NCERT Exemplar Class 11 Physics Solutions Chapter 15 Waves 2021

NCERT Exemplar Class 11 Physics Solutions Chapter 15 Waves

Edited By Safeer PP | Updated on Aug 09, 2022 01:07 PM IST

NCERT Exemplar Class 11 Physics solutions chapter 15 have been created by experts of Physics to help students in their examinations as it is one of the important topics of physics that is repeatedly asked in various examinations. The students can refer NCERT Exemplar Class 11 Physics chapter 15 solutions to understand about waves and its types; longitudinal and transverse waves. The overall chapter deals with waves and its characteristics frequency, amplitude, wavelength etc.

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This Story also Contains

NCERT Exemplar Class 11 Physics Solutions Chapter 15 MCQI
NCERT Exemplar Class 11 Physics Solutions Chapter 15 MCQII
NCERT Exemplar Class 11 Physics Solutions Chapter 15 Very Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 15 Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 15 Long Answer
Main Subtopics of NCERT Exemplar Class 11 Physics Solutions Chapter 15
What will students Learn from NCERT Exemplar Class 11 Physics Solutions Chapter 15?
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
Important Topics To Cover From NCERT Exemplar Class 11 Physics Solutions Chapter 15:
NCERT Exemplar Class 11 Solutions

The students will be provided with the necessary knowledge required to excel in the topic by going through the NCERT Exemplar solutions for Class 11 Physics chapter 15 and avail a thorough understanding of all the concepts.

NCERT Exemplar Class 11 Physics Solutions Chapter 15 MCQI

Question:15.1

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

Answer:

The answer is the option (b) Waves are produced by motorboat on the surface as well as inside water. The waves are both transverse as well as longitudinal.

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. The wavelength of sound waves in the second medium is
$(a) λ$
$(b) \frac{λ}{2}$
$(c) 2 λ$
$(d) 4 λ$

Answer:

The answer is the option (c)

v = u λ, u = \frac{v}{λ}

When a wave passes from one medium to another, its frequency does not change but its velocity and wavelength changes.

\frac{v}{λ} = \frac{2 v}{λ_{2}}

Thus,

λ_{2} = 2 λ

Question:15.3

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

Answer:

The answer is the option (c) Sound is a longitudinal wave with speed

v = \sqrt{\frac{λ P}{ρ}}

.
The density of water vapours is less than the air so, on increasing humidity, the density of medium decreases, in turn, increasing the speed of sound in air by

v \propto \frac{1}{ρ}

.
Hence, c is correct.

Question:15.4

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

Answer:

The answer is the option (c)
Thus, on increasing temperature, the speed also increases, as frequency remains constant during the propagation of the wave. Since

v = u λ

, the wavelength also increases with velocity.

Question:15.5

With the 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:

The answer is the option (b) During propagation of a wave in any medium, only energy is transmitted from one point to another and not matter.

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 a vacuum.

Answer:

The answer is the option (c) A mechanical transverse wave can propagate through solid and on the surface of the liquid as well. A transverse wave cannot reproduce in gases whereas Longitudinal wave can. Also, longitudinal waves are not electromagnetic waves and can never propagate in a vacuum.

Question:15.7

A sound wave is passing through the 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 answer is the option (d) In consecutive compressions and rarefactions, the density of medium particles is maximum and minimum, respectively. Due to the rapid change in density temperature rises while the bulk modulus remains constant. The time of compressions and rarefactions is very small, so heat does not transfer.

