(3)

Velocity of a transverse wave in a stretched string is given by

$v=\sqrt{\frac{T}{\mu }}$                                                      ...(i)

Where $T$ is the tension in the string and $\mu$ is the mass per unit length of the string.

If tension $T$ is doubled, then final speed of the wave in the string will be,

$v\text{'}=\sqrt{\frac{2T}{\mu }}$                                              ...(ii)

From eq. (i) and (ii)

$\frac{v}{v\text{'}}=\frac{1}{\sqrt{2}}$

(4)

[IMAGE 128]

Wavelength of pulse at the lower end,

${\lambda }_1\propto \text{velocity}\left(v_1\right)=\sqrt{\frac{T_1}{\mu }}$

Similarly,   ${\lambda }_2\propto v_2=\sqrt{\frac{T_2}{\mu }}$

$\therefore$   $\frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{(m_1+m_2)g}{m_{2}g}}=\sqrt{\frac{m_1+m_2}{m_2}}$

(3)

Since 4.0 g of a gas occupies 22.4 litres at NTP, so the molecular mass of the gas is $M=4.0\text{g }$${\mathrm{mol}}^{-1}$

As the speed of sound in the gas is $v=\sqrt{\frac{\gamma RT}{M}}$

where $\gamma$ is the ratio of two specific heats, $R$ is the universal gas constant and $T$ is the temperature of the gas.

$\therefore$ $\gamma =\frac{M{v}^2}{RT}$

Here,  $M=4.0$ $\text{g}$ ${\mathrm{mol}}^{-1}=4.0\times {10}^{-3}$ $\mathrm{kg}$ ${\mathrm{mol}}^{-1}$, $v=952$ $\text{m}$ $s^{-1}$

$R=8.3\text{J }$$K^{-1}{\text{mol}}^{-1}$ and 

$T=273$$\text{K}$ (at NTP)

$\therefore$    $\gamma =\frac{(4.0\times {10}^{-3} {\text{kg mol}}^{-1}){(952 {\text{m s}}^{-1})}^2}{(8.3 {\text{J K}}^{-1}{\text{mol}}^{-1})(273 \text{K})}=1.6$

By definition, $\gamma =\frac{C_P}{C_V}$ or $C_P=\gamma C_V$

But $\gamma =1.6$ and $C_V=5.0$ $\text{J}$ $K^{-1}$${\text{mol}}^{-1}$

$\therefore$ $C_P=(1.6)(5.0 {\text{J K}}^{-1}{\text{mol}}^{-1})=8.0$ $\text{J}$ $K^{-1}{\text{mol}}^{-1}$