at $27^{\circ} \mathrm{C}, \mathrm{v}_{1}=\mathrm{n} \lambda,$ at $31^{\circ} \mathrm{C}, \mathrm{v}_{2}=(\mathrm{n}+\mathrm{x}) \lambda$

Now using $\mathrm{v} \propto \sqrt{\mathrm{T}} \quad(\because \mathrm{v}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{M}}})$

$\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}}=\frac{\mathrm{n}+\mathrm{x}}{\mathrm{n}}$

$\Rightarrow \frac{300+\mathrm{x}}{300}=\sqrt{\frac{(273+31)}{(273+27)}}=\sqrt{\frac{304}{300}}=\sqrt{\frac{300+4}{300}}$

$\Rightarrow 1+\frac{\mathrm{x}}{300}=\left(1+\frac{4}{300}\right)^{1 / 2}=\left(1+\frac{1}{2} \times \frac{4}{300}\right)$

$\Rightarrow x=2$           $\left[\because(1+\mathrm{x})^{\mathrm{n}}=1+\mathrm{nx}\right]$