NCERT Solutions for Class 11 Physics Chapter 15 Waves

# NCERT Solutions for Class 11 Physics Chapter 15 Waves

Edited By Vishal kumar | Updated on Sep 07, 2023 01:31 PM IST

## NCERT Solutions Class 11 Physics Chapter 15 – Download Free PDF

NCERT Solutions for Class 11 Physics Chapter 15 Waves hold significant importance for both Class 11 board exams and competitive exams like NEET and JEE Mains. These NCERT solutions waves class 11 are invaluable aids in understanding and mastering the concepts presented in the chapter, encompassing questions from 15.1 to 15.27, with questions 15.22 to 15.27 included in the additional exercise section. Prepared by subject experts, these provide comprehensive support for students. The PDF format of class 11 physics waves ncert solutions by Careers360 facilitates a deeper understanding of the diverse applications of waves in different mediums, making it a valuable resource for students.

The Waves Class 11 chapter 15 solutions discuss questions based on mechanical waves. Electromagnetic waves (discussed in class 12 NCERT) do not necessarily require a medium, that is they can travel through a vacuum. Light, radio waves, X-rays, are examples of electromagnetic waves and they travel in a vacuum at a speed equal to the speed of light. The third kind of wave is Matter waves. They are associated with constituents of matter: electrons, protons, neutrons etc. For example molecules matter waves associated with electrons are employed in electron microscopes. NCERT solutions for Class 11 Physics Chapter 15 Waves help in understanding the formulas studied in the chapter. The class 11 physics ch 15 ncert solutions will help you to understand the concepts studied in the NCERT textbook.

** This chapter has been renumbered as Chapter 14 in accordance with the CBSE Syllabus 2023–24.

Free download class 11 physics waves NCERT solutions PDF for CBSE exam.

## NCERT Solutions for Class 11 Physics Chapter 15: Waves

NCERT Syllabus For Class 11 Physics Chapter 15 Waves: Exercise Question And Solutions

Mass per unit length of the string is

$\\\mu =\frac{M}{l}\\ =\frac{2.50}{20}\\ =0.125\ kg\ m^{-1}$

The velocity of the transverse wave in the string will be

$\\v=\sqrt{\frac{T}{\mu }}\\ =\sqrt{\frac{200}{0.125}}\\ =\sqrt{1600} \\=40\ m\ s^{-1}$

Time taken by the disturbance to travel from one end to the other is

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

Time taken by the stone to the pond can be calculated using the second equation of motion

s = 300 m

u = 0

a = 9.8 m s -2

$\\s=ut_{1}+\frac{1}{2}at_{1}^{2}\\ 300=4.9t_{1}^{2}\\ t_{1}=7.82\ s$

Time taken by the wave to propagate from the pond to the top of the tower is

$\\t_{2}=\frac{300}{340}\\ t_{2}=0.88\ s$

t 1 + t 2 = 8.7 s

The splash is heard at the top of the tower after a time of 8.7 seconds.

Mass per unit length od the wire is

$\\\mu =\frac{M}{l}\\ =\frac{2.10}{12}\\ =0.175\ kg\ m^{-1}$

The speed of a transverse wave in a wire is given by

$\\v=\frac{T}{\mu }\\ 343=\sqrt{\frac{T}{01.175}}\\ T=343^{2}\times 0.175\\ T=2.059\times 10^{4}\ N$

The tension in the wire should be $2.059\times 10^{4}\ N$ such that the speed of the transverse wave in it is equal to 343 m s -1 .

$v=\sqrt{\frac{\gamma P}{\rho }}$

Where $\gamma \ and \ \rho$ are the Bulk's modulus and the density respectively

As we know

$\rho =\frac{M}{V}$

where M is molecular weight of air and V is the volume of 1 mole of air

$\\v=\sqrt{\frac{\gamma P}{\rho }}\\ v=\sqrt{\frac{\gamma PV}{M}}$

From the ideal gas equation PV=nRT

since we are talking about 1 mole we take n = 1

PV=RT

The expression for the speed of sound becomes

$v=\sqrt{\frac{\gamma RT}{M}}$

$\gamma$ , M and R are constant therefore at constant Temperature the speed of sound in the air do not change and it is clear that speed is independent of velocity.