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 $\frac{2}{3}$ of the amplitude of the incident wave
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:

The answer is the option b)

y = - 0.4 \sin 2 π (t + \frac{x}{2})

After reflection of wave changes by phase

180^{\circ}

y_{r} = (\frac{2}{3} \times 0.6) \sin 2 π [π + t + \frac{x}{2}]

y_{r} = - 0.4 \sin 2 π (t + \frac{x}{2})

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:

The answer is the option (b) M= mass of string= 2.5Kg, l=20m
μ=mass per unit length=

\frac{M}{l} = \frac{2.5}{20} = 0.125 k g / m

v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{200}{0.125}} = \sqrt{1600} = \frac{40 m}{s}

Time =

\frac{d i s t a n c e}{s p e e d} = \frac{20 m}{\frac{40 m}{s}} = \frac{1}{2} s e c = 0.5 s e c

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 n′ of the sound as heard by the observer is plotted as a function of time t (Figure). Identify the expected curve.

Answer:

The answer is the option (c) When the observer is at rest and source of sound is moving towards the observer, then the observed frequency

n^{'} = (\frac{v}{v - v_{s}}) n_{0}

where n₀ is the original frequency of the source of sound
v=speed of sound in the medium

n^{'} > n_{0}

v_s=speed of source
When the source is moving away from the observer

n " = \frac{v n_{0}}{v + v_{s}}

n^{^{″}} < n_{0}

Thus, n’ > n” and graph (c) is correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 15 MCQII

Question:15.11

A transverse harmonic wave on a string is described by $y (x, t) = 3.0 \sin (36 t + 0.018 x + \frac{π}{4})$ 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 answer is the option (a), (b), and (c)
The standard form of a wave propagating in a positive direction

y = a \sin (ω t - k x + ϕ)

and

y = 3.0 \sin (36 t + 0.018 x + \frac{π}{4})

A positive sign in the equation shows that the wave is travelling from right to left.

ω = 36; 2 π ν = 36 o r v = \frac{36}{2 π} = \frac{18}{3.14} = 5.7 H z

k = 0.018 = \frac{2 π}{λ}

λ = \frac{2 π}{0.018}

v = ν λ = \frac{18}{π} \times \frac{2 π}{0.018} = \frac{2000 c m}{s} = \frac{20 m}{s}

Distance between two successive crests=

λ = \frac{2 π}{0.018} = \frac{π}{0.009} = 3.14 \times \frac{1000}{9} = \frac{3140}{9} c m = 348.8 c m

Question:15.12

The displacement of a string is given by
$y (x, t) = 0.06 \sin (\frac{2 π x}{3}) \cos (120 π t)$
where x and y are in m and t in s. The length of the string is 1.5m and its mass is $3.0 \times 10^{- 2}$ kg
(a) It represents a progressive wave of frequency 60Hz.
(b) It represents a stationary wave of frequency 60Hz.
(c) It is the result of superposition of two waves of wavelength 3 m, frequency 60Hz each travelling with a speed of 180 m/s in opposite direction.
(d) Amplitude of this wave is constant.

Answer:

The answer is the option (b) and (c)
Equation of stationary wave is

y (x, t) = a \sin (k x) \cos (ω t)

Since the waves are stationary, the amplitude varies from 0 to a=0.06m from nodes to antinodes.
Comparing this equation with the given equation

y (x, t) = 0.06 \sin (\frac{2 π x}{3}) \cos (120 π t)

We get

ω = 120 π = 2 π ν; ν = 60 H z

From equation

\frac{2 π}{3} = k = \frac{2 π}{λ}

λ = 3 m, ν = 60 H z

Speed =

60 \times 3 = 180 m / s

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 answer is the option (c) and (d) Speed of sound wave in the fluid of bulk modulus K and density

ρ

is given by

v = \sqrt{\frac{K}{ρ}}

v \propto \sqrt{K}

(if

ρ

is constant)

v \propto \sqrt{\frac{1}{ρ}}

(if K is constant)

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:

The answer is the option (b) (c) and (d) During propagation of a mechanical wave, each particle is displaced from 0 to its amplitude (which is equal for each particle). Each particle between any 2 successive crests and roughs are in a different phase.
In case of a progressive wave, medium particles oscillate about their mean position with a restoring force

(F \propto (- y))

acting on them like in a simple harmonic motion.
Also,

v = \sqrt{\frac{K}{ρ}}

Where K and

ρ

depend on the nature of the medium.

