**NCERT Solutions Class 11 Physics Chapter 15 Waves**– Here are all the NCERT solutions for Class 11 Physics Chapter 15. This solution contains questions, answers, images, explanations of the complete chapter 15 titled Of Waves taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 15 Waves After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 15 Waves in one place.

## NCERT Solutions Class 11 Physics Chapter 15 Waves

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Physics for Class 11 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 15 Waves , Physics, Class 11.

Class | 11 |

Subject | Physics |

Book | Physics Part I |

Chapter Number | 15 |

Chapter Name |
Waves |

### NCERT Solutions Class 11 Physics chapter 15 Waves

Class 11, Physics chapter 15, Waves solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Question & Answer

Q.1:A string of mass 2.50 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, how long does the disturbance take to reach the other end?

Ans :Mass Of the string, M = 2.50 kg Tension in the string, T = 200 N Length of the string, I = 20.0 m Mass per unit length \(^{\mu=\frac{M}{l}}=\frac{2.50}{20}=0.125 \mathrm{kg} \mathrm{m}^{-1}\) The velocity (v) of the transverse wave in the string is given by the relation: \(\begin{aligned} v &=\sqrt{\frac{T}{\mu}} \\ &=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40 \mathrm{m} / \mathrm{s} \end{aligned}\) \(\therefore \text { Time taken by the disturbance to reach the other end, } t=\overline{v}=\frac{20}{40}=0.50 \mathrm{s}\)

Q.2:A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 \(m s^{-1} ?\left(g=9.8 m s^{-2}\right)\)

Ans :\(\begin{array}{l}{\text { Height of the tower, } s=300 \mathrm{m}} \\ {\text { Initial velocity of the stone, } u=0} \\ {\text { Acceleration, } a=g=9.8 \mathrm{m} / \mathrm{s}^{2}} \\ {\text { Speed of sound in air }=340 \mathrm{m} / \mathrm{s}}\end{array}\) \(\begin{array}{l}{\text { The time }\left(t_{1}\right) \text { taken by the stone to strike the water in the pond can be calculated using }} \\ {\text { the second equation of motion, as: }} \\ {s=u t_{1}+\frac{1}{2} g t_{1}^{2}}\end{array}\) \(\begin{array}{l}{300=0+\frac{1}{2} \times 9.8 \times t_{1}^{2}} \\ {\therefore t_{1}=\sqrt{\frac{300 \times 2}{9.8}}=7.82 \mathrm{s}} \\ {\text { Time taken by the sound to reach the top of the tower, } t=t_{1}+t_{2}} \\ {\text { Therefore, the time after which the splash is heard, } t=t_{1}+t_{2}} \\ {=7.82+0.88=8.7 \mathrm{s}}\end{array}\)

Q.3:A steel wire has a length of 12.0 m and a mass of 2.10 kg. VMnat should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at \(20^{\circ} \mathrm{C}=343 \mathrm{m} \mathrm{s}^{-1}\)

Ans :\(\begin{array}{l}{\text { Length of the steel wire, } l=12 \mathrm{m}} \\ {\text { Mass of the steel wire, } m=2.10 \mathrm{kg}} \\ {\text { Velocity of the transverse wave, } v=343 \mathrm{m} / \mathrm{s}}\end{array}\) Mass per unit length \(\mu=\frac{m}{l}=\frac{2.10}{12}=0.175 \mathrm{kg} \mathrm{m}^{-1}\) For tension T, velocity of the transverse wave can be obtained using the relation: \(\begin{array}{l}{v=\sqrt{\frac{T}{\mu}}} \\ {\therefore T=v^{2} \mu} \\ {=(343)^{2} \times 0.175=20588.575 \approx 2.06 \times 10^{4} \mathrm{N}}\end{array}\)

Q.4:Use the formula \(v=\sqrt{\frac{\gamma P}{p}}\) to explain why the speed of sound in air (a) is independent of pressure, (b) increases With temperature, (c) increases with humidity.