From the equation $\\v=\sqrt{\frac{\gamma RT}{M}}\\$ it is clear that the speed of sound is linearly proportional to the square root of the temperature and therefore it will increase with the increase in temperature.

$v=\sqrt{\frac{\gamma P}{\rho }}$

As the humidity of air increases, the proportion of water molecules(M=18) increases and that of Oxygen(M=32) and Nitrogen(M=28) decreases thus reducing the density of air and as the speed of sound is inversely proportional to the square root of density of air, the speed will increase as the density increases and thus it increases with an increase in humidity.

No, the given function cannot represent a wave as x or t approach infinity the function won't be converging to a constant value and therefore the converse is not true.

No, the given function cannot represent a travelling wave because as x and t become zero the given function won't be converging to a constant value and therefore the converse is not true.

No, the given function cannot represent a travelling wave because as x and t become zero the given function won't be converging to a constant value and therefore the converse is not true.

(a) The wavelength of the reflected sound wave which will be travelling in air is

$\\\lambda _{a}=\frac{v_{a}}{\nu }\\ \lambda _{a}=\frac{340}{10^{6}}\\ \lambda _{a}=3.4\times 10^{-4}\ m$

(b) The frequency of the transmitted sound wave would not change.

The wavelength of the transmitted sound wave which will be travelling in water is

$\\\lambda _{w}=\frac{v_{w}}{\nu }\\ \lambda _{w}=\frac{1486}{10^{6}}\\ \lambda _{w}=1.49\times 10^{-3}\ m$

The wavelength of the sound in the tissue is

$\\\lambda =\frac{V}{\nu }\\ \lambda =\frac{1.7\times 10^{3}}{4.2\times 10^{6}}\\ \lambda =4\times 10^{-4}\ m$

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?

The wave is travelling.

$y(x,t)=Asin(kx+\omega t+\phi )$

The wave is travelling in the negative x-direction i.e. from right to left.

$\\\omega =36\ rad\ s^{-1}\\ k=0.018\ cm^{-1}$

Speed of the wave is

$\\v=\frac{\omega }{k}\\ v=\frac{36\times 10^{-2}}{0.018} \\v=20\ m\ s^{-1}$

where x and y are in cm and t in s. The positive direction of x is from left to right.
(b) What are its amplitude and frequency?

Amplitude A is 3.0 cm.

Frequency is

$\\\nu =\frac{\omega }{2\pi }\\ \nu =\frac{36}{2\pi }\\ \nu =5.73\ Hz$

$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
where x and y are in cm and t in s. The positive direction of x is from left to right.
(c) What is the initial phase at the origin?

The initial phase of the wave at the origin (at x = 0 and t = 0) is $\frac{\pi }{4}$

$y(x,t)=3.0\; \sin (36\; t+0.018\; x+\pi /4)$
where x and y are in cm and t in s. The positive direction of x is from left to right.

(d) What is the least distance between two successive crests in the w ave?

The difference between two consecutive crests is equal to the wavelength of the wave.

$\\\lambda =\frac{2\pi }{k}\\ \lambda =\frac{2\pi\times 10^{-2} }{0.018}\\ \lambda =3.49\ m$

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

for x = 0

$\\y(t)=3.0sin(36t+\frac{\pi }{4})$

The time period of oscillation is T

$T=\frac{\pi }{18}\ s$

To make the y versus t graph we tabulate values of y(t) at different values of t as follows

 $t$ $0$ $\frac{T}{8}$ $\frac{T}{4}$ $\frac{3T}{8}$ $\frac{T}{2}$ $\frac{5T}{8}$ $\frac{3T}{4}$ $\frac{7T}{8}$ $T$ $y(t)$ $\frac{3}{\sqrt{2}}$ $3$ $\frac{3}{\sqrt{2}}$ $0$ $\frac{-3}{\sqrt{2}}$ $-3$ $\frac{-3}{\sqrt{2}}$ $0$ $\frac{3}{\sqrt{2}}$

The graph of y versus t is as follows

For other values of x, we will get a similar graph. Its time period and amplitude would remain the same, it just will be shifted by different amounts for different values of x.

