JEE PYQ: Gravitation Question 22
Question 22 - 2020 (03 Sep Shift 2)
The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $\rho(r) = \rho_0\left(1 - \frac{r^2}{R^2}\right)$. Then the gravitational field is maximum at:
(1) $r = \sqrt{\frac{3}{4}}R$
(2) $r = R$
(3) $r = \frac{1}{\sqrt{3}}R$
(4) $r = \sqrt{\frac{5}{9}}R$
Show Answer
Answer: (4)
Solution
$M(r) = 4\pi\rho_0\int_0^r \left(r^2 - \frac{r^4}{R^2}\right)dr = 4\pi\rho_0\left(\frac{r^3}{3} - \frac{r^5}{5R^2}\right)$. $E = \frac{GM(r)}{r^2} = \frac{4\pi G\rho_0}{r^2}\left(\frac{r^3}{3} - \frac{r^5}{5R^2}\right) = 4\pi G\rho_0\left(\frac{r}{3} - \frac{r^3}{5R^2}\right)$. $\frac{dE}{dr} = 0$: $\frac{1}{3} - \frac{3r^2}{5R^2} = 0$, so $r = \sqrt{\frac{5}{9}}R$.
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