A small point of mass $m$ is placed at a distance $2 R$ from the centre 'O' of a big uniform solid sphere of mass M and radius R. The gravitational force on 'm' due to M is $\mathrm{F}_{1}$. A spherical part of radius $\mathrm{R} / 3$ is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be $\mathrm{F}_{2}$. The value of ratio $\mathrm{F}_{1}$ : $\mathrm{F}_{2}$ is

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Gravitation

A small point of mass

m

is placed at a distance

2 R

from the centre 'O' of a big uniform solid sphere of mass M and radius R. The gravitational force on 'm' due to M is

F_{1}

. A spherical part of radius

R /3

is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of M is found to be

F_{2}

. The value of ratio

F_{1}

F_{2}

(A) $16 : 9$
(B) $11 : 10$
(C) $12 : 11$ ✓ Correct
(D) $12 : 9$

Answer: (C)

Solution

The force on

m

due to the full sphere is

F_{1} = \frac{GM m}{( 2 R ) ^{2}} = \frac{GM m}{4 R ^{2}} .

The removed spherical portion has radius

R /3

, so its mass is

M (\frac{R /3}{R})^{3} = \frac{M}{27}

. From the figure its centre lies at

\frac{2 R}{3}

from O, tangent to the outer surface along the line toward m, putting it at a distance

2 R - \frac{2 R}{3} = \frac{4 R}{3}

from m.

By superposition, the force from the remaining mass equals the force from the full sphere minus the force the removed piece would have exerted:

F_{2} = \frac{GM m}{4 R ^{2}} - \frac{G ( \frac{M}{27} ) m}{( \frac{4 R}{3} ) ^{2}} = \frac{GM m}{4 R ^{2}} - \frac{GM m}{48 R ^{2}} = \frac{12 GM m - GM m}{48 R ^{2}} = \frac{11}{48} \cdot \frac{GM m}{R ^{2}} .

Since

F_{1} = \frac{GM m}{4 R ^{2}} = \frac{12}{48} \cdot \frac{GM m}{R ^{2}}

F_{1} : F_{2} = 12 : 11.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Gravitation

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