KVPY SX SB Exam 19-Nov-2017 Question Paper

Consider a spherical shell of radius r and radial thickness dr. P & P + dP are pressure at its inner and outer surface.
Let ${\mathit{g}}_{\mathit{r}}$ = gravitational acceleration at distance
r (< R) For equilibrium of this shell -
$\left(\mathit{P}+\mathit{d}\mathit{P}\right)\left(4\mathit{\pi }{\mathit{r}}^2\right)+\left(4\mathit{\pi }{\mathit{r}}^2\mathit{d}\mathit{r}\right)\mathit{\rho }{\mathit{g}}_{\mathit{r}}$ $=\left(\mathit{P}\right)\left(4\mathit{\pi }{\mathit{r}}^2\right)$
$\left\{\mathit{\rho }=\frac{3\mathit{M}}{4\mathit{\pi }{\mathit{R}}^3}=\mathrm{Density}\mathrm{of}\mathrm{sphere}\right\}$
⇒ $\mathit{d}\mathit{P}=-\mathit{\rho }{\mathit{g}}_{\mathit{r}}\mathit{d}\mathit{r}$
∵ ${\mathit{g}}_{\mathit{r}}=\frac{4}{3}\mathit{\pi }\mathrm{}\mathit{G}\mathit{\rho }\mathit{r}$
⇒ $\mathit{d}\mathit{P}=-\frac{4}{3}\mathit{\pi }\mathrm{}\mathit{G}{\mathit{\rho }}^2\mathit{r}\mathit{d}\mathit{r}$
{(–) ve sign indicates that pressure is decreasing with radius}
⇒ $\underset{\mathit{P}}{\overset{{\mathit{P}}_0}{\int }}\mathit{d}\mathit{P}=\underset{\mathit{r}}{\overset{\mathit{R}}{\int }}-\frac{4}{3}\mathit{\pi }\mathit{G}{\mathit{\rho }}^2\mathit{r}\mathit{d}\mathit{r}$
⇒ ${\mathit{P}}_0-\mathit{P}=-\frac{4}{3}\mathit{\pi }\mathit{G}{\mathit{\rho }}^2\left(\frac{{\mathit{R}}^2}{2}-\frac{{\mathit{r}}^2}{2}\right)$
⇒ $\mathit{P}={\mathit{P}}_0+\frac{4}{3}\mathit{\pi }\mathit{G}{\mathit{\rho }}^2\left(\frac{{\mathit{R}}^2}{2}-\frac{{\mathit{r}}^2}{2}\right)$
⇒ $\mathit{P}={\mathit{P}}_0+\frac{3\mathit{G}{\mathit{M}}^2}{8\mathit{\pi }{\mathit{R}}^4}\left(1-\frac{{\mathit{r}}^2}{{\mathit{R}}^2}\right)$
Hence$\mathit{P}\propto \frac{1}{{\mathit{R}}^4}$
⇒ Average pressure will also be proportional to $\frac{1}{{\mathit{R}}^4}$
⇒ correct answer is [A]