Question:15.15

The transverse displacement of a string is given by $y (x, t) = 0.06 \sin (\frac{2 π x}{3}) \cos (120 π t)$ . 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 answer is the option (a), (b), and (d).
The given equation

y (x, t) = 0.06 \sin (\frac{2 π x}{3}) \cos (120 π t)

is as per the standard equation of stationary wave, i.e.

y (x, t) = a \sin k x \cos ω t

The frequency of particles within the wave is the same. All the particles between any two consecutive nodes vibrate either the upside or downside having the same phase

120 π t

at a time.
Also,

E \propto A^{2}

where the amplitude of different particles is different between nodes.

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:

The answer is the option (a) and (b) Frequency of source of sound

v_{0} = 400 H z

The velocity of wind from source

v_{w} = \frac{10 m}{s}

Speed of sound in still air

v = \frac{340 m}{s}

Now, the speed of sound with respect to the listener

v + v_{w} = 340 + 10 = \frac{350 m}{s}

As the distance between listener and source does not change, so the frequency of sound does not change as heard by the listener.

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 the same amplitude
d) there is no net transfer of energy across any plane
e) there are some particles which are always at rest

Answer:

The answer is the option (a) (b) (d) and (e)
In a stationary wave, the particles between two nodes vibrate with different amplitude, which increases from node to antinode (0 to maximum) and decreases from anti node to node. The amplitude of particles varies with

λ

. The particles at a node are at rest, and hence there is no net transfer of energy. The particle between two nodes is in the same phase. The motion of particles between two nodes will be either upward or downward crossing the mean position at the same time. Hence, a,b,d and e are correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 15 Very Short Answer

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:

Let the length of wire be L and tension be T
The frequency of nth harmonic is

v = \frac{n}{2 L} \sqrt{\frac{T}{m}}

where m is mass per unit length
Let us assume two cases

v_{1} = \frac{n_{1}}{2 L_{1}} \sqrt{\frac{T_{1}}{m_{1}}}

v_{2} = \frac{n_{1}}{2 L_{2}} \sqrt{\frac{T_{2}}{m_{1}}}

v
In the given question

T_{1}

is same as T and the mass is also the same since the same wire is used.

L_{2} = 2 L_{1}

\frac{v_{1}}{v_{2}} = \frac{n_{1}}{2 L_{1}} \sqrt{\frac{T_{1}}{m_{1}}} \times 2 \times \frac{2 L_{2}}{n_{1}} \sqrt{\frac{m_{1}}{T_{2}}} = \frac{2 n_{1}}{n_{2}}

As tuning fork is same in both harmonics, both the frequencies are equal

2 n_{1} = n_{2}

Thus, when the length of the wire is doubled, the number of harmonics also get doubled for the same frequency.

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:

As the medium, number and frequency of harmonic in open and closed pipes are the same, so the number of nodes and wavelength in both cases will also be the same.
In both end open pipe

L_{1} = 2 \times \frac{λ_{1}}{4} o r λ_{1} = 2 L_{1}

v_{1} = \frac{c}{λ_{1}} = \frac{c}{2 L_{1}}

In one open end pipe

L_{2} = 1 \times \frac{λ_{2}}{4} o r λ_{2} = 4 L_{2}

v_{2} = \frac{c}{4 L_{2}}

v_{1} = v_{2} a n d c_{1} = c_{2} = c

\frac{c}{2 L_{1}} = \frac{c}{4 L_{2}}

4 L_{2} = 2 L_{1}; L_{2} = L_{1} / 2

Question:15.2

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:

v_{A} = 512

v_{o} = v_{A} - v_{B}

On loading frequencies of B decreases

v_{1} = 5

v_{B} = v_{A} \pm 5

v_{B} = 512 \pm 5 = 507 o r 517

Therefore, on loading the frequency of B decreases to 507
Frequency of tuning fork, when unloaded, is 517 Hz.
(When frequency decreases by 10 Hz the number of beats will be same as 512-507=5)

