Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 15 - Waves

Share with your friends

Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 15 Waves are provided here with simple step-by-step explanations. These solutions for Waves are extremely popular among class 11 Science students for Physics Waves Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Physics Ncert Exemplar 2019 Book of class 11 Science Physics Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Physics Ncert Exemplar 2019 Solutions. All Physics Ncert Exemplar 2019 Solutions for class 11 Science Physics are prepared by experts and are 100% accurate.

Page No 105:

Question 15.1:

Water waves produced by a motor boat sailing in water are
(a) neither longitudinal nor transverse.
(b) both longitudinal and transverse.
(c) only longitudinal.
(d) only transverse.

Answer:

When a motor boat travels, its propeller draws water backwards which generates the longitudinal waves, and due to its lateral motion the particles also move up and down which generates the transverse waves. Thus, the water waves generated by the motor boat sailing in water are both longitudinal and transverse.

Hence, the correct answer is option (b).

Page No 105:

Question 15.2:

Sound waves of wavelength λ travelling in a medium with a speed of v m/s enter into another medium where its speed is 2v m/s. Wavelength of sound waves in the second medium is

(a) λ

(b) $\frac{λ}{2}$

(c) 2λ

(d) 4λ

Answer:

As the sound waves travel from one medium to other its velocity and wavelength get changed whereas, its frequency remains constant.

Let the frequency of the sound waves be f.

In the first medium
$v = f λ$ ...(1)

Let the wavelength of the sound waves in the second medium be λ'

$2 v = f λ'$

Substituting the value of the v from equation (1)

$2 f λ = f λ' \Rightarrow λ' = 2 λ$

Hence, the correct answer is option (c).

Page No 106:

Question 15.3:

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

Answer:

Speed of sound wave in air is inversely proportional to the square root of the density of the air. As the humidity of the air increases the air density decreases and hence the speed of the sound wave in air increases.

Hence, the correct answer is option (c).

Page No 106:

Question 15.4:

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

Answer:

As the temperature increases the speed of the sound also increases, and as the speed of wave (v) and the wavelength (λ) are related as:

v = fλ

As the velocity increases, the wavelength also increases.

Hence, the correct answer is option (c).

Page No 106:

Question 15.5:

With propagation of longitudinal waves through a medium, the quantity transmitted is
(a) matter.
(b) energy.
(c) energy and matter.
(d) energy, matter and momentum.

Answer:

With the propagation of the longitudinal waves through a medium only energy is transmitted.

Hence, the correcy answer is option (b).

Page No 106:

Question 15.6:

Which of the following statements are true for wave motion?
(a) Mechanical transverse waves can propagate through all mediums.
(b) Longitudinal waves can propagate through solids only.
(c) Mechanical transverse waves can propagate through solids only.
(d) Longitudinal waves can propagate through vacuum.

Answer:

Mechanical wave needs a material medium for its propagation whether it is longitudinal or transverse wave. Longitudinal wave can propagate through all types of media whereas, the transverse waves can pass through solids only.

Hence, the correct answer is option (c).

Page No 106:

Question 15.7:

A sound wave is passing through air column in the form of compression and rarefaction. In consecutive compressions and rarefactions,
(a) density remains constant.
(b) Boyle’s law is obeyed.
(c) bulk modulus of air oscillates.
(d) there is no transfer of heat.

Answer:

The sound wave propagates as a series of compression and rarefaction, in compression the particles of the air are closer to each other whereas in rarefaction the particles are farther. In case of compression the density is high and in rarefaction the density is low. Thus, density is not constant. Bulk modulus of air does not change when the sound propagates. The propagation of sound is fast process and it is assumed that it is an adiabatic process. The propagation of sound is not isothermal, hence Boyle's law is not obeyed.

