NCERT Solutions for Class 11 Physics Chapter 15 Waves

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:
\(\begin{aligned} v &=\sqrt{\frac{T}{\mu}} \\ &=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40 \mathrm{m} / \mathrm{s} \end{aligned}\)
\(\therefore \text { Time taken by the disturbance to reach the other end, } t=\overline{v}=\frac{20}{40}=0.50 \mathrm{s}\) 

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

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:
\(\begin{array}{l}{v=\sqrt{\frac{T}{\mu}}} \\ {\therefore T=v^{2} \mu} \\ {=(343)^{2} \times 0.175=20588.575 \approx 2.06 \times 10^{4} \mathrm{N}}\end{array}\) 

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,
\(\begin{array}{l}{\text { Mass, } M=\rho V \text { is a constant }} \\ {\gamma \text { and } R \text { are also constants }} \\ {\text { We conclude from equation (iv) that } v \propto \sqrt{T} \text { . }} \\ {\text { Hence, the speed of sound in a gas is directly proportional to the square root of the }} \\ {\text { temperature of the gaseous medium, l..e., the speed of the sound increases with an }} \\ {\text { increase in the temperature of the gaseous medium and vice versa. }}\end{array}\)
\(\begin{array}{l}{\text { (c) Let } v_{m} \text { and } v_{4} \text { be the speeds of sound in moist air and dry air respectively. }} \\ {\text { Let } \rho_{m} \text { and } \rho_{s} \text { be the densities of moist air and dry air respectively. }} \\ {\text { Take the relation: }} \\ {v=\sqrt{\frac{\gamma P}{\rho}}} \\ {\text { Hence, the speed of sound in moist air is: }}\end{array}\)
\(\begin{array}{l}{v_{\mathrm{m}}=\sqrt{\frac{\gamma P}{\rho_{\mathrm{m}}}}} \\ {\text { And the speed of sound in dry air is: }} \\ {v_{\mathrm{d}}=\sqrt{\frac{\gamma P}{\rho_{\mathrm{d}}}}} \\ {\text { On dividing equations ( } i ) \text { and (ii), we get: }}\end{array}\)
\(\begin{array}{l}{\frac{v_{\mathrm{m}}}{v_{\mathrm{d}}}=\sqrt{\frac{\gamma P}{\rho_{\mathrm{m}}} \times \frac{\rho_{\mathrm{d}}}{\gamma P}}=\sqrt{\frac{\rho_{\mathrm{d}}}{\rho_{\mathrm{m}}}}} \\ {\text { However, the presence of water vapour reduces the density of air, l.e., }}\end{array}\)
\(\begin{array}{l}{\rho_{d}v_{d}} \\ {\text { Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous }} \\ {\text { medium, the speed of sound increases with humidity. }}\end{array}\) 

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.
\(\begin{array}{l}{\text { (c) For } x=0 \text { and } t=0, \text { the function }} \\ {\frac{1}{x+v t}=\frac{1}{0}=\infty} \\ {\text { since the function does not converge to a finite value for } x=0 \text { and } t=0, \text { it does not }} \\ {\text { represent a travelling wave. }}\end{array}\)