Question:15.21

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

Answer:

y = 3 \sin ω t + 4 \cos ω t

Let

3 = a \cos ϕ a n d 4 = a \sin ϕ;

\tan ϕ = \frac{4}{3} o r ϕ = t a n^{- 1} \frac{4}{3}

a^{2} (\cos ϕ)^{2} + a^{2} (\sin ϕ)^{2} = 3^{2} + 4^{2}

a^{2} (\cos^{2} ϕ + \sin^{2} ϕ) = 9 + 16

a^{2} = 25

a=5
Now,

y = a \cos ϕ \sin ω t + a \sin ϕ \cos ω t

y = a \sin (ω t + ϕ)

y = 5 \sin (ω t + ϕ)

where

ϕ = t a n^{- 1} \frac{4}{3}

Question:15.22

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

Answer:

Frequency of wire stretched at both ends

v = \frac{n}{2 L} \sqrt{\frac{T}{m}}

As the number of harmonics, length L and tension T is kept the same in both cases

v \propto \frac{1}{\sqrt{m}}

v_{1} / v_{2} = \sqrt{\frac{m_{2}}{m_{1}}}

mass per unit length=mass of wire/length=

\frac{π r^{2} l ρ}{l} = (π r^{2}) ρ

As the material of wire is same

\frac{m_{2}}{m_{1}} = \frac{(π r_{2}^{2}) ρ}{(π r_{1}^{2}) ρ} = \frac{9}{1}

\frac{v_{1}}{v_{2}} = \sqrt{\frac{9}{1}} = 3

Question:15.23

At what temperatures will the speed of sound in air be 3 times its value at $0^{\circ} C$ ?

Answer:

We know that

v \propto T

Let v_T be speed of sound in air at temp T and v₀ at

0^{\circ} C

\frac{v_{T}}{v_{0}} = \sqrt{\frac{T}{T_{0}}}

and

v_{T} = 3 v_{0}

\frac{3 v_{0}}{v_{0}} = \sqrt{\frac{T}{273 + 0}}

T = 2457 K = 2184^{\circ} C

Question:13.24

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

Answer:

Since the frequencies of vibrations are almost equal $n_{1} = n_{2}$
Assuming that $n_{1} > n_{2}$ number of beats per second= $n = n_{2} - n_{1}$
So time period of maxima or beats
$\frac{1}{n} = \frac{1}{n_{2} - n_{1}}$

NCERT Exemplar Class 11 Physics Solutions Chapter 15 Short Answer

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 $2.06 \times 10^{4} N$ is applied?

Answer:

Given: l = 12m
M=2.10 kg

m = \frac{M}{l} = \frac{2.1}{12}

T = 2.06 \times 10^{4} N

v = \sqrt{\frac{T}{m}} = \sqrt{2.06 \times 10^{4} \times \frac{12}{2.10}}

v = 343 m / s

Question:15.26

A pipe of 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz?

Answer:

l = 20 c m = 0.2 m, v = 1237.5 H z,

V = \frac{330 m}{s}

l = \frac{λ}{4}

For fundamental frequency v

ν_{1} = \frac{v}{λ} = \frac{v}{4 l} = \frac{330}{4 \times 20 \times 10^{- 2}} = 412.5 H z

v = 1237.5 H z \frac{v}{v_{1}} = \frac{1237.5}{412.5} = 31

The 3^rd harmonic is excited by 1237.5 Hz

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 beings to move with a speed of 10 m/s towards the platform. What is the frequency of the sound for an observer standing on the platform?

