultrasonic sound travels in water. If λ_w is wavelength of ultrasonic sound in water, then

Question 7.
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s¹? The operating frequency of the scanner is 4.2 MHz.
Answer:
Here, υ= 1.7 km s^_1 = 1.7 x 10³ m s^_1;
υ= 4.2 MHz = 4.2 x 10⁶ Hz

Question 8.
A transverse harmonic wave on a string is described by
y(x, t) = 3.0 Sin (36 t + 0.018 x + π/4)
where x and y are in cm and t in s. The positive direction of x is from left to right.
(a)  Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
Answer:
The equation of the form

represents a harmonic wave of amplitude A, wavelength l and travelling from right to left with a
velocity v.
Now, the given equation for the transverse harmonic wave is

(a) Since the equation (i) and (ii) are of the same form, the given equation also represents a travelling wave propagating from right to left. Further, the coefficient of t gives the speed of the wave. Therefore, v = 2000 cm s^_1 = 20 m s^-1
(b) Obviously, amplitude, A = 3.0 cm

(c) Initial phase at the origin, φ= rad
(d) Least distance between two successive crests in the wave is equal to wavelength.

Question 9.
For the wave described in question no. 8, plot the displacement (y) versus (t) graphs for x = 0,2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer:

For different values of t, we calculate y using equation (i). These values are tabulated below : On plotting y versus t graph, we obtain a sinusoidal curve as shown in above figure.

Similar graphs are obtained for x = 2 cm and x = 4 cm.
The oscillatory motion in travelling wave differs from one point to another only in terms of phase. Amplitude and frequency of oscillatory motion remain the same in all the three cases.

Question 10.
For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m
(b) 0.5 m
(c) λ/2
(d) 3λ/4
Answer:
Given equation of a travelling harmonic wave is
y(x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35) …….(i)
The standard equation of a travelling harmonic wave

Comparing equation (i) and (ii), we get

Question 11.
The transverse displacement of a string (clamped at its two ends) is given by
y(x,f) = 0.06sin cos(120 πt)
Where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 x 10 ² kg. Answer the following:
(a) Does the function represent a travelling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What are the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.
Answer:
Here, the equation for transverse displacement is given by
y(x,f) = 0.06sin cos 120 πt …………(i)
(a) The displacement, which involves harmonic functions of x and f separately, represents a stationary wave and the displacement, which is harmonic function of the form (υt ± x), represents a travelling wave. Hence, the equation given above represents a stationary wave.
(b) When a wave pulse

Question 12.
(i) For the wave on a string described in question no. 11, do all the points on the string oscillate with the same
(a) frequency,
(b) phase,
(c) amplitude? Explain your answers.
(ii) What is the amplitude of a point 0.375 m away from one end?
Answer:
(i) All the points on the string
(a) have the same frequency of oscillation (except at the nodes, where the frequency is zero).
(b) have the same phase between two adjacent nodes,
(c) but not the same amplitude.
(ii) Now, the amplitude of the given wave is
A = 0.06 sin 
At x = 0.375 m, the amplitude is given by
A(at x = 0.375 m) = 0.06 sin  x 0.375
= 0 06 sin  = 0.06 x 0.707 = 0.042 m 4

Question 13.
Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent
(1) a travelling wave,
(2) a stationary wave or
(3) none at all:

(a) y=2cos(3x)sin(10t)
(b) y = 2-√x-vt
(c) y = 3 sin (5x – 0.5 t) + 4 cos (5x – 0.5t)
(d) y = cos x sin t + cos 2x sin 2t

Answer:
(a) This equation has two harmonic functions of each x and t separately, so it represents stationary wave.
(b) This function does not represent any wave as it contains no harmonic function.
(c) It represents progressive/travelling harmonic wave as the arguments of cosine and sin functions are same.
(d) This equation is the sum of two functions cos x sin t and cos 2x sin 2t each represen­ting a stationary wave. Therefore it represents superposition of two stationary waves

Question 14.
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 x 10^-2 kg and its linear mass density is 4.0 x 10 ² kg m^-1. What is
(a) the speed of a transverse wave on the string, and
(b) the tension in the string?
Answer:
Here, υ = 45 Hz, M = 3.5 x 10^-2 kg;
mass/length =μ = 4.0 x  10^-2  kg m^-1

Question 15.
A meter-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
Answer:
As there is a piston at one end of the tube, it behaves as a closed organ pipe, which produces odd harmonics only. Therefore, the pipe is in resonance with the fundamental note and the third harmonic (79.3 cm is about 3 times 25.5 cm]
In the fundamental mode,  = l₁= 25.5 cm
λ = 4 x 25.5 = 102 cm = 1.02 m.
Speed of sound in air,
υ = υλ= 340 x 1.02 = 346.8 m s^-1.

Question 16.
A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is speed of sound in steel?
Answer:

A rod clamped in the middle has antinodes (A) at its ends and node (N) at the point of clamping. In fundamental mode, thus the length of the rod is
l= or 2l
Where l = length of rod and also λ = wavelength of the wave
Here, l = 100 cm
υ = 2.53 kHz = 2.53 x 10 ³ Hz
∴ λ= 2 x 100 = 200 cm
If υ be the speed of sound in steel, then
υ = υ λ = 2.53 x 10³ x 200
= 506 x 10³ cm s^_1
= 5.06 x 10³ ms^-1
υ= 5.06 km s^_1.

Question 17.
A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s^_1
Answer:
Here, L = 20 cm = 0.2 m, υ„ = 430 Hz, υ= 340 m s^_1
The frequency of n^th normal mode of vibration of closed pipe is


Question 18.
Two sitar strings A and B playing the note ‘Ga’ are slight out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Answer:
We know that  υ ∝ √T
where u = frequency, T = tension
The decrease in the tension of a string decreases its frequency.
So let us assume that original frequency υ_A of A is more than the frequency υ_B of B.
Thus υ_A – υ_B = ± 6 Hz (given)
and υ_A = 324 Hz
∴  324 – υ_B  = ± 6
or υ_B = 324 ± 6 = 318 Hz or 330 Hz.
On reducing tension of A, Δυ= 3 Hz
If υ_B  = 330 Hz and on decreasing tension in A, will be reduced i.e. no. of beats will increase, but this is not so because no. of beats becomes 3.
∴ must be 318 Hz because on reducing the tension in string A, its frequency may be reduced to 321 Hz, thus giving 3 beats with υ_B = 318 Hz.

Question 19.
Explain why (or how):
(a) in a sound wave, a displacement node is a pressure antinode and vice versa,
(b) bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”,
(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,
(d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
(e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Answer:
(a) In a sound wave, a node is a point where the amplitude of oscillation i.e. displacement is zero as here a compression and a rarefaction meet and the pressure is maximum, so it is called pressure antinode. While an antinode is a point where the amplitude of oscillation is maximum i.e. displacement is maximum but pressure is minimum. So this point is called pressure node.
Hence displacement node coincides with pressure antinode and displacement antinode with pressure node.
(b) Bats emit ultrasonic waves of large frequencies (small wavelength) when they fly. These ultrasonic waves are received by them after reflection from the obstacle. Their ears are so sensitive and trained that they not only get the information of the distance of the obstacle but also that of the nature of the reflecting surface.
(c) The quality of the sound produced by an instrument depends upon the number of overtones. Since the number of overtones is different in the cases of sounds produced by violin and sitar therefore we can distinguish through them.
(d) Solids possess both the volume elasticity and the shear elasticity. Therefore they can support both longitudinal and transverse waves.
On the other hand, gases have only the volume elasticity and no shear elasticity, so only longitudinal waves can propagate in gases.
(e) A sound pulse is a combination of waves of different wavelengths. In a dispersive medium, the waves of different wavelengths travel with different speeds in different directions e. with different velocities. So the shape of the pulse gets distorted i.e. a plane wave front in a non- dispersive medium does not remain a plane wave front in a dispersive medium.

Question 20.
A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air.
(i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of 10 m s^-1.
(b) recedes from the platform with a speed of 10 m s^-1?
(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms¹.
Answer:
Here, frequency of source of sound,υ = 400 Hz; v = 340 m s^_1 = speed of sound
Speed of source = υ_s = 10m s^_1
(i) (a) When the train approaches the platform, the apparent frequency as heard by the observer on the platform will be

(b) When the train recedes from the platform, the apparent frequency as heard by the observer is given by according to the formula :

(ii) The speed of sound in each case remains same i.e.340 ms^-1.

