Question

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 $r(r<R)$, then the correct option(s) is (are)

• A. $P(r=0)=0$
• B. $\frac{P(r=3R/4)}{P(r=2R/3)}=\frac{63}{80}$
• C. $\frac{P(r=3R/5)}{P(r=2R/5)}=\frac{16}{21}$
• D. $\frac{P(r=R/2)}{P(r=R/3)}=\frac{20}{27}$

Answer 1

Answer: (B), (C)

$\frac{P(r=3R/5)}{P(r=2R/5)}=\frac{16}{21}$


Consider a small patch of thickness $dr$ at a distance $r$ from the center.

so, the prassure is given by, $P=\frac{dF}{dA}$

$dF=[P-(P+dP\left)\right]A$
$\Rightarrow \frac{Gm}{r^2}dm=-\left(dP\right)A$
$\Rightarrow {\int }_0^{P}dP=-{\int }_R^r\frac{G\frac{M{r}^3}{R^3}4a^{2}dr\rho }{r^2\left(4\pi r^2\right)}$
$\therefore P=\frac{G\rho M}{2R}(1-\frac{r^2}{R^2})$

for option A:
if $r=0$,
$P$ can't be zero.
Hence option A is wrong.

for option B:
$\frac{P(r=3R/4)}{P(r=2R/3)}=\frac{R^2-9R^2/16}{R^2-4R^2/9}=\frac{63}{18}$
Hence option B is correct.

for option C:
$\frac{P(r=3R/5)}{P(r=2R/5)}=\frac{R^2-9/25R^2}{R^2-4/25R^2}=\frac{16}{21}$
Hence option C is correct.

for option D:
$\frac{P(r=R/2)}{P(r=R/3)}=\frac{R^2-1/4R^2}{R^2-1/9R^2}=\frac{27}{32}$
Hence option D is wrong.