Answer:

$v_{0} = 400 H z$ and

$V_{s} = \frac{10 m}{s}$ given
Velocity of sound in air
$V_{a} = \frac{330 m}{s}$
Apparent frequency by the observer standing on the platform
$v^{'} = \frac{V_{a}}{V_{a} - V_{s}} v_{0}$
$v^{'} = 330 \times \frac{400}{330 - 10} = 412.5 H z$

Question:15.28

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

Answer:

The displacement of medium particles at a distance 10,20,30,40 and 50cm are always at rest. (property of nodes in stationary waves)
At

t = \frac{T}{4} a n d \frac{3 T}{4}

all particles are at rest which happens in a stationary wave when the particle crosses its mean position.
So, the graph of wave shows a stationary wave. Nodes are at x=10, 20,30, 40 cm and distance between successive nodes is

\frac{λ}{2} = (30 - 20); λ = 20 c m

Question:15.29

The pattern of standing waves formed on a stretched string at two instants of time are shown in the figure. The velocity of two waves superimposing to form stationary waves is 360 m/s 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:

Frequency of wave

ν = 256 H z

T = \frac{1}{ν} = \frac{1}{256} s e c o n d = 0.00390 = 3.9 \times 10^{- 3} s e c o n d s

In stationary wave, a particle passes through its mean position after every $\frac{T}{4}$ time

T = \frac{T}{4} = 3.9 \times \frac{10^{- 3}}{4} = 0.975 \times 10^{- 3} s e c = 9.75 \times 10^{- 4} s e c

Point does not vibrate at A, B, C,D and E. The point A’ and C’ are at maximum displacement

Between A’ and C’ =

λ = \frac{V}{v} = \frac{360}{256} = 1.41 m

Question:15.30

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

Answer:

(a) Pipe partially filled with water behaves like one end open organ pipe. The first harmonic is heard at L=17 cm

L = \frac{λ}{4}; λ = 4 L = 0.68 m

V = \frac{v}{λ} = 512 \times \frac{0.68 m}{s} = \frac{348.16 m}{s}

(b)

\frac{V_{0}}{V_{T}} = \sqrt{\frac{T_{0}}{T}}

\frac{V_{0}}{348.16} = \sqrt{\frac{273}{293}}

V_{0} = 348.16 \times 0.96526 = \frac{336 m}{s}

(c) Water and mercury in the tube reflect sound into air column to form stationary wave and reflection is more in mercury compared to water because of its higher density. The intensity of sound heard in mercury will be larger, but the reading remains the same as the medium in tube and tunic fork does not change.

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:

Let the number of loops in the string be n
Length of each loop is

\frac{λ}{2}

L = n \frac{λ}{2} o r λ = \frac{2 L}{n}

V = \frac{v}{λ} a n d λ = \frac{V}{v} = \frac{2 L}{n}

v = \frac{V n}{2 L}

Where V in a stretched string is

\sqrt{\frac{T}{m}}

v = \frac{n}{2 L} \sqrt{\frac{T}{m}}

For n=1,

v_{1} = \frac{1}{2 L} \sqrt{\frac{T}{m}} = v_{0}

For n=2,

v_{2} = \frac{2}{2 L} \sqrt{\frac{T}{m}} = 2 v_{0}

Similarly for n=3 and so on

v_{1} : v_{2} : v_{3} : v_{4} = n_{1} : n_{2} : n_{3} : n_{4} = 1 : 2 : 3 : 4

NCERT Exemplar Class 11 Physics Solutions Chapter 15 Long Answer

Question 32:

The earth has a radius of 6400 km. The inner core of the 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in a 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 in solid parts and of 5 km/s in liquid parts of the earth. An earthquake occurs at someplace 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:

r₁=1000km, r₂=3500km, r₃=6400km, d₁=1000km
d₂=3500-1000=2500km
d₃=6400-3500=2900km
Solid distance diametrically=