Hence, the correct answer is option (d)

Page No 106:

Question 15.8:

Equation of a plane progressive wave is given by $y = 0.6 \sin 2 π (t - \frac{x}{2}) .$ On reflection from a denser medium its amplitude becomes 2/3 of the amplitude of the incident wave. The equation of the reflected wave is

(a) $y = 0.6 \sin 2 π (t + \frac{x}{2})$

(b) $y = - 0.4 \sin 2 π (t + \frac{x}{2})$

(c) $y = 0.4 \sin 2 π (t + \frac{x}{2})$

(d) $y = - 0.4 \sin 2 π (t - \frac{x}{2}) .$

Answer:

According to the question the reflected wave's amplitude will be 2/3 times of the incident wave. Thus, the amplitude of the reflected wave will be $\frac{2}{3} \times 0.6 = 0.4 units$

As the wave is reflected from a denser medium a phase difference of π is introduced in it.

$y = 0.4 \sin [2 π (t - \frac{x}{2}) + π] = - 0.4 \sin [2 π (t - \frac{x}{2})]$

Hence, the correct answer is option (d).

Page No 107:

Question 15.9:

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

Answer:

Given,
Tension in the string, T = 200 N
Mass of the string, M = 2.50 kg
Length of the string, L = 20.0 m

Linear mass density of the string, $μ = \frac{M}{L} = \frac{2.5}{20} = 0.125 kg / m$

Velocity of the disturbance on a string is given by

$v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{200}{0.125}} = 40 m / s$

Time taken by the disturbance to reach the other end

$t = \frac{L}{v} = \frac{20}{40} = 0.5 s$

Hence, the correct answer is option (b).

Page No 107:

Question 15.10:

A train whistling at constant frequency is moving towards a station at a constant speed V. The train goes past a stationary observer on the station. The frequency nÎ„ of the sound as heard by the observer is plotted as a function of time t (Fig 15.1) . Identify the expected curve.

Answer:

In the given question, firstly the train is approaching a stationary observer then it passes by and starts moving away.

Let the speed of the sound in air be v
When the train approaches the observer, apparent frequency is given by

$n' = (\frac{v}{v - V}) n$

When the train moves away from the observer, apparent frequency is given by

$n' = (\frac{v}{v + V}) n$

Option (c) shows most suitable graph.

Hence, the correct answer is option (c).

Page No 108:

Question 15.11:

A transverse harmonic wave on a string is described by y (x, t) = 3.0 sin (36t + 0.018x + π/4) where x and y are in cm and t is in s. The positive direction of x is from left to right.
(a) The wave is travelling from right to left.
(b) The speed of the wave is 20 m/s.
(c) Frequency of the wave is 5.7 Hz.
(d) The least distance between two successive crests in the wave is 2.5 cm.

Answer:

The general equation of a travelling wave is given by:

y(x, t) = A sin (ωt − kx + Φ)

Where,

A = Amplitude
k = wave number
ω = angular frequency

The above equation is for that wave which travels in left to right direction.

The equation given in the question is: y (x, t) = 3.0 sin (36t + 0.018x + π/4)

It is the equation of the travelling wave which moves in right to left direction, for the coefficients of t and x have the same signs.

The speed of the wave is given by:

$v = \frac{ω}{k}$
Here,
ω = 36 s^-1
k = 0.018 cm^-1

$v = \frac{36}{0.018} = 2000 cm / s$

Frequency of the wave is given by

$f = \frac{ω}{2 π} = \frac{36}{2 \times 3.14} = 5.73 Hz$

The least distance between two successive crests in the wave is the wavelength and that can be calculated as:

$λ = \frac{2 π}{k} = \frac{2 π}{0.018} = 348.8 cm$

Hence, the correct answers are option (a), (b) and (c).

Page No 108:

Question 15.12:

The displacement of a string is given by y (x, t) = 0.06 sin (2πx/3) cos (120πt) 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 × 10^–2 kg.
(a) It represents a progressive wave of frequency 60 Hz.
(b) It represents a stationary wave of frequency 60 Hz.
(c) It is the result of superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m/s in opposite direction.
(d) Amplitude of this wave is constant.

Answer:

The general equation of a standing wave with node at x = 0 is given as:

y(x, t) = 2A sin kx cos ωt ...(1)

The given equation is for a standing wave with a node at x = 0.