Question 21.
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^-1. What are the frequency, wavelength, and speed of sound for an observer standing on the station platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s^-1? The speed of sound in still air can be taken as 340 m s^-1.
Answer:
Here, v = 340 m s^-1; υ = 400 Hz
(a) Speed of the wind, v_w = 10 m s^-1
As the direction of blow of wind (yard to station) is the same as the direction of sound, therefore, for the observer standing on the platform; velocity of sound,
υ’ =υ + υ_w = 340 + 10 = 350 m s^-1
As there is no relative motion between source of the sound and the observer, the frequency of sound will remain unchanged.
Thus, frequency of sound = 400 Hz
Wavelength of sound,

(b) Speed of the observer, υ₀ = 10 m s^-1 (towards yard). When the observer moves towards the source of sound, the apparent frequency,

The wavelength of sound waves is not affected due to the motion of the observer and hence the wavelength of the sound waves will remain unchanged.
Speed of sound relative to the observer
= 340 + 10 = 350 m s^-1
Therefore, situations (a) and (b) are not equivalent.

Question 22.
A travelling harmonic wave on a string is described by y (x, t) = 7.5 sin (0.0050 + 12t + π/4)
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacement and velocity as the x = 1 cm point at f = 2 s, 5 s and 11s.
Answer:



Question 23.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium.
Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation?
If the pulse rate is 1 after every 20 s, that is the whistle is blown for a split of second after every 20 s, is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:
(a) A narrow sound pulse such as a short pip by a whistle does not have a definite wavelength or frequency. However, being a sound wave, it has a definite speed.
(b) If a short pip is produced after every 20 s, then frequency of the note produced by the whistle cannot be called 1/20 or 0.05 Hz. We may call 0.05 Hz as the frequency of repetition of the short pip.

Question 24.
One end of a long string of linear mass density 8 x 10 ³ kg m^-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t=0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer:
Here, μ = 8.0 x 10^-3 kg m^-1, u = 256 Hz,
T= 90 kg = 90 x 9.8 = 882 N
Amplitude of wave, A =0 cm = 0.05 m.
As, the wave propagating along the string is a transverse travelling wave, the velocity of the wave is given by

As, the wave is propagating along positive x direction the equation of the wave is
y(x-t) =A sin(ωt-kx)
= 0.05 sin (1.61 x 10³ t- 4.84 x)
Here x, y, are in meter and t is in second.

Question 25.
A SONAR system fixed in a submarine operator at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h^-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s^-1.
Answer:
Here, μ = 40.0 kHz ;
Speed of sound wave in water, υ = 1, 450 m s^_1
The frequency of the waves from SONAR system will undergo a change in the following two steps:
(1) Frequency of waves from SONAR system as received by the enemy submarine moving towards the system :

(2) Frequency of waves from the enemy’s submarine as received by SONAR system : The enemy submarine will reflect the waves of frequency υ’ = 42.756 kHz and will thus act as a source of waves moving with a speed; υ_s = 100 m s^_1 toward the SONAR system (listener). If υ” is the apparent frequency as received by SONAR system, then

Question 26.
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of S wave is about km s^-1, and that of P wave is 8.0 km s^-1. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?
Answer:
Let υ₁,υ₂ be the velocities of S waves and P waves, and t₁ ,t₂ be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earthquake from the seismograph, then
l = υ₁t₁ = υ₂t₂       -(i)
Now, υ₁= 4 km s^_1 and υ₂ = 8 km s^_1
∴  4t₁ =8 t₂ or t₁ = 2t₂               …(ii)
Also, t₁ – t₂ = 4 min = 240 s.
Using (ii), 2t₂ – t₂ = 240 s ;t₂ = 240 s
t₁= 2 x t₂ = 2 x 240 = 480 s
Now, from (i) l = υ₁t₁ = 4 x 480 = 1920 km.
Hence earthquake occurs 1920 km away from the seismograph.

Question 27.
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly towards a flat wall surface, the bat is moving at 0.03 times the speed of the sound in air. What frequency does the bat hear reflected off the wall?
Answer:
Here, υ = 40 kHz,
speed of sound in air,υ = 340 m s^_1
∴  v’ = speed of bat
= 0.03 x υ = 0.03 x 340 = 10.20 m s^-1

The bat moving towards wall acts as a moving source and for the sound waves reflected from the wall, it acts as a moving observer. Thus the source and observer are approaching each other with same speed. i.e. υ_s = υ₀ = υ’ = 10.2 m s^-1
Thus apparent frequency of the reflected sound waves heard by the bat is given by

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 15 Waves and we hope this detailed NCERT Solutions are helpful.

NCERT Exercises

Question 1.
Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its center of mass.
(d) An arrow released from a bow.
Answer:
(a) It is not a periodic motion as the swimmer completes only one trip but if he makes more than one trip and time for each trip is same, the motion can be categorised as periodic.
(b) It is a periodic motion because a freely suspended bar magnet if once displaced from N-S direction and released, it oscillates about this position. Hence it is simple harmonic motion also.
(c) It is also a periodic motion.
(d) It is not a periodic motion.

Question 2.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) The rotation of earth about its axis.
(b) Motion of an oscillating mercury column in a U-tube.
(c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) General vibrations of a polyatomic molecule about its equilibrium position.
Answer:
(a) It is periodic but not simple harmonic motion because it is not to and fro motion about a fixed point.
(b) It is simple harmonic motion.
(c) It is simple harmonic motion.
(d) It is periodic but not simple harmonic motion. A polyatomic molecule has a number of natural frequencies and its general motion is the resultant of simple harmonic motions of a number of different frequencies. The resultant motion is periodic but not simple harmonic motion.

Question 3.
Figure depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

Answer:
(a) Does not represent periodic motion, as the motion neither repeats nor comes to mean position.
(b) Represents periodic motion with period equal to 2 s.
(c) Does not represent periodic motion, because it is not identically repeated.
(d) Represents periodic motion with period equal to 2 s.

Question 4.
Which of the following functions of time represent
(a) simple harmonic,
(b) periodic but not simple harmonic, and
(c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):

(a) Sin ωt-cos ωt
(b) Sin³ ωt
(c) 3 cos( -2ωt)
(d) Cos ωt +Cos 3ωt+cos 5ωt
(e) exp(-ω² t²)
(f) 1+ωt+ω²t²

Answer:
The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent simple harmonic motion, if it can be written

(d)    cos ωt+ cos 3ωt +cos 5 ωt
It represents periodic but not simple harmonic motion. Its time period is
 .It can be noted that each term represents a periodic function with a different angular frequency. Since period is the least interval of time after which a function repeats its
value, cos ωt has a period T = , cos3 ωt has a period , cos5 ωt has period , the last two forms repeat after any integral multiple of their period. Thus each term in the sum repeats itself after T, and hence the sum is a periodic function with a period .
(e) exp (-ω² t²) : It is an exponential function which decreases monotonically with increasing time and tends to zero as t →∞ and thus never repeats itself. Therefore it represents non-periodic motion.
(f)  1 + ωt+ (-ω² t² ): It represents non-periodic motion (physically unacceptable because the function tends to infinity as t →∞).

Question 5.
A particle is in linear simple harmonic motion between two points, A and 6 10 cm apart. Take the direction from  A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Answer:

Refer figure, here A and B represent the two extreme positions of a simple harmonic motion. For velocity, the direction from A to B is taken to be positive. For acceleration and the force, the direction is taken positive if directed along AP and negative if directed along BP.
(a) Harmonic motion is momentarily at rest being its extreme position of motion. Therefore, its velocity is zero, acceleration is +ve because it is directed along AP. Force is also +ve, since the force is directed along AP, i.e., +ve direction.
∴     0, + ,+
(b) At the end B, velocity is zero. Here, acceleration and force are negative as they are directed along BP, i.e., along negative direction.
∴   0, -, –
(c) At the mid point of AB going towards A, the particle is at its mean position P, with a tendency to move along PA, i.e., -ve direction. Hence, velocity is -ve. Both, acceleration and force are zero.
∴   -, 0,0
(d) At 2 cm away from B going towards A, the particle is at Q, with a tendency to move along QP, which is negative direction.Here velocity, acceleration and force, all are negative.
∴   -, -, –
(e) At 3 cm away from A going towards B, the particle is at R, with a tendency to move along RP, which is positive direction. Here, velocity, acceleration and force, all are positive.
∴ +,  +, +
(f) At 4 cm away from B going towards A, the ,particle is at S with a tendency to move along SP which is negative direction,Here, velocity, acceleration and force, all are negative.
∴   -, -, –

Question 6.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) o = -200x²
(c) a = -10 x
(d) o=100x³
Answer:
A particle is said to be executing simple harmonic motion, if the acceleration ‘a’ produced in it satisfies the following two conditions:
(1) ‘a’ is directly proportional to the displacement (say y) from the mean position, i.e. a ∝ y.
(2) ‘a’ is directed towards mean position i.e. acts opposite to the direction in which y increases.
Mathematically a = -ω²y …(i)
where ω = angular frequency.