2 d_{1} + d_{3} = 21000 + 2900 = 7800 k m

Time taken by wave produced by earthquake in solid part=

\frac{7800}{8} s

Liquid part along diametrically

= 2 d_{2} = 2 \times 2500

Time taken by seismic wave in liquid part=

2 \times \frac{2500}{5}

Total time=

\frac{7800}{8} + \frac{5000}{5} = 1975 s e c = 32 m i n a n d 55 s e c

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:

We know that

c = \sqrt{\frac{3 P}{ρ}}

for molecules

c = \sqrt{\frac{3 R T}{M}}

\frac{p}{ρ} = \frac{P T}{M}

where M is the molar mass of gas

\frac{P}{ρ} = \frac{\frac{R T}{V}}{\frac{M}{V}}

v = \frac{γ P}{ρ} = \frac{γ R T}{M}

PV=nRT When n=1

P = \frac{R T}{V}

c v = \sqrt{\frac{\frac{3 R T}{M}}{\frac{γ R T}{M}}} = \sqrt{} \frac{3}{γ}

c / v = \sqrt{\frac{\frac{3 R T}{M}}{\frac{γ R T}{M}}} = \sqrt{} \frac{3}{γ}

γ = \frac{C_{p}}{C_{v}} = \frac{7}{5}

adiabatic constant for diatomic gas

\frac{c}{v} = \sqrt{\frac{15}{7}}

=constant

Question:15.34

Given below are some functions of x and t to represent the displacement of an elastic wave.
$a) y = 5 \cos (4 x) \sin (20 t)$
$b) y = 4 \sin (5 x - \frac{t}{2}) + 3 \cos (5 x - \frac{t}{2})$
$c) y = 10 \cos [(252 - 250) π t] \cos [(252 + 250) π t]$
$d) y = 100 \cos (100 π t + 0.5 x)$
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
Give reasons for your answers

Answer:

(a) A wave travelling along (-x) direction has +kx as in

d) y = 100 \cos (100 π t + 0.5 x)

(b ) A stationary wave equation is

a) y = 5 \cos (4 x) \sin (20 t)

Beats involve $(v_{1} + v_{2})$ and $(v_{1} - v_{2})$ So they can be represented by

y = [\cos (252 - 250) π t]

i.e. option c

$y = 4 \sin (5 x - \frac{t}{2}) + 3 \cos (5 x - \frac{t}{2})$

Let

4 = a \cos ϕ a n d 3 = a \sin ϕ

;

\tan ϕ = \frac{3}{4} o r ϕ = t a n^{- 1} \frac{3}{4}

a^{2} \cos ϕ^{2} + a^{2} \sin ϕ^{2} = 3^{2} + 4^{2}

a^{2} (\cos^{2} ϕ + s i n^{2} ϕ) = 9 + 16

a^{2} = 25

a=5
Now,

y = a \cos ϕ \sin (5 x - \frac{t}{2}) + a \sin ϕ \cos (5 x - \frac{t}{2})

y = a \sin (5 x - \frac{t}{2} + ϕ)

y = 5 \sin (5 x - \frac{t}{2} + ϕ)

Which represents the progressive wave in +x direction as the sign of kx or (5x) and

ω t (\frac{1}{2 t})

are opposite so it travels in+x direction.

Question:15.35

In the given progressive waves $y = 5 \sin (100 π t - 0.4 π x)$ where y and x are in m, t is in s. What is the
a) amplitude
b) wavelength
c) frequency
d) wave velocity
e) particle velocity amplitude

Answer:

Given the progressive wave equation

y = a \sin ω t - k x + ϕ

y = 5 \sin (100 π t - 0.4 π x)

(a) Amplitude a=5m

(b) k = \frac{2 π}{λ} = 0.4 π

Wavelength

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

(c) ω = 2 π v; v = \frac{ω}{2 π}, w h e r e ω = 100 π

v = \frac{100 π}{2 π} = 50 H z

Wave velocity

V = v λ = 50 \times 5 = \frac{250 m}{s}

(e) Particle velocity in the direction of amplitude at a distance x from source

y = 5 \sin 100 π t - 0.4 π x

\frac{d y}{d t} = 5 \times 100 π \cos (100 π t - 0.4 π x)