Here, ω = 120 π

That means the frequency of the wave is $f = \frac{ω}{2 π} = \frac{120 π}{2 π} = 60 Hz$

The equation (1) is the equation that is the result of superposition of two progressive waves:

y(x,t) = A sin (kx − ωt) and y(x,t) = A sin (kx + ωt)

Thus, the given equation y (x, t) = 0.06 sin (2πx/3) cos (120πt) is the result of the superposition of the waves:

y(x,t) = 0.03 sin (2πx/3 − 120πt) and y(x,t) = 0.03 sin (2πx/3 + 120πt)

These waves have the amplitude of 0.03 m that is 3 cm and the frequency of 60 Hz. The speeds of these waves can be calculated as:

$v = \frac{120 π}{2 π / 3} = 180 m / s$

The amplitude of the standing wave depends on x, it cannot be constant for all the particles.

Hence, the correct answer is option (b).

Page No 108:

Question 15.13:

Speed of sound waves in a fluid depends upon
(a) directly on density of the medium.
(b) square of Bulk modulus of the medium.
(c) inversely on the square root of density.
(d) directly on the square root of bulk modulus of the medium.

Answer:

The speed of sound waves in the fluid is given by

$v = \sqrt{\frac{B}{ρ}}$

Where,

B = Bulk modulus of the fluid

ρ = density of the fluid

Hence, the correct answers are options (c) and (d).

Page No 108:

Question 15.14:

During propagation of a plane progressive mechanical wave
(a) all the particles are vibrating in the same phase.
(b) amplitude of all the particles is equal.
(c) particles of the medium executes S.H.M.
(d) wave velocity depends upon the nature of the medium.

Answer:

When a plane progressive mechanical wave travels through a medium all the particles oscillate (perform S.H.M), but the particles are not in same phase. The amplitude of all the particles is equal, and the speed of the wave depends on the nature of the medium.

Hence, the correct answers are options (b), (c) and (d).

Page No 108:

Question 15.15:

The transverse displacement of a string (clamped at its both ends) is given by y (x, t) = 0.06 sin (2πx/3) cos (120πt).
All the points on the string between two consecutive nodes vibrate with
(a) same frequency
(b) same phase
(c) same energy
(d) different amplitude.

Answer:

The given equation is that of a standing wave which has a node at x = 0.

Between the two nodes all the points have same phase but different amplitudes. The amplitudes are the function of x.

The frequency of oscillation of all the particles in a standing wave is same.

Hence, the correct answers are option (a), (b) and (d).

Page No 109:

Question 15.16:

A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m/s. Given that the speed of sound in still air is 340 m/s,
(a) the frequency of sound as heard by an observer standing on the platform is 400 Hz.
(b) the speed of sound for the observer standing on the platform is 350 m/s.
(c) the frequency of sound as heard by the observer standing on the platform will increase.
(d) the frequency of sound as heard by the observer standing on the platform will decrease.

Answer:

As there is no relative motion between the source and the observer, thus the frequency received by the observer is same as the original frequency of the source that is 400 Hz. The direction of wind is same as the direction of the propagation of sound, thus the effective speed of sound for the observer standing on the station becomes 340 + 10 = 350 m/s.

Hence, the correct answers are options (a) and (b).

Page No 109:

Question 15.17:

Which of the following statements are true for a stationary wave?
(a) Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
(b) All the particles cross their mean position at the same time.
(c) All the particles are oscillating with same amplitude.
(d) There is no net transfer of energy across any plane.
(e) There are some particles which are always at rest.

Answer:

A stationary wave is the one which is formed by the superposition of two identical waves moving in opposite directions and there is no transfer of energy. In this standing wave the particles have different amplitudes that depends on their positions, there are many points which are always at rest they are known as nodes. All the particles between two consecutive nodes are in same phase and the particles which lie just outside the region between the two consecutive nodes are 180 degrees out of phase from them. But, all the particles reach the mean position at same time.

Hence, the correct answers are options (a), (b), (d) and (e).

Page No 109:

Question 15.18:

A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire?

