KVPY SX SB Exam 19-Nov-2017 Question Paper

The energy will propagate in form of spherical blast wave which is longitudinal in nature.
Hence velocity of propagate of disturbance$\mathit{v}=\sqrt{\frac{\mathit{\gamma }\mathit{P}}{\mathit{\rho }}}$
Where P = Pressure
∵ PV = nRT
$\mathit{v}=\sqrt{\frac{\mathit{\gamma }\mathit{P}}{\mathit{V}\mathit{\rho }}}$
Where $\mathit{V}=\frac{4\mathit{\pi }{\mathit{r}}^3}{3}$
& v = velocity of propagation = $\frac{\mathit{d}\mathit{r}}{\mathit{d}\mathit{t}}$
⇒ $\frac{\mathit{d}\mathit{r}}{\mathit{d}\mathit{t}}=\sqrt{\frac{\mathit{\gamma }\mathit{R}\mathit{T}}{\left(\frac{4}{3}\mathit{\pi }{\mathit{r}}^3\right)}\mathit{\rho }}$
⇒ $\frac{\mathit{d}\mathit{r}}{\mathit{d}\mathit{t}}=\mathit{k}={\mathit{r}}^{3∕2}$
k = constant = $\sqrt{\frac{3\mathit{\gamma }\mathit{R}\mathit{T}}{4\mathit{\pi }\mathit{\rho }}}$
⇒ ${\mathit{r}}^{3∕2}$dr = kdt
$\int {\mathit{r}}^{3∕2}\mathit{d}\mathit{r}=\int \mathit{k}\mathit{d}\mathit{t}$
${\mathit{r}}^{5∕2}=\mathit{k}\mathit{t}\Rightarrow \mathit{r}={\left(\mathit{k}\mathit{t}\right)}^{2∕5}$
⇒ $\mathit{r}\propto {\mathit{t}}^{2∕5}$
Hence correct answer is (B)