Q.15.10 (a) For the travelling harmonic wave

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 $(a)\; 4\; m$

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =k\Delta x$

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

$k=2\pi \times 0.008\ cm^{-1}$

Phase difference for two points separated by a distance of 4 m would be

$\\\Delta \phi =2\pi \times 0.0080\times 4\times 100\\ \Delta \phi=6.4\pi \ rad$

Q.15.10 (b) For the travelling harmonic wave

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

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =k\Delta x$

$y(x,t)=2.0\; \cos 2\pi (10t-0.0080\; x+0.35)$

$k=2\pi \times 0.008\ cm^{-1}$

Phase difference for two points separated by a distance of 0.5 m would be

$\\\Delta \phi =2\pi \times 0.0080\times 0.8\times 100\\ \Delta \phi=0.8\pi \ rad$

Q.15.10 (c) For the travelling harmonic wave

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 $(c)\lambda /2$

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =\frac{2\pi }{\lambda }\Delta x$

Phase difference for two points separated by a distance of $\frac{\lambda }{2}$ would be

$\\\Delta \phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{2}\\ \Delta \phi =\pi\ rad$

Q.15.10 (d) For the travelling harmonic wave

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 $(d)3\lambda /4$

The phase difference between two points separated by a distance of $\\\Delta x$ is given by

$\\\Delta \phi =\frac{2\pi }{\lambda }\Delta x$

Phase difference for two points separated by a distance of $\frac{3\lambda }{4}$ would be

$\\\Delta \phi =\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}\\ \Delta \phi =\frac{3\pi }{2}\ rad$

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$ .

(a) Does the function represent a travelling wave or a stationary wave?

The given function is of the following form

$y(x,t)=2Asin\left (kx \right )cos(\omega t)$

which is the general equation representing a stationary wave and therefore the given function represents a stationary wave.

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$ .

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

We know that when two waves of the same amplitude, frequency and wavelength travelling in opposite directions get superimposed we get a stationary wave.

$\\y_{1}=asin(kx-\omega t)\\ \\y_{2}=asink(\omega t+kx)$

$\\y_{1}+y_{2}=asin(kx-\omega t)+asink(kx+\omega t)\\ =asin(kx)cos(\omega t)-asin(\omega t)cos(kx)+asin(\omega t)cos(kx)+asin(kx)cos(\omega t)\\ =2asink(kx)cos(\omega t)$

Comparing the given function with the above equations we get

$\\k=\frac{2\pi }{3}\\ \lambda =\frac{2\pi }{k}\\ \lambda =3\ m$

$\\\omega =120\pi \\ \nu =\frac{\omega }{2\pi }\\ \nu =60\ Hz$

$\\v=\nu \lambda \\ v=60\times 3=180\ m\ s^{-1}$

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is $3.0\times 10^{-2}\; kg$ .

(c) Determine the tension in the string

The mass per unit length of the string is

$\\\mu =\frac{M}{l}\\ \mu =\frac{3\times 10^{-2}}{1.5}\\ \mu =0.02\ kg\ m^{-1}$

Speed of a transverse wave is given by

$\\v=\sqrt{\frac{T}{\mu }}\\ T=\mu v^{2}\\ T=0.02\times (180)^{2}\\ T=648\ N$

(i) (a) All the points vibrate with the same frequency of 60 Hz.

(b) They all have the same phase as it depends upon time.

(c) At different points, the amplitude is different and is equal to A(x) given by

$A(x)=0.06sin(\frac{2\pi }{3}x)$

(ii)

$\\A(0.375)=0.06sin(\frac{2\pi }{3}\times 0.375)\\ A(0.375)=0.06\times sin(\frac{\pi }{4})\\ A(0.375)=0.06\times \frac{1}{\sqrt{2}}\\ A(0.375)=0.042m$

$(a)\; y=2\; \cos (3x)\sin (10t)$

$(b)\; y=2\sqrt{x-vt}$

$(c)y=3\sin (5x-0.5t)+ 4\; \cos (5x-0.5t)$

$(d)\; y=\cos x\; \sin t+\cos 2x\: \sin \: 2t$

(a) It represents a stationary wave.

