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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2025 · 22 January · Morning Shift · Physics · Gravitation

A small point of mass m is placed at a distance 2 R from the centre...

A small point of mass mm is placed at a distance 2R2 R from the centre ' O′O^{\prime} of a big uniform solid sphere of mass M and radius R . The gravitational force on ' m ' due to M is F1\mathrm{F}_1. A spherical part of radius R/3\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 F2\mathrm{F}_2. The value of ratio F1:F2\mathrm{F}_1: \mathrm{F}_2 is 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

Options

  1. A12:1112: 11
  2. B11:1011: 10
  3. C12:912: 9
  4. D16:916: 9

Correct answer

A. 12:1112: 11

Step-by-step solution

F1=GMm(2R)2.....(1) F2=GMm(2R)2−(G(M27)m(4R3)2)F2=1148GMmR2....(2) F1:F2=12:11\begin{aligned} & \mathrm{F}_1=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2} .....(1)\\ & \mathrm{~F}_2=\frac{\mathrm{GMm}}{(2 \mathrm{R})^2}-\left(\frac{\mathrm{G}\left(\frac{\mathrm{M}}{27}\right) \mathrm{m}}{\left(\frac{4 \mathrm{R}}{3}\right)^2}\right) \\ & \mathrm{F}_2=\frac{11}{48} \frac{\mathrm{GMm}}{\mathrm{R}^2} ....(2)\\ & \mathrm{~F}_1: \mathrm{F}_2=12: 11 \end{aligned}
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