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:

(A) $r=\frac{1}{\sqrt{3}} \mathrm{R}$

(D) $r=\sqrt{\frac{3}{4}} \mathrm{R}$

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Gravitation

The mass density of a planet of radius R varies with the distance r from its centre as

ρ (r) = ρ_{0} (1 - \frac{r ^{2}}{R ^{2}})

. Then the gravitational field is maximum at:

(A) $r = \frac{1}{3} R$
(B) $r = R$
(C) $r = \frac{5}{9} R$ ✓ Correct
(D) $r = \frac{3}{4} R$

Answer: (C)

Solution

The field at radius

r

is set by the mass enclosed within it,

E = \frac{G m _{enc}}{r ^{2}}

. Integrating the given density,

m_{enc} = 4 π ρ_{0} \int_{0}^{r} (1 - \frac{r ^{'2}}{R ^{2}}) r^{'2} d r^{'} = 4 π ρ_{0} (\frac{r ^{3}}{3} - \frac{r ^{5}}{5 R ^{2}})

, so

E = 4 π G ρ_{0} (\frac{r}{3} - \frac{r ^{3}}{5 R ^{2}})

. Setting

\frac{d E}{d r} = 0

gives

\frac{1}{3} - \frac{3 r ^{2}}{5 R ^{2}} = 0

, hence

r^{2} = \frac{5 R ^{2}}{9}

and the field is greatest at

r = \frac{5}{9} R

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Single correct option

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