Maximum velocity of the particle is at its mean position

\cos 100 π t - 0.4 π x = 1

100 π t - 0.4 π x = 0

\frac{d y}{d t} = 5 \times 100 π = 500 π m / s

Question:15.36

For the harmonic travelling wave $y = 2 \cos 2 π (10 t - 0.0080 x + 3.5)$ 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) $\frac{3 λ}{4}$
e) what is the phase difference between the oscillation of a particle located at x = 100 cm at t = Ts and t = 5s?

Answer:

y = 2 \cos 2 π (10 t - 0.0080 x + 3.5)

y = 2 c o s (2 π 10 t - 0.016 π x + 7 π)

Wave is propagated in +x direction because

ω

t and kx are in opposite sign in
standard equation

y = a \cos ω t - k x + ϕ

a = 2, ω = 20 π, k = 0.016 π a n d ϕ = 7 π

apath difference=4m=400cm
phase difference

Δ ϕ = \frac{2 π}{λ} \times p = \frac{2 π}{λ} \times 400 = k \times 400 = 0.016 π \times 400 = 0.64 π r a d

(b) path difference=0.5 m

Δ ϕ = \frac{2 π}{λ} \times p = k \times p = 0.016 π \times 50 = 0.8 π r a d

(c ) path difference=

λ_{2}

Δ ϕ = \frac{2 π}{λ} \times p = \frac{2 π}{λ} \times \frac{λ}{2} = π r a d i a n

d) Δ ϕ = \frac{2 π}{λ} \times p = \frac{2 π}{λ} \times \frac{3 λ}{4} = \frac{3}{2} π r a d i a n

(e)

T = \frac{2 π}{ω} = \frac{2 π}{20 π} = \frac{1}{10} s e c

At x=100 , t=T

ϕ_{1} = 20 π T - 0.016 π (100) + 7 π = 20 π \times \frac{1}{10} - 1.67 π + 7 π = 7.4 π

At t=5s

ϕ_{2} = 20 π (5) - 0.016 π 100 + 7 π = 100 π - 1.67 π + 7 π = 105.4 π

ϕ_{2} - ϕ_{1} = 98 π r a d i a n

The questions and numericals might seem a bit difficult and tricky at times but the students can use NCERT Exemplar Class 11 Physics solutions chapter 15 pdf download for step-by-step solutions of all the problems.

Also, Read NCERT Solution subject wise -

Also, check NCERT Notes subject wise -

Class 11 Physics NCERT Exemplar Solutions Chapter 15 includes the following topics:

Main Subtopics of NCERT Exemplar Class 11 Physics Solutions Chapter 15

15.1 Introduction
15.2 Transverse and longitudinal waves
15.3 Displacement relation in a progressive wave
15.4 The speed of a travelling wave
15.5 The principle of superposition of waves
15.6 Reflection of waves
15.7 Beats
15.8 Doppler Effect

What will students Learn from NCERT Exemplar Class 11 Physics Solutions Chapter 15?

Students will be able to collect information on wave motion and their characteristics and relation between them. They can learn about the various sound waves and how to distinguish between them. NCERT Exemplar Class 12 Physics chapter 15 solutions provides solutions to the questions regarding the speed of sound and also about resonance with the help of a tuning fork. The students will get a better understanding of waves by solving the questions and their explanations given in NCERT Exemplar Class 11 Physics solutions chapter 15 on topics like the speed of sound, nodes and antinodes of sound waves etc. the lesson also enlightens students about the basic characteristics of transverse and longitudinal waves and how to calculate the displacement of a progressive wave.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

Chapter 2 Units and Measurement

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 Systems of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Important Topics To Cover From NCERT Exemplar Class 11 Physics Solutions Chapter 15:

Although the entire chapter is very important, due to various questions being asked repeatedly in the exams but here are a few important topics that should be focused upon:

The definitions and their respective formulae of all the characteristics of a progressive wave should be focused on by the students. Many questions and numerical value questions in the exercise are based on them.
The students should be well known with the concept of the principle of superposition of waves that are given in detail in the NCERT Exemplar Class 11 Physics Solutions Chapter 15.
Resonance and beats are some of the important topics covered with the help of examples for the students to understand in a better way. Doppler’s Effect and its related numericals must be practised thoroughly.