Ans :\(\begin{array}{l}{M=\sqrt{\frac{\gamma P}{\rho}}} \\ {v \text { here, }} \\ {\text { Density, } \rho=\frac{\text { Mass }}{\text { Volume }}=\frac{M}{V}} \\ {M=\text { Molecular weight of the gas }} \\ {V=\text { Volume of the gas }}\end{array}\) \(\begin{array}{l}{\text { Hence, equation ( } i \text { ) reduces to: }} \\ {v=\sqrt{\frac{\gamma P V}{M}}}\end{array}\) \(\begin{array}{l}{\text { Now from the ideal gas equation for } n=1 :} \\ {P V=R T} \\ {\text { For constant } T, P V=\text { Constant }} \\ {\text { since both } M \text { and } y \text { are constants, } v=\text { Constant }} \\ {\text { Hence, at a constant temperature, the speed of sound in a gaseous medium is }} \\ {\text { independent of the change in the pressure of the gas. }}\end{array}\) \(\begin{array}{l}{\text { (b) Take the relation: }} \\ {v=\sqrt{\frac{\gamma P}{\rho}}} \\ {\text { For one mole of an ideal gas, the gas equation can be written as: }} \\ {P V=R T}\end{array}\) \(\begin{array}{l}{P=\frac{R T}{V} \ldots(i i)} \\ {\text { Substituting equation (ii) in equation (i), we get: }} \\ {v=\sqrt{\frac{\gamma R T}{V \rho}}=\sqrt{\frac{\gamma R T}{M}}}\end{array}\) where, \(\begin{array}{l}{\text { Mass, } M=\rho V \text { is a constant }} \\ {\gamma \text { and } R \text { are also constants }} \\ {\text { We conclude from equation (iv) that } v \propto \sqrt{T} \text { . }} \\ {\text { Hence, the speed of sound in a gas is directly proportional to the square root of the }} \\ {\text { temperature of the gaseous medium, l..e., the speed of the sound increases with an }} \\ {\text { increase in the temperature of the gaseous medium and vice versa. }}\end{array}\) \(\begin{array}{l}{\text { (c) Let } v_{m} \text { and } v_{4} \text { be the speeds of sound in moist air and dry air respectively. }} \\ {\text { Let } \rho_{m} \text { and } \rho_{s} \text { be the densities of moist air and dry air respectively. }} \\ {\text { Take the relation: }} \\ {v=\sqrt{\frac{\gamma P}{\rho}}} \\ {\text { Hence, the speed of sound in moist air is: }}\end{array}\) \(\begin{array}{l}{v_{\mathrm{m}}=\sqrt{\frac{\gamma P}{\rho_{\mathrm{m}}}}} \\ {\text { And the speed of sound in dry air is: }} \\ {v_{\mathrm{d}}=\sqrt{\frac{\gamma P}{\rho_{\mathrm{d}}}}} \\ {\text { On dividing equations ( } i ) \text { and (ii), we get: }}\end{array}\) \(\begin{array}{l}{\frac{v_{\mathrm{m}}}{v_{\mathrm{d}}}=\sqrt{\frac{\gamma P}{\rho_{\mathrm{m}}} \times \frac{\rho_{\mathrm{d}}}{\gamma P}}=\sqrt{\frac{\rho_{\mathrm{d}}}{\rho_{\mathrm{m}}}}} \\ {\text { However, the presence of water vapour reduces the density of air, l.e., }}\end{array}\) \(\begin{array}{l}{\rho_{d}v_{d}} \\ {\text { Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous }} \\ {\text { medium, the speed of sound increases with humidity. }}\end{array}\)

Q.5:You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t)where x and t must appear in the combination x - v t or x + v t, i.e. y = f (x ± vt). Is the converse true? Examine if the following functions for V can possibly represent a travelling wave: \(\begin{array}{l}{\text { (a) }(x-v t)^{2}} \\ {\quad \log \left[\frac{x+y t}{x_{0}}\right]} \\ {\frac{1}{(x+v t)}}\end{array}\)

Ans :No; (a) Does not represent a wave (b)Represents a wave (c) Does not represent a wave The converse of the given statement is not true. The essential requirement for a function to represent a travelling wave is that it should remain finite for all values Of x and t. Explanation: \(\begin{array}{l}{\text { (a) For } x=0 \text { and } t=0, \text { the function }(x-v t)^{2} \text { becomes } 0 \text { . }} \\ {\text { Hence, for } x=0 \text { and } t=0, \text { the function represents a point and not a wave. }}\end{array}\) \(\begin{array}{l}{\text { (b) For } x=0 \text { and } t=0, \text { the function }} \\ {\log \left(\frac{x+y t}{x_{0}}\right)=\log 0=\infty}\end{array}\) Since the function does not converge to a finite value for x = O and t =O, it represents a travelling wave. \(\begin{array}{l}{\text { (c) For } x=0 \text { and } t=0, \text { the function }} \\ {\frac{1}{x+v t}=\frac{1}{0}=\infty} \\ {\text { since the function does not converge to a finite value for } x=0 \text { and } t=0, \text { it does not }} \\ {\text { represent a travelling wave. }}\end{array}\)

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