Physics · Gravitation

JEE Main 2025 — 22 January, Morning Shift — Question 54

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 $\mathrm{F}_{2}$ . The value of ratio

$F_{1}: F_{2}$ is

Option A:
$16: 9$
Option B:
$11: 10$
Option C:
$12: 11$
Correct
Option D:
$12: 9$

Answer: C

Step-by-step solution

$\mathrm{F}_{1}=\frac{\mathrm{GMm}}{(2 \mathrm{R})^{2}}$

$F_{2}=\frac{G M m}{(2 R)^{2}}-\left(\frac{G\left(\frac{M}{27}\right) m}{\left(\frac{4 R}{3}\right)^{2}}\right)$ $\mathrm{F}_{2}=\frac{11}{48} \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$ $\mathrm{F}_{1}: \mathrm{F}_{2}=12: 11$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 22 January, Morning Shift
Subject: Physics
Chapter: Gravitation
Topic: Gravitational Field and Gravity