(b) The given function does not represent a wave as we can see at certain values of x and t the function would not be defined.

(c) It represents a travelling wave.

(d) It represents a stationary wave. It is a superposition of two stationary waves which ultimately results in another stationary wave.

Length of the string is l given by

$\\l=\frac{M}{\mu }\\ l=\frac{3.5\times 10^{-2}}{4\times 10^{-2}}\\ l=0.875m$

Since the wire is vibrating in the fundamental mode

$\\l=\frac{\lambda }{2}\\ \lambda =2l\\ \lambda =2\times 0.875\\ \lambda =1.75m$

Speed of the string (v) is

$\\v=\nu l\\ v=45\times 1.75\\ v=78.75\ ms^{-1}$

Tension in the string is given by

$\\T=\mu v^{2}\\ T=4\times 10^{-2}\times (78.75)^{2}\\ T=248.0625\ N$

The pipe behaves as a pipe open at one end and closed at one end. Such a pipe would produce odd harmonics i.e.

$\nu =(2n-1)\frac{v}{4l_{n}}$

Two consecutive modes of vibration are given in the question

For l 1 = 25.5 cm

$\nu =(2n-1)\frac{v}{4l_{1}}$

For l 2 = 79.3 cm

$\\\nu =(2(n+1)-1)\frac{v}{4l_{2}}\\ \nu =(2n+1)\frac{v}{4l_{2}}$

Since at both these modes the system resonates with the same frequency we have

$\\(2n-1)\frac{v}{4l_{1}} =(2n+1)\frac{v}{4l_{2}}\\ \frac{2n-1}{2n+1}=\frac{25.5}{79.3}\\ \frac{2n-1}{2n+1}\approx \frac{1}{3}\\ n=1$ (our approximation is correct since the edge effects may be neglected)

$\\\nu =\frac{v}{4l_{1}}\\ v=340\times 4\times 0.255\\ v=346.8\ ms^{-1}$

When the rod is clamped at the middle at is vibrating in the fundamental mode, a node is formed at the middle of the rod and antinodes at the end. i.e.

$\\\frac{L}{2}=\frac{\lambda }{4}\\ \lambda =2L\\ \lambda =0.2m$

where L is the length of the rod.

Speed of sound in steel is

$\\v=\nu \lambda \\ v=2.53\times 10^{3}\times 0.2\\ v=5060ms^{-1}$

Let the n th harmonic mode of the pipe get resonantly excited by a 430 Hz source.

$\\\frac{(2n-1)v}{4l}=\nu \\ 2n-1=\frac{430 \times 4\times 0.2}{340}\\ n=1$

The pipe resonates with a 430 Hz source in the fundamental mode when one end is open.

Let the m th harmonic mode of the pipe get resonantly excited by a 430 Hz source when both ends are open.

$\\\lambda =\frac{2l}{m}\\ m=\frac{2l\nu }{v}\\ m=\frac{2\times 0.2\times 430}{340}\\ m=0.5$

Since m is coming out to be less than 1 the same source will not be in resonance with the pipe if both ends are open.

$\\\nu _{A}=324Hz\\ Beat\ frequency(b)=6Hz\\ \nu _{B}=\nu _{A}\pm b\\ \nu _{B}=318Hz \\or \\\nu _{B}=330Hz$

Since frequency increases with an increase in Tension, the frequency of string A must have decreased. Therefore $\nu _{B}=318Hz$ .(If it were 330 Hz the beat frequency would have increased with decrease in Tension in string A)

Q.15.19 Explain why (or how) :

In the propagation of a sound wave the pressure increases at points where displacement decreases, Therefore maximum pressure at points of minimum displace and vice-versa i.e. a displacement node is a pressure antinode and vice versa.

Q.15.19 Explain why (or how) :

Bats emit ultrasonic waves and when these waves strike the obstacles and get reflected back to the bats they ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”.