(a) a =7x does not satisfy eqn. (i), so it does not represent simple harmonic motion.
(b) a = -200.x², does not satisfy equation (i) hence it does not represent simple harmonic motion.
(c) a = -10x, here x = It satisfies equation (i), so it represents simple harmonic motion.
(d) a = 100x³ does not satisfy equation (i), hence it does not represent simple harmonic motion.

Question 7.
The motion of a particle executing simple harmonic motion is described by the displacement function.
x(t) = A cos(ωt+ φ)
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm s^-1, what are its amplitude and initial phase angle? The angular frequency of the particle is Π s^-1 If instead of the cosine function, we choose the sine function to describe the simple harmonic motion : x = B sin ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.
Answer:
Here, at t = 0, x = 1 cm and υ= co cm s^_1;
ω= Π S^_1
Given : x = A cos(ωt + φ) … (i)
Since at t = 0, x = 1, we get
1 = A cos(Π x 0 +φ) = A cosφ
The instantaneous particle velocity is given by


Question 8.
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
Answer:
Here, m = 50 kg,
Maximum extension, y = 20 – 0 = 20 cm = 0.2 m
Maximum force, F=mg = 50 x 9.8 = 490 N

Question 9.
A spring having a spring constant 1200 N m^-1 is mounted on a horizontal table as shown in figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine
(i) the frequency of oscillations of the mass,
(ii)maximum acceleration of the mass,and
(iii) the maximum speed of the mass.
Answer:
Here, m = 3.0 kg; k = 1200 N m^_1,
A = 2 cm = 0.02 m
(i) The frequency of oscillations of the attached mass is

Question 10.
In previous question let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t= 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
In what way do these functions for simple harmonic motion differ from each other, in frequency, in amplitude or the initial phase?
Answer:
From previous solution

(a) As time is noted from the mean position, hence using
x = A sin ωt, we get x = 2 sin 20t
(b) At maximum stretched position, the mass is at the extreme right position, with an  initial phase of  rad. Then,

Question 11.
Figures (a) and (b) correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:
In figure
(a) T = 2 s; A = 3 cm
At t = 0, OP makes angle  with x-axis, i.e. = radian. While moving clockwise, here φ = +  Thus the x-projection of OP at time t will give us the equation of simple harmonic motion, given by

or x = -3 sin ωt (where x is in cm)
In figure
(b) T = 4s;A = 2m At t = 0, OP makes an angle π  with the positive direction of x-axis, i.e.  φ= π While moving anticlockwise, here φ=+π
Thus the x-projection of OP at time t will give us the equation of simple harmonic motion given by

Question 12.
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t=0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s)
(a) x=-2sin 
(b) x=cos 
(c) x=3sin 
(d) x=2 cos πt
Answer:

(d) Here, x = 2 cos πt
A = 2 cm, φ = 0 and ω= π rad s^-1
Therefore, at t = 0, the particle is at the point P on the right extreme position as shown in figure (iv).

Question 13.
Figure (i) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (ii) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in figure (ii) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in figure (i) and the two masses in figure (ii) are released, what is the period of oscillation in each case?
Answer:
(a) Maximum extension of the spring :
(i) Suppose the maximum extension produced in the spring is y. Then,
F = ky   (in magnitude)
or y = F/k
(ii) In this case, each mass relative to the other behaves as if the other mass is fixed. In other words, force F on each mass acts as the force of reaction developed due to force F on the other mass. Therefore, in this case also, maximum extension is given by
y = F/k
(b) Period of oscillation:
In figure
(i), T = 2n T=2π√m/k
To calculate the period of oscillation, the spring in figure
(ii) can be considered as to be equivalent to the two springs, each of length 1/2 and joined at the point O, the center of the string as shown in the figure.

If k’ is force constant of each half, then k’ = 2k (Because, if a spring is cut to half of its length, its force constant becomes double. Therefore,
T = 2π√m/2k

Question 14.
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad min^-1, what is its maximum speed?
Answer:
Stroke of piston = 2 times the amplitude.
Let A = amplitude,
Stroke = 1 m (given)
∴ 1 =2A or A =  m
Now, v_max =  = 200 x  = 100 m/min
= ms^-1=  ms^-1

Question 15.
The acceleration due to gravity on the surface of moon is 1.7 m s^-2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 m s^-2)
Answer:

Question 16.
Answer the following questions:
(a) Time period of a particle in S.H.M. depends on the force constant k and mass m of the particle :
T = 2π√ . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π√  Think of a qualitative argument to appreciate this result.
(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time under gravity?
(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
Answer:
(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.
(b) The effective restoring force acting on the bob of simple pendulum in displaced position is
F = -mgsinB. When θ is small, sin θ = θ . Then the expression for time period of simple pendulum is given by
T = 2π√ 
When θ is large sin θ < θ. If the restoring force mgsinθ is replaced by mgQ, this amounts to effective reduction in the value of ‘g’ for large angles and hence an increase in the value of time period T.
(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
(d) We know that gravity disappears under free fall, so frequency is zero.

Question 17.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
Answer:
When the car moves along a circular track, the bob of the simple pendulum will possess centripetal acceleration,
a_c=υ²/R  (radially inward along horizontal)
The acceleration due to gravity (g) acts vertically downwards.
Therefore, effective acceleration of the pendulum,

Question 18.
A cylindrical piece of cork of base area A and height h floats in a liquid of density p,. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

where p is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer:
Let X be the equilibrium position of a cylinder floating in a given liquid.
A = area of cross-section of the cylindrical piece of cork.
h = height of the cylindrical piece of cork,
ρ = density of the material of the cylindrical cork

ρ_c= density of the liquid in which it floats.
l = length of the cylindrical piece of cork dipping in the liquid upto point P in position X.
W = weight of the cylindrical cork.
W₁ = weight of the liquid displaced by the cork.
V = its volume m = mass of cork = Ahρ.
∴ V= Ah
W=mg = (Vρ)g = (Ahρ)g
W₁ = Area of cross-section of cork x length of cylinder dipping in liquid x density of liquid x g = Alρ₁g
According to the law of flotation,

Question 19.
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the liquid column of mercury in the (U-tube executes simple harmonic motion.
Answer:
The suction pump creates the pressure difference, thus mercury rises in one limb of the U-Tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be explained as:

Consider the mercury contained in a vertical U-tube upto the level P and Q in its two limbs.
Let ρ = density of the mercury.
L = total length of the mercury column in both the limbs.
A = internal cross-sectional area of U-tube.
m = mass of mercury in U-tube = LAρ.
Let the mercury be depressed in left limb to P’ by small distance y, then it rises by the same amount in the right limb to position Q’
∴‍ Difference in levels in the two limbs
= P’Q’ = 2y
∴ Volume of mercury contained in the column of length
2y = A x 2y
m’ = A x 2y x ρ x g
If W = weight of liquid contained in the column of length 2y
Then W=m’g = A x 2y x ρ x g
This weight produces the restoring force (F) which tends to bring back the mercury to its equilibrium position.
∴ F = -2Ayρg = -(2Aρg)y
If a = acceleration produced in the liquid column, then

where h = height of mercury in each limb. Now from (i), it is clear that a ∝  y and -ve sign shows that it acts opposite to y, so the motion of mercury in U-tube is simple harmonic in nature having time period (T) given by

Question 20.
An air chamber of volume V has a neck area of cross section A into which a ball of mass m just fits and can move up and down without any friction as shown in figure. Show that when the ball is pressed down a little and released  it executes SHM. Obtain an expression for the time period of oscillations assuming pressure- volume variations of air to be isothermal

Answer:
Consider an air chamber of volume V with a long neck of uniform area of cross section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, the pressure of air below the ball inside the chamber is equal to the atmospheric pressure. Increase the pressure on the ball by a little amount P, so that the ball is depressed to position D, where CD = y.

