$M=4.0 \mathrm{g} \mathrm{mol}^{-1}$
As the speed of the sound in the gas is
$v=\sqrt{\frac{\gamma R T}{M}}$
where $\gamma$ is the ratio of two specific heats, $R$ is the universal gas constant and $T$ is the temperature of the gas.
$\therefore \gamma=\frac{M v^{2}}{R T}$
Here, $M=4.0 \mathrm{g} \mathrm{mol}^{-1}=4.0 \times 10^{-3} \mathrm{kg} \mathrm{mol}^{-1}$
$v=952 \mathrm{ms}^{-1}, R=8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$ and $T=273 \mathrm{K}(\mathrm{at} \mathrm{NTP})$
$\because \quad \gamma=\frac{\left(4.0 \times 10^{-3} \mathrm{kg} \mathrm{mol}^{-1}\right)\left(952 \mathrm{ms}^{-1}\right)^{2}}{\left(8.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)(273 \mathrm{K})}=1.6$
By definition, $\gamma=\frac{C_{p}}{C_{v}}$ or $C_{p}=\gamma C_{v}$
But $\gamma=1.6$ and $C_{v}=5.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$
$\therefore \quad C_{p}=(1.6)\left(5.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\right)=8.0 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$
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[$A$] The time $\mathrm{T}_{A 0}=\mathrm{T}_{\mathrm{OA}}$
[$B$] The velocities of the two pulses (Pulse $1$ and Pulse $2$) are the same at the midpoint of rope.
[$C$] The wavelength of Pulse $1$ becomes longer when it reaches point $A$.
[$D$] The velocity of any pulse along the rope is independent of its frequency and wavelength.
| List $-I$ | List $-II$ |
| $(A)$ Moment of inertia of solid sphere of radius $(R)$ about any tangent | $(I)$ $\frac{5}{3} MR ^{2}$ |
| $(B)$ Moment of inertia of hollow sphere of radius $(R)$ about any tangent | $(II)$ $\frac{7}{5} MR ^{2}$ |
| $(C)$ Moment of inertia of circular ring of radius $(R)$ about its diameter. | $(III)$ $\frac{1}{4} MR ^{2}$ |
| $(D)$ Moment of inertia of circular disc of radius $(R)$ about any diameter. | $(IV)$ $\frac{1}{2} MR ^{2}$ |
Question: Choose the correct answer from the options given below