Q.15.19 Explain why (or how) :

We can distinguish between the two notes with the same frequency as the harmonics they emit are different.

Q.15.19 (d) Explain why (or how) :

Transverse waves produce shear, gases don't have shear modulus and cannot sustain shear and therefore can only propagate longitudinal waves. Solids have both shear and bulk modulus of elasticity and can propagate both transverse and longitudinal waves.

As we know a pulse contains waves of different wavelengths, these waves travel at different speeds in a dispersive medium and thus the shape of the pulse gets distorted.

$\nu _{o}=\left ( \frac{v\pm v_{o}}{v\pm v_{s}} \right )\nu$

where $\\\nu _{o}$ is the frequency as observed by the observer, $\nu$ is the frequency of the source, v is the speed of the wave, v o is the speed of the observer and v s is the speed of the source.

(i) (a) When the source is moving towards the observer and the observer is stationary.

$\\\nu _{o}=\left ( \frac{v}{v- v_{s}} \right )\nu \\ \nu _{o}=\frac{340}{340-10}\times 400\\ \nu _{o}=412Hz$

(b)

$\\\nu _{o}=\left ( \frac{v}{v+ v_{s}} \right )\nu \\ \nu _{o}=\frac{340}{340+10}\times 400\\ \nu _{o}=389Hz$

(ii) The speed of the sound does not change as it is independent of the speed of observer and source and remains equal to 340 ms -1 .

Speed of the wind v w = 10 m s -1

Speed of sound in still air v a = 340 m s -1

Effective speed with which the wave reaches the observer = v = v w + v a = 10 + 340= 350 m s -1

There is no relative motion between the observer and the source and therefore the frequency heard by the observer would not change.

The wavelength of the sound as heard by the observer is

$\\\lambda =\frac{v}{\nu }\\ \lambda =\frac{350}{400}\\ \lambda =0.875m$

The above situation is not identical to the case when the air is still and the observer runs towards the yard as then there will be relative motion between the observer and the source and the frequency observed by the observer would change.

$\nu _{o}=\left ( \frac{v\pm v_{o}}{v\pm v_{s}} \right )\nu$

The frequency as heard by the observer is

$\\\nu _{o}=\frac{340+10}{340}\times 400\\ \nu _{o}=411.76Hz$

$\\\lambda=\frac{340}{400}\\ \lambda=0.85m$

$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$
(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?

$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$

The displacement of oscillation of a point at x = 1 cm and t = 1 s is

$\\y(1,1)=7.5sin(0.0050\times 1+12\times 1+\frac{\pi }{4})\\ y(1,1)=7.5sin(12.79)\\ y(1,1)=7.5sin(\frac{12.266\times180^{o} }{\pi })\\ y(1,1)=7.5sin(733.18^{o})\\ y(1,1)=1.71cm$

The general expression for the velocity of oscillation is

$\\v_{y}(x,t)=\frac{\mathrm{dy(x,t)} }{\mathrm{d} t}\\=\frac{d}{dt}\left [ sin\left ( 7.5sin(0.0050x+12t+\frac{\pi }{4} \right ) \right ]\\ =90cos(0.0050x+12t+\frac{\pi }{4})$

$\\v_{y}(1,1)=90cos(0.0050\times 1+121+\frac{\pi }{4})\\ =90cos(12.79)\\ =90cos(733.18^{o})\\ =87.63\ cm\ s^{-1}$

$y(x,t)=7.5\: \sin (0.0050\: x+12t+\pi /4)$

k=0.005 cm -1

$\omega =12rad/s$

The velocity of propagation of the wave is

$\\v=\nu \lambda \\ v=\frac{\omega }{2\pi }\times \frac{2\pi }{k}\\ v=\frac{\omega }{k}\\ v=\frac{12}{0.005\times 100}\\ v=24ms^{-1}$

The velocity of oscillation of point at x = 1 cm and t = 1 cm is not equal to the propagation of the wave.

## NCERT solutions for class 11 physics chapter 15 waves additional exercise

(b) Locate the points of the string which have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s.

