A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own…

1. $\mathrm{P}\left(\mathrm{r}=0\right)=0$
2. $\frac{\mathrm{P}\left(\mathrm{r}\mathrm{ }=\mathrm{ }\frac{3\mathrm{R}}{4}\right)}{\mathrm{P}\left(\mathrm{r}\mathrm{ }=\mathrm{ }\frac{2\mathrm{R}}{3}\right)}=\frac{63}{80}$
3. $\frac{\mathrm{P}\left(\mathrm{r}\mathrm{ }=\mathrm{ }\frac{3\mathrm{R}}{5}\right)}{\mathrm{P}\left(\mathrm{r}\mathrm{ }=\frac{2\mathrm{R}}{5}\right)}=\frac{16}{21}$
4. $\frac{\mathrm{P}\left(\mathrm{r}\mathrm{ }=\mathrm{ }\frac{\mathrm{R}}{2}\right)}{\mathrm{P}\left(\mathrm{r}\mathrm{ }=\frac{\mathrm{R}}{3}\right)}=\frac{20}{27}$

Solution

$\mathrm{d}\mathrm{E}=\left[\mathrm{P}-\left(\mathrm{P}+\mathrm{d}\mathrm{P}\right)\right]\mathrm{A}$
$\Rightarrow \frac{\mathrm{G}\mathrm{m}}{{\mathrm{r}}^2}\mathrm{d}\mathrm{m}=-\left(\mathrm{d}\mathrm{P}\right)\mathrm{A}\mathrm{ }$
$\Rightarrow \underset{0}{\overset{\mathrm{P}}{\int }}\mathrm{d}\mathrm{P}=-\underset{\mathrm{E}}{\overset{\mathrm{r}}{\int }}\frac{\mathrm{G}\frac{\mathrm{M}{\mathrm{r}}^3}{{\mathrm{R}}^3}4\mathrm{a}{\mathrm{r}}^2\mathrm{d}\mathrm{r}\mathrm{\rho }}{{\mathrm{r}}^2\left(4\mathrm{\pi }{\mathrm{r}}^2\right)}$
$\therefore \mathrm{P}=\frac{\mathrm{G}\mathrm{\rho }\mathrm{M}}{2\mathrm{R}}\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right)$ ;

Asked in: JEE Advanced 2015 (Paper 2)

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