Answer:

As the wire's length is increased to twice and still it is in resonance this implies that the number of loops that are formed also gets doubled. This implies that the wire now vibrates in the second harmonic.

Page No 109:

Question 15.19:

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

Answer:

Let the speed of sound be v.

The first harmonic of the open organ pipe is given by:

$f = \frac{v}{2 L} = 480 Hz$ ...(1)

Let the length of the pipe closed at one end be L'

The first harmonic of the closed organ pipe is given by:

$f' = \frac{v}{4 L'} = 480 Hz$ ...(2)

From equations (1) and (2)

$\frac{v}{4 L'} = \frac{v}{2 L} \Rightarrow L' = \frac{L}{2}$

Page No 109:

Question 15.20:

A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?

Answer:

Frequency of tuning fork A, f_A = 512 Hz

As the beat frequency is 5 Hz, so the frequency of tuning fork B can be either 507 Hz or 517 Hz.

When the tuning fork B is loaded with the wax its frequency decreases.

If the frequency of tuning fork B is 517 Hz, in order to maintain the beat frequency to 5 Hz, the frequency of tuning fork B gets reduced to 507 Hz.

If the frequency of tuning fork B is 507 Hz, on loading it with the wax the beat frequency will increase anyway.

Hence, the frequency of tuning fork B when unloaded is 517 Hz.

Page No 109:

Question 15.21:

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

Answer:

The given function is the superposition of two simple harmonic displacements:

$y_{1} = 3 \sin ω t y_{1} = 4 \cos ω t = 4 \sin (ω t + \frac{π}{2})$

Let, A₁ = 3 cm and A₂ = 4 cm, and the phase difference between them, $ϕ = \frac{π}{2}$

The resultant amplitude can be calculated as:
$A = \sqrt{A_{1}^{2} + A_{1}^{2} + 2 A_{1} A_{2} \cos ϕ} \Rightarrow A = \sqrt{3^{2} + 4^{2} + 2 (3) (4) \cos \frac{π}{2}} = \sqrt{9 + 16} = 5 cm$

Page No 109:

Question 15.22:

A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?

Answer:

Let the length of the wire be l, radius of the wire be r, mass per unit length be μ, density of the wire be ρ, tension in the wire be T, and n be the number of harmonic.

The frequency of vibration of the wire is given by

$f = \frac{n}{2 l} \sqrt{\frac{T}{μ}}$

We can express the mass per unit length in terms of density of the material as:

$μ = \frac{mass}{length} = \frac{volume \times density}{length} = \frac{(π r^{2} l) \times ρ}{l} = π r^{2} ρ$

Thus, frequency of vibration becomes:
$f = \frac{n}{2 l} \sqrt{\frac{T}{π r^{2} ρ}}$

When the radius (r) of the wire is made thrice (3r),

The frequency of vibration is given as:

$f' = \frac{n}{2 l} \sqrt{\frac{T}{π {(3 r)}^{2} ρ}} = \frac{n}{2 l (3)} \sqrt{\frac{T}{π r^{2} ρ}}$

Hence, $f' = \frac{f}{3}$
Frequency becomes one-third of the initial frequency.

Page No 110:

Question 15.23:

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

Answer:

The speed of sound is proportional to square root of the absolute temperature (temperature in kelvins).

Let T kelvins be the temperature at which the speed of sound becomes 3 times its value at 0 °C (i.e. 273.15 K).

$\frac{v_{T}}{v_{0}} = \sqrt{\frac{T}{273.15}} = 3 \frac{T}{273.15} = 9 \Rightarrow T = 2458.35 K$

In degree celsius, temperature is given by:
t = T − 273.15 = 2458.35 − 273.15 = 2185.2 ^oC

Page No 110:

Question 15.24:

When two waves of almost equal frequencies n₁ and n₂ reach at a point simultaneously, what is the time interval between successive maxima?