NCERT Exemplar Class 11 Solutions

NCERT Exemplar Class 11 Mathematics Solutions

NCERT Exemplar Class 11 Chemistry Solutions

NCERT Exemplar Class 11 Physics Solutions

NCERT Exemplar Class 11 Biology Solutions

Check Class 11 Physics Chapter-wise Solutions

Chapter 1	Physical world
Chapter 2	Units and Measurement
Chapter 3	Motion in a straight line
Chapter 4	Motion in a Plane
Chapter 5	Laws of Motion
Chapter 6	Work, Energy and Power
Chapter 7	System of Particles and Rotational motion
Chapter 8	Gravitation
Chapter 9	Mechanical Properties of Solids
Chapter 10	Mechanical Properties of Fluids
Chapter 11	Thermal Properties of Matter
Chapter 12	Thermodynamics
Chapter 13	Kinetic Theory
Chapter 14	Oscillations
Chapter 15	Waves

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Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Are oscillation and waves the same thing?

No, they are completely different topics.

2. Do we need to go through a complete chapter or just the main topics?

One should always go through the complete chapter for better understanding of the concept.

3. Is it beneficial for compititive exam as well?

Yes, the chapter is important for the board as well as for competitive exams like NEET and JEE Main

Also Read

NCERT Solutions for Class 11 Biology Chapter 12 Respiration in Plants

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

NCERT Solutions for Class 11 Physics Chapter 15 Waves

NCERT Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

NCERT Solutions For Class 11 Biology Chapter 9 Biomolecules

NCERT Class 11 Physics Chapter 2 Notes Units and Measurement - Download PDF

NCERT Class 11 Biology Chapter 22 Notes Chemical Coordination And Integration- Download PDF Notes

NCERT Class 11 Biology Chapter 19 Notes Excretory Products And Their Elimination- Download PDF Notes

NCERT Class 11 Biology Chapter 20 Notes Locomotion And Movement- Download PDF Notes

NCERT Class 11 Biology Chapter 5 Notes Morphology Of Flowering Plants- Download PDF Notes

NCERT Class 11 Biology Chapter 15 Notes Plant Growth And Development- Download PDF Notes

NCERT Exemplar Class 11 Chemistry solutions Chapter 11 P-Block Elements

NCERT Exemplar Class 11 Chemistry Solutions Chapter 14 Environmental Chemistry

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NCERT Exemplar Class 11 Biology Solutions Chapter 10 Cell Cycle and Cell Division

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 11 Physics Solutions Chapter 15 Waves

NCERT Exemplar Class 11 Physics Solutions Chapter 15 MCQI

NCERT Exemplar Class 11 Physics Solutions Chapter 15 MCQII

NCERT Exemplar Class 11 Physics Solutions Chapter 15 Very Short Answer

NCERT Exemplar Class 11 Physics Solutions Chapter 15 Short Answer

NCERT Exemplar Class 11 Physics Solutions Chapter 15 Long Answer

Main Subtopics of NCERT Exemplar Class 11 Physics Solutions Chapter 15

What will students Learn from NCERT Exemplar Class 11 Physics Solutions Chapter 15?

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

Important Topics To Cover From NCERT Exemplar Class 11 Physics Solutions Chapter 15:

NCERT Exemplar Class 11 Solutions

Check Class 11 Physics Chapter-wise Solutions

Also Check NCERT Books and NCERT Syllabus here

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