The wavelength of the given wave is

$\\\lambda =\frac{2\pi }{k}\\ \lambda =\frac{2\pi }{0.005 cm^{-1}}\\ \lambda =1256cm\\ \lambda =12.56m$

The points with the same displacements and velocity at the same instant of time are separated by distances $n\lambda$ .

The points of the string which have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s would be at a distance of

$\pm \lambda ,\pm 2\lambda ,\pm 3\lambda ...$ from x = 1cm.

$\lambda =12.56m$

Therefore all points at distances $\pm 12.56m,\pm 25.12m,\pm 37.68m$ from the point x=1cm would have the same transverse displacements and velocity as the $x=1\: cm$ point at $t=2\: s,$ 5 s and 11 s.

(a) The pulse does not have a definite frequency or wavelength however the wave has definite speed given the medium is non-dispersive.

(b) The frequency of the note produced by the whistle is not 0.05 Hz. It only implies the frequency of repetition of the pip of the whistle is 0.05 Hz,

$y(x,t)=Asin(\omega t\pm kx+\phi )$

A=0.05 m

Tension in the string is T=mg

$\\T=90\times 9.8\\ T=882N$

The speed of the wave in the string is v

$\\v=\sqrt{\frac{T}{\mu }}\\ v=\sqrt{\frac{882}{8\times 10^{-3}}}\\ v=332ms^{-1}$

Angular frequency of the wave is

$\\\omega =2\pi \nu \\ \omega =2\pi \times 256\\ \omega =1608.5rad/s$

$\\k=\frac{2\pi }{\lambda }\\ k=\frac{2\pi\nu }{v}\\ k=4.84m^{-1}$

Since at t=0, the left end (fork end) of the string x=0 has zero transverse displacements (y=0) and is moving along the positive y-direction, the initial phase is zero. $\left (\phi =0\ rad \right )$

Taking the left to the right direction as positive we have

$y(x,t)=0.05sin(1608.5t-4.84x)$

Here t is in seconds and x and y are in metres.

Frequency of SONAR $(\nu )$ =40 kHz

Speed of enemy submarine v o =360 km h -1 = 100 m s -1

$\\\nu _{o}=(\frac{v+v_{o}}{v})\nu \\ =\frac{1450+100}{1450}\times 40\times 10^{3}\\ =42.76\ kHz$

This is the frequency which would be observed and reflected by the enemy submarine but won't appear the same to the SONAR(source) as again there is relative motion between the source(enemy submarine) and the observer(SONAR)

The frequency which would be received by the SONAR is

$\\\nu'_{o} =\left ( \frac{v}{v-v_{s}} \right )\nu _{o}\\ =\frac{1450}{1450-100}\times 42.76\times 10^{3}\\ =45.92\ kHz$

Let us assume the earthquake occurs at a distance s.

$\\\Delta t=4\times 60=240s\\ \Delta t=\frac{s}{v_{s}}-\frac{s}{v_{p}} \\\Delta t=\frac{s}{4}-\frac{s}{8}\\ 240=\frac{s}{8}\\ s=1960km$

The origin of the earthquake is at a distance of 1960 km.

Apparent frequency striking the wall and getting reflected is

The frequency emitted by the bats is $\nu =40kHz$

Speed of sound is v

Speed of bat is 0.03v

$\\\nu'=(\frac{v}{v-v_{s}})\nu \\ =\frac{v}{v-0.03v}\times 40kHz\\ =41.237kHz$

Frequency of sound as heard by the bat

$\\\nu''=(\frac{v+v_{o}}{v})\nu'\\ =\frac{v+0.03v}{v}\times \nu' \\=1.03\times 41.237kHz \\ =42.474kHz$

## NCERT Solutions for Class 11 Physics Chapter 15 Waves Free PDF Download

The ncert solutions waves class 11 pdf includes in-depth explanations of all the crucial topics and subtopics, making it easier for students to comprehend the concepts.

The NCERT solutions for Waves Physics class 11 are crucial in establishing the groundwork for the Class 12 CBSE board exams, which determine students' admission to higher education institutes. Emphasis on the fundamental concepts and theoretical aspects of this chapter is critical for building a strong foundation for the Class 12 CBSE board exam. Moreover, Class 11 Physics Ch 15 NCERT solutions have been developed in accordance with the latest CBSE and NCERT guidelines and syllabus.