There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Δy

Clearly, F  ∝ y, Negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, it will start executing linear SHM in the neck of chamber with C as mean position.
In SHM, the restoring force, F = -ky Comparing (i) and (ii), we have k = BA²/V, which is the spring factor.Now, inertia factor = mass of ball = m.
As,

Question 21.
You are riding an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of
(a) the spring constant k and
(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
Answer:
(a) Here mass supported by each wheel, M = 750 kg and x =15 cm = 0.15 m
If k is spring constant of the spring, then restoring force developed on being compressed through a distance x
F = -kx
If M is mass supported by each wheel, then
∴  k x = Mg

If b is damping constant for the spring and shock absorber system, then damped amplitude of oscillation is given by
A=A₀e-^bt/2M
where T is period of oscillation of the spring, A₀ the initial amplitude of the oscillations and  M, the mass supported by it. From the above relation we have
A=A₀e-^bt/2M
As the amplitude of oscillation decreases by 50% during one complete oscillation,


Question 22.
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Answer:
Consider a particle of mass m executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by y = A sin ωt

Question 23.
A circular disc of mass 10 kg is suspended by a wire attached to its center. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = -αθ, where J is the restoring couple and 0 the angle of twist)
Answer:
Here, m = 10 kg; R = 15 cm ; T = 1.5 s Moment of inertia of the disc about wire
I =  MR² = –x 10 x (0.15)² = 0.1125 kg m²
When the wire is twisted by rotating the disc and then released, restoring torque will be set up. If a is angular acceleration produced, then restoring torque
τ = lα                                                         …(i)
The torsional spring constant k of the wire is defined by the relation
τ = -kθ                                                       …(ii)
where θ  is the angle of twist From the equations (i) and (ii) we have
lα= -kθ  or α= -k/l  θ
Since k/I is constant, it follows that angular acceleration is directly proportional to the angle of twist (angular displacement). Hence, the motion executed by the disc is simple harmonic in nature and the period of torsional oscillation is given by

Question 24.
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of body when the displacement is
(a) 5 cm,
(b) 3 cm and
(c) 0 cm.
Answer:
Here, r = 5 cm = 0.05 m; T = 0.2 s;

Question 25.
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the center with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω ,x₀ and v₀.
Answer:
Let the displacement of the particle at any time t be represented by
x = Acos(ωt + φ_O)   …(i)
where A = amplitude, φ_O ⁼ initial phase If ν be the velocity of the particle at time t,

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NCERT Exercises

Question 1.
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the
diameter of an oxygen molecule to be 3 Å
Answer:
Here, the diameter of an oxygen molecule
d = 3 A = 3 x 10^-10 m
The radius of an oxygen molecule

Question 2.
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22.4 liters.
Answer:
Here, R = 8.31 J mol^-1 K ^-1
T = 0°C = 273 K; n= 1 mole
P = 1 atm = 0.76 of Hg column
= 0.76 x 13.6 x 10³ x 9.8 = 1.013 x 10⁵ N m²

Question 3.
Figure shows plot of PV/T versus P for 1.00 x 10^-3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: T, > T₂ or T, < T₂?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00 x 10^-3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H₂ = 2.02 u, of 0₂ = 32.0 u, ff = 8.31 J mol^-1 MK^-1.)
Answer:
(a) Since the dotted plot is parallel to P-axis, it tells that value of  remains constant even when P is changed.
(b) Curve at temperature T, is more close to the dotted plot than the curve at the temperature T₂. Since the behaviour of a real gas approaches the perfect gas behaviour, as the temperature is increased therefore T₁ > T₂.
(c) If the amount of gas under consideration is 1 mole, then the value of PV/T, where the curves meet PV/T- axis will be R (=8.31 J mole^-1 K^-1). For oxygen, molecular mass is 32.0 g i.e. 32.0 x 10^-3 kg. Since mass of oxygen gas under considerations is 1.0 ^x 10^-3 kg, the value of PV/T, where the curves meet the PV/T-axis is given by

(d) If we obtained similar plots for 1.0 x 10^-3 kg of hydrogen, the value of PV/T, where the curves meet the y-axis will not be the same. Since molecular mass of hydrogen is 2.02 x 10^-3 kg, the mass of hydrogen that will yield same value of PV/T i.e., 0.26 J K^-1 will be

Question 4.
An oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol^-1 K^-1,
molecular mass of 0₂ = 32 u).
Answer:
Here, initially in the oxygen cylinder
Volume V₂ = 30 liters = 30 x 10^-3 m³
Pressure P₁= 15 atm = 15 x 1.013 x 10⁵ Pa
Temperature = 27 + 273 = 300 K
If the cylinder contains,let n₂ moles of the oxygen gas then P₁V₁ = n₁RT,

Initial mass of oxygen in the cylinder
m_x = n₁ molecular weight of 0₂
= 18.253 x 32 = 584.1 g
Finally in the oxygen cylinder, let n₂ moles of oxygen be left
Final volume V₂ = 30 x 10^-3 m³
Final pressure P₂ = 11 atm = 11 x 1.013 x 10⁵ Pa
Final temperature T₂ = 17 + 273 = 290 K

Question 5.
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35°C?
Answer:
When the air bubble is at depth of 40 m
Volume of air bubble V₂ = 1.0 cm³ = 1.0 x 10^-6 m³
Pressure of air bubble P₁ = 1 atm + 40 m depth of water = 1 atm + h₁ρg
= 1.013 x 10⁵ + 40 x 10³ x 9.8 = 4.93 x 10⁵ Pa
Temperature of air bubble T₁ = 12 + 273 = 285 K
When the air bubbles reaches at the surface of lake, pressure of air bubbles
P₂ = 1 atm = 1.013 x 10⁵ Pa
Temperature of air bubble T₂ = 35°C = 35 + 273 = 308 K
Volume of air bubbles = ?

Question 6.
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27°C and 1 atm pressure.
Answer:
Here, volume of air molecules, V = 25.0 m³
Temperature of air molecules
T = 27°C = 27 + 273 = 300 K
Pressure of air molecules
P = 1 atm = 1.013 x 10^s Pa
Now, PV = nRT

Question 7.
Estimate the average thermal energy of a helium atom at
(i) room temperature (27°C),
(ii) the temperature on the surface of the Sun(6000 K),
(iii) the temperature of 10 million kelvin (the typical core temperature in case of a star).
Answer:
The average kinetic energy of the gas at a temperature T

Question 8.
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v_ms the largest?
Answer:
All the three vessels (at the same temperature and pressure) have same volume. So, in accordance with the Avogadro’s law, the three vessels will contain equal number  of respective molecules, being equal to Avogadro’s number N = 6.023 x 10²³.

at a given temperature therefore, rms speed of molecules will not be the same in the three cases. As neon has the smallest mass, therefore, rms speed will be the largest in case of neon.

Question 9.
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20°C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Answer:
Here, atomic mass of argon M₁= 39.9
Atomic mass of helium He = 4.0
Let C₁ and C₂ be the rms velocities of argon and helium at temperature T₁K and T₂ K respectively.
We know that rms speed is given by

Question 10.
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17°C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).
Answer:
Here, Pressure of nitrogen = 2 atm
= 2 x 1.013 x 10⁵ N m-² = 2.026 x 10⁵ Nm²
Temperature of nitrogen = 17° C
= 17 + 273 = 290 K
Molecular diameter d = 2 x 1= 2Å= 2x 10^-10 m

Question 11.
A meter long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer:
When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end, figure
(a) The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.

∴ P₁ = 76 cm and V₁ = 15 cm³
When the tube is held vertically, 15 cm air gets another 9 cm of air filled in the right hand side (in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure, figure
(b) Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively.
The pressure of air = 76 – (76 – h) = h cm of mercury
∴ V₂ = (24 + h) cm³ and P₂ = h cm.

Since h cannot be negative (because more mercury cannot flow into the tube), therefore h = 23.8 cm. Thus, in the vertical

Question 12.
From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^-1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^-1. Identify the gas.
Answer:
According to Graham’s law of diffusion

Question 13.
A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
n₂ = n₁ exp [-mg (h₂ – h₁)/k_BT]
where n₂,n₁ refer to number density at heights h₂ and /i, respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
n₂ = n₁ exp [-mg N_A (ρ – ρ’) [h₂– h₁)/(ρRT)]
where ρ is the density of the suspended particle, and ρ’ that of surrounding medium.
[N_A is Avogadro’s number, and R is the universal gas constant.]
Answer:
According to the law of atmospheres,

where n₂, n₂ refer to number density of particles at heights h₂ and h₁ , respectively.
If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we will have to take effective weight of the suspended particles.
Let V = average volume of a suspended particle, r = density of suspended particle, r’ = density of liquid, m = mass of equal volume of liquid displaced.
According to Archimede’s principle, effective weight of one suspended particle = actual weight – weight of liquid displaced = mg- m’g

Question 14.
Given below are the densities of some solids and liquids. Give rough estimates of the sizes of their atoms

Answer:

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NCERT Exercises

Question 1.
A geyser heats water flowing at the rate of 3.0 liters per minute from 27°C to 77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 x 10⁴ J g^_1 ?
Answer:
Here, mass of water heated,
m = 3000 g min^-1
Volume of water heated = 3.0 lit min¹
Rise in temperature ΔT = (77 – 27)°C = 50°C
Specific heat of water c = 4.2 J s^-1 °C^-1
Amount of heat used ΔQ = mc ΔT

Question 2.
What amount of heat must be supplied to 2.0 x 10^-2 kg of nitrogen (at room temperature)to raise its temperature by 45°C at constant pressure? (Molecular mass of N₂ = 28;R = 8.3 J mol^-1 K¹.)
Answer:
Mass of nitrogen gas,
m = 2 x 10² kg = 20 g
Molecular mass of nitrogen gas = 28 g
Rise in temperature, ΔT = 45°C
Heat supplied = ?

