A spherical body of radius  R consists of a fluid of constant density and is in equilibrium under its own gravity.  If  $P\left(r\right)$ is the pressure at  $r\left(r<R\right),$ then the correct option(s) is (are)

$\text{ The acceleration due to gravity at }x\text{ from center }=g=\frac{Gm}{x^2}=\frac{G\rho \frac{4}{3}\pi x^3}{x^2}=\frac{4}{3}G\rho \pi x$

$\text{ On the element }\mathrm{dm}\text{ of thickness }\mathrm{dx},\text{ the inward gravitational force is balanced by the outward pressure. }$

$-dP\left(A\right)=\left(dm\right)g$

$\Rightarrow -dPA=\left(\rho Adx\right)\left(\frac{4}{3}G\rho \pi x\right)$

$Integrating,$

$\begin{array}{l}\Rightarrow P=k\left(R^2-r^2\right)\\ \text{ Note: Assume pressure }P=0\text{ at }r=R\end{array}$

$Now, check all options$