Q 11-07-172JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $\rho(r) = \rho_0\left(1 - \dfrac{r^{2}}{R^{2}}\right)$. Then the gravitational field is maximum at:
Answer: (D) $r = \sqrt{\dfrac59}R$
Mass inside $r$: $M(r) = \displaystyle\int_0^r 4\pi x^{2}\rho_0\left(1 - \frac{x^{2}}{R^{2}}\right)dx = 4\pi\rho_0\left(\frac{r^{3}}{3} - \frac{r^{5}}{5R^{2}}\right)$.
$$g(r) = \frac{GM(r)}{r^{2}} = 4\pi G\rho_0\left(\frac r3 - \frac{r^{3}}{5R^{2}}\right)$$
$\dfrac{dg}{dr} = 0 \Rightarrow \dfrac13 = \dfrac{3r^{2}}{5R^{2}} \Rightarrow r = \sqrt{\dfrac59}R$ (inside the planet, and $g$ falls off outside).
Solution by Sreeraj P, M.Sc Physics