For nitrogen (N₂) which is a diatomic gas, molar specific heat at constant pressure
C_p =  R where R = 8.3 J mol^-1 K^_1
Therefore, heat supplied to the gas
ΔQ = nC_pΔT = x  x R x 45 =  x x8.3 x 45
ΔQ = 933.8 J = 934 J

Question 3.
Explain why
(a) Two bodies at different temperatures T₁) and T₂ if brought in thermal contact do not necessarily settle to the mean temperature (T₁ + T₂)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Answer:
(a) In thermal contact, heat flows from the body at higher temperature to the body at lower temperature till temperatures become equal. The final temperature can be the mean temperature (T₁+ T₂)/2 only when thermal capacities of the two bodies are equal.
(b) This is because heat absorbed by a substance is directly proportional to the specific heat of the substance.
(c) During driving, the temperature of air inside the tyre increases due to motion. According to Charle’s law, P T. Therefore, air pressure inside the tyre increases.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Answer:
As the gas is completely insulated, the process is adiabatic.

Question 5.
In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)
Answer:
ln first case, the state of the gas changes adiabatically from A to
Therefore, ΔQ = 0, ΔW= – 22.3 J
If ΔH is change in internal energy of the system then,
ΔQ = ΔH + ΔW
0 = ΔH + ΔW
ΔH = – ΔW = – (- 22.3 J) = 22.3 J
In the second case, when the state A is taken to state B, the heat absorbed by the system
ΔQ = 9.35 cal = 9.35 x 4.2 J = 39.27 J = 39.3 J
ΔW = ?
Applying first law of thermodynamics ΔQ = ΔH + ΔW
ΔW =ΔH – ΔQ = 39.3 – 22.3 = 17.0 J

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock. The cylinder A contains a gas at standard temperature and pressure, while the cylinder B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface?
Answer:
(a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmosphere pressure will become two times. Therefore, pressure will decrease to one-half,i.e.. 0.5 atmosphere.
(b) There will be no change in the internal energy of the gas as no work is done on/ by the gas.
(c) Also, there will be no change in temperature of the gas as gas does no work in expansion.
(d) No, because the process called free expansion is rapid and cannot be controlled. The intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

Question 7.
A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Answer:
Useful work done per min (output)
= 5.4 x 10⁸ J
Heat absorbed per min (input) = 3.6 x 10⁹ J

= 0.15 = 0.15 x 100% = 15%
Heat energy wasted per minute = Heat energy
wasted per useful work done per minute
= 3.6 x 10⁹ – 5.4 x 10⁸ = 3.06 x 10⁹ J min^-1

Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Answer:
Given that
Heat supplied ΔQ = 100 W = 100 J s
Useful work done ΔW = 75 J s^-1
Change internal energy ΔU = ?
According to first law of thermodynamics,
change in internal energy is given by
ΔU = ΔQ – ΔW
ΔU = 100 – 75 = 25 J s-¹ = 25 W

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in figure.

Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.
Answer:
Total work done by gas equals to area enclosed.
W = Area DEED
=  x EF x DF
=   x (5 – 2) x (600 – 300) =450J

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.
Answer:
Here, T₁ = 36°C = 36 + 273 = 309 K
T₂ = 9°C = 9 + 273 = 283 K
Coefficient of performance of the refrigerator

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 12 Thermodynamics and we hope this detailed NCERT Solutions are helpful.

NCERT Exercises

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer:
Here, Triple point of neon, T, = 24.57 K
Triple point of CO_2, T₂ = 216.55 K
Relation between kelvin scale and Celsius scales
T_c=T_k– 273.15
where
T_c = Temperature on Celsius scale
T_k = Temperature on kelvin scale
For neon T_c = 24.57 – 273.15 = -248.58°C
For CO₂, T_c = 216.55 – 273.15 = -56.60°C
Relation between kelvin and Fahrenheit scales
F=C+32
For neon T_F = x – 248.58 + 32 = -415.44°F
For CO₂T_F = x – 56.60° + 32 = – 69.88°F 5

Question 2.
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_e?
Answer:
Here, triple point of water on scale
A = 200 A
triple point of water on scale B = 350 B triple point of water on kelvin scale = 273.16 K
According to question
200 A = 350 B = 273.16 K

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R₀ (1 + α (T-T₀)] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
Answer:
Here, R₀ = 101.6 Ω; T₀ = 273.16 K
Case (i) R₁ = 165.5 Ω; T, = 600.5 K
Case (ii) R₂ = 123.4 Ω; T₂ = ?
Using the relation R = R₀ [1 + α(T – T₀)]
Case (i) 165.5 = 101.6 [1 + a (600.5 – 273.16)]


Question 4.
Answer the following :
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by
t_c =T-273.15
Why do we have 273.15 in this relation, and not 273.16?
(d) What is the temperature of the triple­ point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
Answer:
(a) This is because the triple point of water has a unique value i.e. 273.16 K at a unique point, where exists unique values of pressure and volume. On other hand, the melting point of ice and boiling point of water do not have unique set of values as they change with the change in pressure and volume.
(b) The kelvin absolute scales also have the fixed points as the Celsius scales have. The other fixed point is absolute zero. It corresponds to the temperature, when the volume and pressure of a gas will become zero.
(c) Triple point of water on Celsius scale is 01°C and on kelvin scale is 273.16 and the size of degree on the two scale is same, so t_c – 0.01 = T- 273.16
∴ t_c = T- 273.15
(d) The unit interval size of Fahrenheit scale is 212 – 32 = 180 divisions Also we know that the unit interval size of absolute scale is 100.
∴ Triple point of water on an absolute scale having 180 divisions is given by

Question 5.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers
A and B?
(b) What do you think is the reason behind the slightly difference in answers of thermometers A and 6? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Answer:
Let T be the melting point of sulphur.
The triple point of water, T_tr = 273.16 K
For thermometer A : P_tr = 1.25 x 10⁵ Pa
P = 1.797 x 10⁵ Pa

Question 6.
A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0°C? Coefficient of linear expansion of steel = 1.20 x 10-⁵ K-¹
Answer:
Here, length of the steel tape at 27°C is 100 cm.
∴ L = 100 cm, T = 27°C
The length of steel tape at 45°C is V = L + ΔL
L’ = L + αLΔT (∵AL = aLΔT)
L’ = L + αL(T₂ – T₁)
L’ = 100 + (1.20 x 10^-5) x 100 x (45° – 27°)
= 100.0216 cm
Length of 1 cm mark at 27°C on this scale, at
45°c=
Length of 63 cm measured by this tape at 45°C will be = x 63 = 63.0136 cm.
When temperature is 27°C, the size of 1 cm mark on the steel tape will be exactly 1 cm as the steel tape has been calibrated at 27°C. Therefore, length of the steel of the rod at 27°C = 63 x 1 = 63 cm

Question 7.
A large steel wheel is to be fitted on to a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 x 10 ⁵ K^-1.
Answer:
Here, T₁  = 27°C = 27 + 273 = 300 K
Let L₁ and L₂ be the linear dimensions of steel at temperatures T₁ and T₂ respectively.
Now L, = 8.70 cm, I₂ = 8.69 cm,

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of
copper = 1.70 x 10^-5 K^-1.
Answer:
Here, Coefficient of linear expansion of copper, α = 1.70 x 10^-5 °C^-1
ΔT = 227 -27 = 200°C
Therefore, coefficient of superficial expansion of copper


Question 9.
A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = 2.0 x 10^-5 K^-1; Young’s modulus of brass = 0.91 x 10¹¹pa.
Answer:
Here l₁ = 1.8 m, t₁= 27°C, t₂ = -39°C
∴t = t₂– t1= 39 – 27 = -66°C
_l2 = length at t₂°C
For brass, α = 2 x 10^-5 K
∴ Y = 0.91 x 10¹¹ Pa
diameter of wire,  d = 2.0 mm = 2.0 x 10^-3 m
If A be the area of cross-section of the wire,

Negative sign indicates that the force is inwards due to the contraction of the wire.