Answer:

Frequencies of the waves are n₁ and n₂.
Due to difference in the frequencies of the waves reaching at a point, beats will be produced.
Beat frequency = n₁ − n₂.
Time interval between successive maxima means the time period of the beats,

Hence, $T = \frac{1}{|n_{1} - n_{2}|}$

Page No 110:

Question 15.25:

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

Answer:

Given,

Tension in the wire, T = 2.06 × 10⁴N
Length of the wire, L = 12 m
Mass of the wire, M = 2.10 kg
Mass per unit length, $μ = \frac{M}{L} = \frac{2.10}{12} = 0.175 kg / m$

Speed of transverse wave on the string, $v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{2.06 \times 10^{4}}{0.175}} \approx 343 m / s$

Page No 110:

Question 15.26:

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

Answer:

Length of the pipe l = 20 cm = 0.20 m

Frequency f = 1237.5 Hz

Velocity of sound in air v = 330 m/s

$fundamental harmonic f_{1} = \frac{v}{λ} = \frac{v}{4 l} = \frac{330}{4 \times 0.20} = 412.5 Hz$

nth harmonic corresponds to the frequency f_n=1237.5 Hz

$f_{n} = n f_{1} n = \frac{1237.5}{412.5} = 3$

So the third harmonic is excited by the source frequency of 1237.5 Hz

Page No 110:

Question 15.27:

A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train begins to move with a speed of 10 m s^–1 towards the platform. What is the frequency of the sound for an observer standing on the platform?
(sound velocity in air = 330 m s^–1)

Answer:

When the source (train) is moving towards the observer (platform) the apparent frequency is given by the formula $(\frac{v}{v - v_{s}}) f_{s}$
â€‹Velocity of sound in air v = 330 m s^–1

Frequency of whistle f_s= 400 Hz

Speed of the train v_s= 10 m/s

Apparent frequency when source is moving $(\frac{v}{v - v_{s}}) f_{s} = (\frac{330}{330 - 10}) 400 = 412.5 Hz$

Page No 110:

Question 15.28:

The wave pattern on a stretched string is shown in Fig. 15.2. Interpret what kind of wave this is and find its wavelength.

Answer:

It can be seen from the graph that there are some points on the graph which are always at rest.

The points on positions x = 10,20,30,40 never move.

These are forming nodes which characterize a stationary wave.

â€‹The distance between two successive nodes is equal to λ/2

$ λ = 2 \times (node to node distance) = 2 \times (20 - 10) = 20 cm$

So the wave formed is a stationary wave with wavelength equal to 20 cm.

Page No 111:

Question 15.29:

The pattern of standing waves formed on a stretched string at two instants of time are shown in Fig. 15.3. The velocity of two waves superimposing to form stationary waves is 360 ms^–1 and their frequencies are 256 Hz.

(a) Calculate the time at which the second curve is plotted.
(b) Mark nodes and antinodes on the curve.
(c) Calculate the distance between A′ and C′.

Answer:

Given, frequency of the wave, f = 256 Hz
and velocity of the wave, v = 360 ms^–1 .

(a) The time period of the wave is given by,
$T = \frac{1}{f} = \frac{1}{256} = 3.90 \times 10^{- 3} s$

In the second curve a stationary wave is formed.
Time taken to move the particle from amplitude position (point at maximum distance from the mean position) to mean position is equal to one-fourth of the time period.
So, $t = \frac{T}{4} = \frac{3.90 \times 10^{- 3}}{4} = 9.76 \times 10^{- 4} s$
(b) Nodes are the point, where the displacement of the particle is zero.
Nodes are at the points A, B, C, D, E
Antinodes are the points where the displacement of the particles is maximum, so A^' and C^' are the anti-nodes.

(c) A^' and C^' are anti nodes, so the distance between them is equal to the wavelength of the wave,

$λ = \frac{v}{f} = \frac{360}{256} = 1.41 m$
â€‹

Page No 111:

Question 15.30:

A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20°C, calculate

(a) speed of sound in air at room temperature
(b) speed of sound in air at 0°C
(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Answer:

(a) If a pipe partially filled with water whose upper surface of the water acts as a reflecting surface of a closed organ pipe.