## NCERT Solutions for Class 11 Physics Chapter Wise

 Chapter 1 Physical world Chapter 2 Chapter 3 Motion in a straight line Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Waves

### Chapter 15: Waves

• #### Wave velocity

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Wave velocity(W) = Frequency(f) * wavelength (λ) [SI unit- m/s]

• #### velocity of the transverse wave in a stretched string

where:

v = Velocity of the wave (in meters per second, m/s), T = Tension in the string (in Newtons, N) and μ = Linear mass density of the string (in kilograms per meter, kg/m)

• #### Velocity of a longitudinal wave in an elastic medium

Where:

E = Young's modulus of the material (in Pascals, Pa), ρ = Density of the material (in kilograms per cubic meter, kg/m³)

## Importance of NCERT Solutions for Class 11 Physics Chapter 15 Waves:

• In NCERT Class 11 Physics chapter 15 Waves, you will study mathematical and physical aspects of the longitudinal, transverse and physical wave.
• The solutions of NCERT Class 11 waves help students in self-preparation of the chapter. In both board exam and competitive exams like JEE Main and NEET beats and the Doppler effects are important topics.
• There are questions based on these topics in the CBSE NCERT solutions for Class 11 Physics chapter 15 Waves.

## Key features of Class 11 Physics Waves NCERT Solutions

1. Comprehensive Coverage: These class 11 physics ch 15 ncert solutions cover all the important topics and questions presented in the chapter, ensuring a thorough understanding of wave phenomena.

2. Exercise and Additional Exercise Solutions: Detailed solutions are provided for exercise questions (15.1 to 15.21) and additional exercise questions (15.22 to 15.27), allowing for comprehensive practice and self-assessment.

3. Clarity and Simplicity: The solutions are explained in clear and simple language, making complex wave concepts more accessible to students.

4. Problem-Solving Skills: By practising with these solutions and class 11 physics chapter 15 waves notes, students develop strong problem-solving skills, which are essential in physics and other subjects.

5. Exam Preparation: These wave class 11 solutions are designed to help students prepare effectively for their exams, including board exams and competitive exams like NEET and JEE Mains.

## Subject wise NCERT Exemplar solutions

1. What are the features of NCERT solutions for Class 11 Physics Chapter 15?

Waves is one of the important chapters of Class 11 NCERT syllabus. The questions given in the NCERT books are important to understand the concepts well. There are two exercise and 27 questions discussed in the Waves Class 11 NCERT chapter. For more questions refer to NCERT exemplar.

2. What are the topics covered in Waves Class 11 NCERT chapter?

The main topics covered in the NCERT Class 11 textbook chapter 15 are

• Transverse and longitudinal waves
•  Displacement relation in a progressive wave
• The speed of a traveling wave
• The principle of superposition of waves
•  Reflection of waves
• Beats
•  Doppler effect
3. Write Some Characteristics of a Wave.

Some characteristics of a wave include

Amplitude: The maximum displacement of a wave from its equilibrium position.

Wavelength: The distance between two consecutive points on a wave that are in phase.

Frequency: The number of complete waves passing a point in one second.

Period: The time taken for one complete wave to pass a given point.

Speed: The rate at which the wave is travelling through the medium.

Phase: The position of a point on a wave relative to a fixed point in time.

Direction of propagation: The direction in which the wave is travelling.

4. Define transverse wave motion?

In transverse wave motion, the particles of the medium move perpendicular to the direction in which the wave travels, creating crests and troughs. Examples include light waves, electromagnetic waves, and water waves on the surface of a liquid.

5. Is waves class 11 easy chapter?

Yes, the "Waves" chapter in Class 11 Physics is generally considered to be relatively easier compared to some other chapters in the subject. It introduces fundamental concepts like frequency, wavelength, amplitude, and period, and these concepts are typically straightforward to understand.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 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.

With increase of temperature, which of these changes?

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

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 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