Question 10.
A brass rod of length 50 cm and diameter mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’developed at the junction? The ends of the rod are free to expand
(Coefficient of linear expansion of brass = 2.0 x 10^-5 °C^-1, steel = 1.2 x 10^-5 °C^-1).
Answer:
For brass rod : α′ = 2.0 x 10^-5 °C^-1 ;
l′₁ = 50 cm; ΔT = 250 – 40 = 210°C
The length of the brass rod at 250°C is given by
l′₂ =l′₁ (1+ α′ ΔT)
= 50(1 + 2.0 x 10^-5 x 210) = 50.126 cm
For steel rod: α’ = 1.2 x 10 ⁵ °C ^-1
l′₁= 50 cm;
ΔT’ = 250 – 40 = 210°C
The length of the steel rod at 250°C is given by
l′₂ +l′₁ { (1 + α’ ΔT’)
= 50(1 + 1.2 x 10^-5 x 210)
= 50.126 cm
Therefore, the length of the combined rod at 250°C
= l₂ + l′₂ = 50.21 + 50.126 = 100.336 cm
As the length of the combined rod at 40°C
= 50 + 50 = 100 cm
The change in length of the combined rod at  250°C = 100.336 -100.0 = 0.336 cm
No thermal stress is developed at the junction since the rod freely expand.

Question 11.
The coefficient of volume expansion of glycerine is 49 x 10^-5 C^-1. What is the fractional change in its density for a 30°C rise in temperature?
Answer:
Here, γ= 49 x 10^-5 °C^-1; ΔT = 30°C
Let there be m grams of glycerine and its initial volume be V. Suppose that the volume of the glycerine becomes V after a rise of temperature of 30°C then,

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg^-1 °C^-1.
Answer:
Here, P = 10 kW = 10⁴ W,
mass, to = 8.0 kg = 8 x 10³ g
rise in temperature, ΔT = ?,
time, t = 2.5 min = 2.5 x 60 = 150 s
Sp. heat, c = 0.91 J g^-1 °C^-1
Total energy, = P x t = 10⁴ x 150 = 15 x 10⁵ J
As, 50% of energy is lost,

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt?
(Specific heat of copper = 0.39 J g^-1 °C^_1; heat of fusion of water = 335 J g¹).
Answer:
Here, mass of copper block, m₁ = 2.5 kg
Specific heat of copper,
C = 0.39 Jg^-1 K-¹ = 0.39 x 10³ J kg^{-1 0}C
Temperature of furnace, ΔT = 500°C Latent heat of fusion,
L = 335 Jg^-1 = 335 x 10³ J kg^-1
If Q be the heat absorbed by the copper block,then

Question 14.
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
Answer:
Here mass of metal, m = 0.20 kg = 200 g
Fall in temperature of metal
ΔT = 150 – 40 = 110°C
If c is specific heat of the metal, then heat lost by the metal,
ΔQ = mcΔT = 200 x c x 110
Volume of water = 150 c.c.
∴ Mass of water, m’ = 150 g
Water equivalent of calorimeter,
ω= 0.025 kg = 25 g
Rise in temperature of water and calorimeter, ΔT’ = 40-27 = 13°C
Heat gained by water and calorimeter,
ΔQ’ = (m’ + w)ΔT’
= (150 + 25) x 13 = 175 x 13
As, ΔQ = ΔQ’
From (i) and (ii), 200 x c x 110 = 175 x 13

If some heat is lost to the surroundings, value of c so obtained will be less than the actual value of c.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.

The measured molar specific heats of these gases are markedly different from these for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal mol^-1 K¹. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
Answer:
The gas which are listed in the above table are diatomic gases and not mono-atomic gases. For diatomic gases, molar specific heat

which agrees fairly well with all observations listed in the table except for chlorine. A monoatomic gas molecule has only the translational motion. A diatomic gas molecule, apart from translation motion, the vibrational as well as rotational motion is also possible. Therefore, to raise the temperature of 1 mole of a diatomic gas through 1°C, heat is to be supplied to increase not only translational energy but also rotational and vibrational energies. Hence, molar specific heat of a diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. shows that for chlorine molecule, at room temperature vibrational motion also occurs along with translational and rotational motions, whereas other diatomic molecules at room temperature usually have rotational motion apart from their translation motion. This is the reason that chlorine has somewhat larger value of molar specific heat.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO₂?
(c) What are the critical temperature and pressure for CO₂? What is their significance?
(d) Is CO₂ solid, liquid or gas at

(a) -70°C under 1 atm,
(b) -60°C under 10 atm,
(c) 15°C under 56 atm?

Answer:

(a) The solid, liquid and vapour phases of C0₂ can exist in equilibrium at its triple point O,   corresponding to which P_tr = 5.11 atm and T_tr = – 56.6°C
(b) From the vaporisation curve (I) and the fusion curve (II), it follows that both the boiling and fusion points of CO₂ decrease with the decrease of pressure.
(c) For CO₂, P_c =0 atm and T_c = 31.1°C Above its critical temperature, CO₂ gas can not be liquified, however large pressure may be applied.
(d)

(a) -70°C under 1 atm : This point lied in vapour region. Therefore, at -70°C under 1 atm, CO₂ is vapour.
(b) -60°C under 10 atm : this point lies in solid region. Therefore, CO₂ is solid at -60°C under 10 atm.
(c) 15°C under 56 atm : This point lies in liquid region. Therefore, CO₂ is liquid at 15°C under 56 atm.

Question 17.
Answer the following questions based on the P-Tphase diagram of CO₂ as given Q. 16:
(a) CO₂ at 1 atm pressure and temperature -60°C is t:c ipressed isothermally. Does it go through a liquid phase?
(b) What happens when CO₂ at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO₂ at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure.
(d) CO₂ is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
Answer:
(a) No. CO₂ at 1 atm pressure and -60°C is vapour. If it is compressed isothermally when pressure is increased without changing the temperature, it will go to solid phase directly without going through the liquid phase.
(b) CO₂ at 4 atm pressure and at room temperature (say 25°C) is vapour. If it is cooled at constant pressure, it will again condense to solid without going through the liquid phase (the horizontal line through the initial point intersects only the sublimation curve III).
(c) CO₂ at 10 atm pressure and at -65°C is solid. As CO₂ is heated at constant pressure, it will go to liquid phase and then to the vapour phase. It is because, the horizontal line through the initial point intersects both the fusion and vapourisation curves. The fusion and boiling points can be known from the points, where the horizontal line at P-T diagram at 10 atm (initial point) intersects the respective curves.
(d) It will not exhibit any clear phase transition to the liquid phase. However, CO₂ gas will depart more and more from the ideal gas behaviour as its pressure increases.

Question 18.
A child running a temperature of 101°F is given an antipyrine (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g^-1.
Answer:
Here, fall in temperature = ΔT

Question 19.
A ‘thermocole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6h.The outside temperature is 45°C, and co-efficient of thermal conductivity of thermocole is 0.01 J s^-1 K^-1. [Heat of fusion of water = 335 x 10^J J kg^-1]
Answer:
Here, length of each side,
Z = 30 cm = 0.3 m
Thickness of each side, Ax = 5 cm = 0.05 m
total surface area through which heat enters into the box,
A = 6l² = 6 x 0.3 x 0.3 = 0.54 m²
Temp, diff., ΔT = 45 – 0 = 45°C,
K = 0.01 Jm^-1m^-1 °C^-1
time, At = 6 hrs = 6 x 60 x 60 s
Latent heat of fusion, L = 335 x 10³ J kg^-1
Let m be the mass of ice melted in this time

Question 20.
A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of kg min^-1 when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s^-1 m^-1 K¹; Heat of vaporisation of water = 2256 x 10³ J kg¹.
Answer:
Here, K = 109 J s^-1 m¹ K¹; A = 0.15 m²; d = 1.0 cm = 10^-2 m; T₂ = 100°C, t = 1 min = 60 s Let T₁ be the temperature of the part of boiler in contact with the stove. Therefore, amount of heat flowing per minute through the base of the boiler,

Question 21.
Explain why:
(a) a body with large reflectivity is a poor emitter.
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water
Answer:
(a) We know that a + r + t = 1 Where a, r and t are absorbance, reflectance and transmittance respectively of the surface of the body, t is also called emittance (e). Also according to Krichhoff’s law
e α a that is good absorber are good emitters and hence poor reflectors and vice-versa i.e. If r is large (i.e. large reflectively) a is smaller and hence e is smaller i.e. poor emitter.
(b) The thermal conductivity of brass is high e. brass is a good conductor of heat. So when a brass tumbler is touched, heat quickly flows from human body to the tumbler. Consequently, the tumbler appears colder. On the other hand, wood is a bad conductor of heat. So heat does not flow from the human body to the wooden tray, thus it appears relatively hotter.
(c) Let T be the temperature of the hot iron in the furnace. Thus according to Stefan’s law, heat radiated per second per unit area (E) is given by E = σT⁴ . When the body is placed in open at temperature T₀, then the heat radiated/sec/area (E’) is given by
E’ = σ(T₄-T⁴₀)
Clearly E’ < E, so the optical pyrometer gives too low a value for the temperature of a red hot iron piece in open.
(d) Gases are generally insulators. The Earth’s atmosphere acts like an insulating blanket around it and does not allow heat to escape out but reflects it back to the Earth. If this atmosphere is absent, then the Earth would naturally be colder as all its heat would have escaped out.
(e) This is because steam has much higher heat capacity (540 cal g^_1) than the heat capacity of water (80 cal g^_1) at the same temperature. Thus heating systems based on circulation of steam are more efficient than those based on circulation of hot water.