If the length of the air column is varied until its natural frequency equals the frequency of the fork, then the column resonates and emits a loud note.
The frequency of tuning fork, f= 512 Hz.

For observation of first maxima of intensity,

Length of the air column $l = \frac{λ}{4}$

$λ = 4 l = 4 \times 0.17 = 0.68 m$

Speed of the sound = $v = f \times λ = 512 \times 0.68 = 348.16 {ms}^{- 1}$

(b) velocity of the sound is directly proportional to $\sqrt{T}$

$\frac{Velocity of the sound at Room te 0 Cmperature}{velocity of sound at 0^{0} C} = \frac{v_{20}}{v_{0}} = \sqrt{\frac{293}{273}} \Rightarrow v_{0} = v_{20} \times \sqrt{\frac{273}{293}} = 348.16 \times 0.97 = 338.01 {ms}^{- 1}$

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

â€‹

Page No 111:

Question 15.31:

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

Answer:

from the figure above it can be seen t hat the string vibrates with its ends fixed.

In fig (a) the string vibrates in 1 loop and in (b) it vibrates with two loops similarly 3 and 4 loops in figure (c) and (d).

The frequency of vibration is given by $\frac{velocity of the wave}{wavelength of the wave}$

in each of the cases the velocity of the wave is same only the wavelength of the wave changes

In figure (a) the wavelength and the length of the string are related by the relation $L = \frac{λ}{2}$

So the value of the frequency of vibration for loop one is $f_{1} = \frac{velocity of the wave}{wacvelength of the wave} = \frac{v}{2 L}$

Similarly for the next loop in figure (b) relationship between wavelength and length is given by λ = L

So the frequency f₂can now be written as f₂= $\frac{v}{λ} = \frac{v}{L} = \frac{2 v}{2 L}$

f₂can be written as f₂= 2f₁

Similarly for figure (c) the relationship between wavelength and length is given by $L = \frac{3 λ}{2}$

$\Rightarrow λ = \frac{2 L}{3}$

Frequency f₃can now be written as $f_{3} = \frac{3 v}{2 L} = 3 f_{1}$

Similarly for the figure(d) the frequency f₄= 4f₁

Now the ratio of the frequencies of the loops is given by 1:2:3:4

Page No 111:

Question 15.32:

The earth has a radius of 6400 km. The inner core of 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in molten state. Then again from 3500 km to 6400 km the earth is solid. Only longitudinal (P) waves can travel inside a liquid. Assume that the P wave has a speed of 8 km s^–1 in solid parts and of 5 km s^–1 in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth if wave travels along diameter?

Answer:

To calculate the time taken by the wave to reach the other end of the earth if it is generated at one of the ends

r₁= 1000 km

r₂= 3500 km

r₃= 6400 km

The total distance along the diameter where the material is in solid form is $2 (1000 + 2900) = 7800 km$

The total distance along the diameter where the material is in liquid form 5000 km

Total time taken by the wave to reach the other end = $(\frac{7800}{8} + \frac{5000}{5}) = 1975 s = 32 minutes 55 seconds$

Page No 112:

Question 15.33:

If c is r.m.s. speed of molecules in a gas and v is the speed of sound waves in the gas, show that c/v is constant and independent of temperature for all diatomic gases.

Answer:

R.m.s. Speed of the molecule is given by $c = \sqrt{\frac{3 RT}{M}} = \sqrt{\frac{3 P}{ρ}}$

Where M is the molar mass of the gas.

velocity v of the wave is given by $\sqrt{\frac{γP}{ρ}}$

Taking the ratio of the two $\frac{c}{v} = \frac{\sqrt{\frac{3 P}{ρ}}}{\sqrt{\frac{γP}{ρ}}} = \sqrt{\frac{3}{γ}}$

For the diatomic gas $γ = \frac{7}{5}$

The ratio comes out to be $\sqrt{\frac{3}{\frac{7}{5}}} = \sqrt{\frac{15}{7}} = constant$

Page No 112:

Question 15.34:

Given below are some functions of x and t to represent the displacement of an elastic wave.
(a) y = 5 cos (4x) sin (20t)
(b) y = 4 sin (5x – t/2) + 3 cos (5x – t/2)
(c) y = 10 cos [(252 – 250) πt] cos [(252 + 250)πt]
(d) y = 100 cos (100πt + 0.5x)
State which of these represent
(a) a travelling wave along –x direction
(b) a stationary wave
(c) beats
(d) a travelling wave along +x direction.
Given reasons for your answers.

