JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 22)
The mass density of a planet of radius R varies with the distance r from its centre as
$$\rho $$(r) = $${\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$.
Then the gravitational field is maximum at :
$$\rho $$(r) = $${\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$.
Then the gravitational field is maximum at :
$$r = {1 \over {\sqrt 3 }}R$$
r = R
$$r = \sqrt {{3 \over 4}} R$$
$$r = \sqrt {{5 \over 9}} R$$
Explanation
_3rd_September_Evening_Slot_en_22_1.png)
dm = $$\rho $$dv
$$\int {dm} $$ $$= \int {{\rho _0}} 4\pi {r^2}dr$$
$$ \Rightarrow $$ M = $$ \int {{\rho _0}} 4\pi {r^2}dr$$
Also E = $${{GM} \over {{r^2}}}$$
$$ \Rightarrow E{r^2} = 4\pi G\,\int\limits_0^r {{\rho _0}} \left( {1 - {{{r^2}} \over {{R^2}}}} \right){r^2}dr$$
$$ \Rightarrow E = 4\pi G{\rho _0}\left( {{{{r}} \over 3} - {{{r^3}} \over {5{R^2}}}} \right)$$
For maximum E, $${{dE} \over {dr}} = 0$$
$${1 \over 3} - {{3{r^2}} \over {5{R^2}}}$$ = 0
$$ \therefore $$ $$r = \sqrt {{5 \over 9}} R$$
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