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
Answer:
According to Newton’s law of cooling, the rate of loss of heat ∝ cooling difference in temperature.
Here, average of 80°C and 50°C = 65°C Temperature of surroundings = 20°C
∴ Difference = 65 – 20 = 45°C
Under these conditions, the body cools 30°C in 5 minutes.

Now that you are provided all the necessary information regarding NCERT Solutions for Class 11 Physics Physics Chapter 11 Thermal Properties of matter and we hope this detailed NCERT Solutions are helpful.

NCERT Exercises

Question 1.
Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.
Answer:
(a) The height of the blood column in the human body is more for the feet as compared to that for the brain that is why the blood exerts more pressure at the feet than at the brain.
(b) We know that the density of air is maximum near the surface of earth and decreases rapidly with height and at a height of about 6 km, it decreases to nearly half of its value at the sea level. Beyond 6 km height, the density of air decreases very slowly with height. Due to this reason, the atmospheric pressure at a height of about 6 km decreases to nearly half of its value at sea level.
(c) Since due to the applied force on the liquid, according to Pascal’s law, the pressure is transmitted equally in all directions inside the liquid. Thus there is no fixed direction for the pressure due to liquid. Hence hydrostatic pressure is a scalar quantity.

Question 2.
Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape.
Answer:
(a) When a small quantity of liquid is poured on solid, three interfaces namely liquid-air, solid-air and solid- liquid are formed. The surface tensions corresponding to these three interfaces i.e. T_LA,T_SA and T_SL respectively are related to the angle of contact of a liquid with a solid as:

In case of mercury-glass, T_SA < T_SL therefore, from
(i) cos θ is negative or θ > 90° i.e. θ is obtuse. But on the other hand, in case of water- glass, T_sa >T_SLi so from (i) cos θ is positive i.e. θ is less than 90° or acute.

(b) For equilibrium of a liquid drop on the surface of a solid, the equation
T_SA = T_SL + T_LA cos θ …(i)
must be satisfied. For mercury-glass, angle of contact is obtuse. In order to achieve this obtuse value of angle of contact, the mercury tends to form a drop.
But in case of water-glass, the angle of contact is acute, so equation (i) is not satisfied. In order to achieve this acute value of angle of contact, the water tends to spread.

(c) Surface tension of liquid is defined as the force acting per unit length on either side of an imaginary line drawn tangentially on the surface of the liquid at rest. Since this force is independent of the area of the liquid surface, therefore surface tension is also independent of the area of the liquid surface.

(d) We know that the cloth has narrow spaces in the form of fine capillaries. The rise of liquid in a capillary tube is given by

i.e. h α cos θ. It follows that if 0 is small, cos θ will be large and detergent will rise more in the narrow spaces in the cloth. Now as detergents having small angles of contact can penetrate more in cloth.

(e) In the absence of external forces, only force acting on the liquid drop is due to surface tension. A drop of liquid tends to acquire minimum surface area due to the property of surface tension. Since for a given volume of liquid, surface area is minimum for a sphere, so the liquid drop will always assume a spherical shape.

Question 3.
Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally… with temperatures (increases/decreases)
(b) Viscosity of gases … with temperature, whereas viscosity of liquids … with temperature (increases/decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to …., while for fluids it is proportional to …. (shear strain/rate of shear strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows…….(conservation of mass/Bernoulli’s principle)
(e) For the model of a plane in a wind tunnel, turbulence occurs at a … speed than for turbulence for an actual plane (greater/ smaller)
Answer:
(a) decreases
(b) increases; decreases
(c) shear strain; rate of shear strain
(d) conservation of mass of Bernoulli’s principle
(e) greater.

Question 4.
Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it.
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers.
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection.
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel.
(e) A spinning cricket ball in air does not follow a parabolic trajectory
Answer:
(a) If we blow over a piece of paper, velocity of air above the paper becomes more than that below it. As K.E. of air above the paper increases, so in accordance with
Bernoulli’s theorem  ²=constant,
its pressure energy and hence its pressure decreases. Due to greater value of pressure below the piece of paper = atmospheric . pressure, it remains horizontal and does not fall.
On the other hand if we blow under the paper, the pressure on the lower side decreases. The atmospheric pressure above the paper will Therefore bend the paper downwards. So the paper will not remain horizontal.
(b) This can be explained from the equation of continuity i.e. = a₂υ₂. As we try to close a water tap with our fingers, the area of cross-section of the outlet of water jet is reduced considerably as the openings between our fingers provide constriction (i.e., regions of smaller area).
Thus velocity of water increases greatly and fast jets of water come through the openings between our fingers.
(c) According to Bernoulli’s theorem, we know that
…………..(i)
Here, the size of the needle controls the velocity of flow and the thumb pressure controls pressure.
Now P occurs with power one and velocity v occurring with power 2 in equation (i),hence the velocity has more influence. That is why the needle of syringe has a better control over the flow rate.
(d) When a fluid is flowing out of a small hole in a vessel, it acquires a large velocity and hence possesses large momentum. Since no external force is acting on the system, a backward velocity must be attained by the vessel (according to the law of conservation of momentum). As a result of it, backward thrust is experienced by the vessel.
(e) This is due to Magnus effect : Let a ball moving to the right be given a spin at the top of the ball. The velocity of air at the top is higher than the velocity of air below the ball. So according to Bernoulli’s theorem, the pressure above the ball is less than the pressure below the ball. Thus there is a net upward force on the spinning ball, so the ball follows a curved path. This dynamic lift due to spinning is known as Magnus effect.

Question 5.
A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?
Answer:

Question 6.
Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m^-3. Determine the height of the wine column for normal atmospheric pressure.
Answer:
P= Normal atmospheric pressure = 1.013 x 10⁵ Pa
Let h be the height of the French wine column which earth’s atmosphere can support.
∴ If P’ be the pressure corresponding to height h of wine column,
Then, P’ = hρ_ωg
where ρ_ω = density of wine = 984 kg m ^-3
Now according to given statement
P’ = P
or hρ_wg = P

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.
Answer:
Here, depth of water column, h = 3 km = 3 x 10³ m
density of water, ρ = 10³ kg m ^-3
If P be the pressure exerted by this water column at this depth, then
P = hρg = 3 x 10³ x 10³ x 9.8
= 29.4 x 10⁶ Pa = 30 x 10⁶ Pa = 3 x 10⁷ Pa
As the structure is put on the sea, sea water will exert upward thrust of 3 x 10⁷ Pa. Maximum stress which the vertical off-shore structure can withstand = 10⁹ Pa. (Given) Then 3 x 10⁷ Pa < 10⁹ Pa Thus we conclude that the structure is suitable as the stress applied by it is much lesser than the maximum stress it can withstand.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear?
Answer:
The maximum force, which the bigger piston can bear,
F = 3000 kg f = 3000 x 9.8 N

Since the liquid transmits pressure equally, therefore the maximum pressure the smaller piston can bear is 6.92 x 10⁵ Pa.

Question 9.
A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?
Answer:
Here, for water column in one arm of U-tube :
h₁ = 10.0 cm
ρ₁ = 1 g cm^-3
For spirit column in other arm of U-tube,
h₂ = 12.5 cm
p₂ = ?