Answer:

(a) A travelling wave along -x direction must have +kx in its argument.

This is in the equation y = 100 cos (100πt + 0.5x)

y = 100 cos (100πt + 0.5x) represents the wave travelling along -x direction.

(b) y = 5 cos (4x) sin (20t) is a stationary wave as it has nodes and anti- nodes at specific values of x.

(c) y
y = 10 cos [(252 – 250) πt] cos [(252 + 250)πt] represents beats as it involves the frequencies such as $(f_{1} - f_{2}) a n d (f_{1} + f_{2})$ in the form of

(252 – 250) πt and (252 + 250)πt

â€‹ (d)
y = 4 sin (5x – t/2) + 3 cos (5x – t/2) â€‹
$y = A \cos ϕ \sin (5 x - \frac{t}{2}) + A \sin ϕ \cos (5 x - \frac{t}{2}) = A \sin (5 x - \frac{t}{2} + ϕ) Where A = \sqrt{4^{2} + 3^{2}} = 5 y = 5 \sin (5 x - \frac{t}{2} + ϕ)$

Therefore, y = 4 sin (5x – t/2) + 3 cos (5x – t/2) â€‹ represents the wave travelling along +x direction.
â€‹

Page No 112:

Question 15.35:

In the given progressive wave y = 5 sin (100πt – 0.4πx ) where y and x are in m, t is in s. What is the
(a) amplitude
(b) wave length
(c) frequency
(d) wave velocity
(e) particle velocity amplitude.

Answer:

(a) Amplitude A is given by 5 m.

(b) Relation between wavelength and wavevector is given by

$k = 0.4 π \frac{2 π}{λ} = 0.4 π λ = 5 m$

(c) frequency $f = \frac{ω}{2 π} = \frac{100 π}{2 π} = 50 Hz$

(d) Wave velocity v = f λ = $50 \times 5 = 250 {ms}^{- 1}$

(e) Particle velocity amplitude

$v = \frac{d y}{d t} = \frac{d}{d t} (5 \sin (100 π t - 0.4 π x)) v = 500 π \cos (100 π t - 0.4 π x)$

Velocity amplitude is given by $500 π$ ms^-1.

Page No 112:

Question 15.36:

For the harmonic travelling wave y = 2 cos 2π (10t – 0.0080x + 3.5) where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of

(a) 4 m

(b) 0.5 m

(c) $\frac{λ}{2}$

(d) $\frac{3 λ}{2}$ (at a given instant of time)

(e) What is the phase difference between the oscillation of a particle located at x = 100cm, at t = T s and t = 5 s?

Answer:

(a) Path difference for the travelling wave   $y = 2 \cos 2 π (10 t - 0.0080 x + 3.5)$

   $Phase difference Δϕ = \frac{2 π}{λ} \times Path difference = k \times Path difference Δϕ = 0.016 π \times 400 = 6.4 π radians$

(b) Path difference P = 50 cm

   $Phase difference Δϕ = \frac{2 π}{λ} \times Path difference = k \times Path difference Δϕ = 0.016 π \times 50 = 0.8 π radians$

(c) Phase difference $Δϕ = \frac{2 π}{λ} \times \frac{λ}{2} = π radians$

(d)  Phase difference $Δϕ = \frac{2 π}{λ} \times \frac{3 λ}{4} = \frac{3 π}{2} radians$

(e)
$A t t = T, ϕ = \frac{2 π}{T} = 20 π rad T t = 5 s, ϕ' = \frac{2 π}{0.1} \times 5 = 100 π rad ∆ ϕ = ϕ' - ϕ = 80 π rad$

View NCERT Solutions for all chapters of Class 11