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms?
(Specific gravity of mercury = 13.6)
Answer:
On pouring 15 cm of water and spirit each into the respective arms of the U-tube, the mercury level will rise in the arm containing the spirit.
ρ_w= density of mercury.
Let us select two points A and B lying in the same horizontal plane as shown. Thus according to Pascal’s law,



Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Answer:
No, Bernoulli’s theorem is used only for stream-line flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer:
No, it does not matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation, provided the atmospheric pressure at the two points where Bernoulli’s equation is applied are significantly different.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 x 10^-3 kg s^-1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 x 10³ kg m³ and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].
Answer:
Here, l = 1.5 m; r = 1 cm = 0.01 m;
q = 0.83 Pa s
mass of the glycerine flowing per sec,
m = 4.0 x 10^-3 kg s^-1
density of glycerine, p = 1.3 x 10³ kg m³ Therefore, volume of glycerine flowing per second,


Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s^-1 and 63 m s^-1 respectively. What is the lift on the wing if its area is
2.5 m²? Take the density of air to be 1.3 kg m^-3.
Answer:
Let υ₁ and υ₂ be the speeds on the upper and lower surfaces of the wings of the aeroplane respectively,
P₁ and P₂ be the pressures on the upper and lower surfaces of the wings respectively.
Here, υ₁= 70 m s^-1; υ₂ = 63 m s^-1; p = 1.3 kg m³
The level of the upper and lower surfaces of the wings from the ground may be taken same.
∴ h₁-h₂
area of wing, A = 2.5 m²
Thus from Bernoulli’s Theorem,

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Answer:
Figure (a) is incorrect. According to equation of continuity, i.e., aυ= a constant,
where area of cross-section of tube is less, the velocity of liquid flow is more. So the velocity of liquid flow at a constriction of tube is more than the other portion of tube.
According to Bernoulli’s Theorem,
 ²= a constant,
i.e., where υ is more, P is less and vice versa.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² on one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1, what is the speed of ejection of the liquid through the holes?
Answer:
Here, cross-section of the tube,
a₁ = 8.0 cm² = 8.0 x 10^-4 m²;
The speed of liquid in the tube

Question 17.
A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Answer:
We know that a soap film has two free surfaces, so total length of the film to be supported, Z = 2 x 30 cm
or l= 60 cm = 0.60 m
Let T = surface tension of the film
If F = total force on the slider due to surface tension, then
P = T x 2l = T x 0.6 N
W= 1.5 x 10^-2 N
In equilibrium position, the force F on the slider due to surface tension must be balanced by the weight (W) supported by the slider.

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 x 10^-2 N. What is the weight supported by a film of the same liquid at the same temperature in figure (b) and (c)? Explain your answer physically.

Answer:
(a) Here, length of the film supporting the weight l = 40 cm = 0.4 m
Total weight supported (i.e. force) = 4.5 x 10^-2 N. Film has two free surfaces,
∴ Surface tension,

Since the liquid is same for all the cases (a), (b) and (c), and temperature is also same, therefore surface tension for cases (b) and (c) will also be the same = 5.625  x 10^-2. In figure (b), and (c), the length of the film supporting the weight is also the same as that of (a), hence the total weight supported in each case is 4.5 x 10-² N.

Question 19.
What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 x 10^-1 N m¹. The atmospheric pressure is 1.01 x 10^s Pa. Also give the excess pressure inside the drop.
Answer:
Here, radius of drop,
R = 3.0 mm = 3.0 x 10^-3 m;
Surface tension of mercury,
T = 4.65 x 10^-1 Nm¹
Pressure outside the mercury drop,
P₀ = atmospheric pressure = 1.01 x  Pa
If P_i is pressure inside the drop, then excess of pressure inside the mercury drop,

Question 20.
What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 x 10^-2 Nm¹? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 x 10⁵ Pa).
Answer:
Here, surface tension of the soap solution,
T = 2.5 x 10² Nm¹
density of the soap solution, p = 1.2 x 10³ kg m³;
radius of the soap bubble,

Question 21.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m, compute the force necessary to keep the door closed.
Answer:
For compartment containing water,
h₁ = 4 m, ρ₁ = 10³ kg m^-3.
The pressure exerted by water at the door provided at bottom,

To keep the door closed, the force equal to 55 N should be applied horizontally on the door from compartment containing water to that containing acid.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in figure, (a). When a pump removes some of the gas, the manometer reads as in figure (b). The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b) in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (ignore the small change in the volume of the gas).

Answer:
Here, atmospheric pressure, P = 76 cm of mercury.
(a) In figure (a) pressure head, h = +20 cm
∴ Absolute pressure = P + h = 76 + 20 = 96 cm of mercury
Absolute pressure = P + h = 76 + (-18) = 58 cm of mercury
Gauge pressure = h = -18 cm of mercury
(b) Here 13.6 cm of water added in right limb 13.6 , is equivalent to  = 1 cm of mercury column.
i.e. h’ = 1 cm of mercury column.
Now pressure at A,
P_a = P + h’ = 76 + 1 = 77 cm
Let the difference in mercury levels in the two limbs be h_v then pressure at b,
P_b = 58 + h₁,
As, P_a = P_b’
∴ 77 = 58 + h₁, or h₁ = 77 – 58 = 19 cm of mercury column.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessels the same in the two cases? If so, why do the vessel filled with water to that same height give different readings on a weighing scale?
Answer:
Since the pressure depends upon the height of water column and the height of the water column in the two vessels of different shapes is the same, hence there will be same pressure due to water on the base of each vessel. As the base area of each vessel is same, hence there will be equal force acting on the two base areas due to water pressure. The water exerts force on the walls of the vessel also. In case, the walls of the vessel are not perpendicular to base, the force exerted by water on the walls has a net non-zero vertical component which is more in first vessel than that of second vessel. That is why, the two vessels filled with water to same vertical height show different readings on a weighing machine.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein?
(Density of blood = 1.06 x 10³ kg m^-3)
Answer:
Here, gauge pressure, P = 2000 Pa
density of whole blood, ρ = 1.06 x 10³ kg m³
g = 9.8m s^-2
Let h = height at which blood container must be placed = ?
Using the relation, P = hρg, we get

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy.
(a) How does the pressure changes as the fluid moves along the tube if dissipative forces are present? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Answer:
(a) While deriving Bernoulli’s equation, we say that, decrease in pressure energy per second = increase in K.E./sec. + increase in P.E./sec. Assume that viscous forces are absent. Thus as the fluid flows from lower to upper edge, there is a fall of pressure energy due to the fall of pressure. If dissipating forces are present, then a part of this pressure energy will be used in overcoming these forces during the flow of fluid. So there shall be greater drop of pressure as the fluid moves along the tube.
(b) Yes, the dissipative forces become more important as the fluid velocity increases. According to Newton’s law of viscous drag,
we know that F = -ηA 
Clearly as v increases, velocity gradient increases and hence viscous drag i.e. dissipative force also increases.

Question 26.
(a) What is the largest average velocity of blood flow in an artery of radius 2 x 10^-3 m if the flow must remain laminar?
(b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 x 10^-3 Pa s). Density of blood is 1.06 x 10³ kg/m³
Answer:
Here, p = 2 x 10^-3 m;
D = 2r = 2 x 2 x 10^-3 = 4 x 10^-3 m;
η= 2.084 x 10^-3 Pa s,
For flow to be laminar, N_R = 2000

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25 m². If the speed of the air is 180 km h^-1 over the lower wing and 234 km h^-1 over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m³).
Answer:
Here, υ₁ = 180 km h^-1 = 50 m s^-1, υ₂ = 234 km h^-1 = 65 m s^-1
area of the each wing, A = 25 m²
and density of air, ρ= 1 kg m ³
According to Bernoulli’s theorem,

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius
2.0 x 10 ⁵ m and density 1.2 x 10³ kg m³? Take the viscosity of air at the temperature of the experiment to be 1.8×10^-5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Answer:
Here, r = 2x 10^-5m, q = 1.8 x 10^-5 Pa s,
p = 1.2 x 10³ kg m^-3
When the buoyancy of the drop due to air is neglected, the terminal speed of the drop is given by

Question 29.
Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N m^-1. Density of mercury = 13.6 x 10³ kg m^-3.
Answer:
Here, angle of contact, θ = 140°
r = 1 mm = 10^-3 m
Surface tension, T = 0.465 N m^_1
density of mercury, ρ = 13.6 x 10³ kg m^-3

= – 5.34 x 10^-3 m = – 5.34 mm.
Here negative sign shows that the mercury level is depressed in the tube relative to the mercury surface outside

Question 30.
Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 x 10^-2N m^-1. Take the angle of contact to be zero and density of water to be 10 x 10³ kg m^-3 (g = 9.8 m s^-2).
Answer:
Here, T = 7.3 x 10^-2 Nm^-1
ρ = 1.0 x 10³ kg m^-3; θ = 0°
For narrow tube, 2r₁ = 3.00 mm = 3 x 10^-3 m or r₁ = 1.5 x 10-³ m
For wider tube, 2r₂ = 6.00 mm = 6 x 10^-3 m or r₂ = 3 x 10-³ m
Let h₁ h₂ be the heights to which water rises in narrow tube and wider tube respectively

Question 31.
(a) It is known that density r of air decreases with height r as………….
whereρ₀ = 1.25 kg m^-3 is the density at sea level, and y₀ is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of g remains constant.
(b) A large He balloon of volume 1425 m³ is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y₀ = 8000 m and ρ_0He  = 0.18 kg m³]
Answer:
(a) We know that the rate of decrease of density ρ of air is directly proportional to density ρ i.e.

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