JEE Main 2025PhysicsGravitationKeplers Law And Universal Law Of GravitationmediumMCQ

JEE Main 2025Gravitation Question with Solution

From: JEE Main 2025 (Online) 22nd January Morning Shift

Question

A small point of mass is placed at a distance from the centre ' ' of a big uniform solid sphere of mass M and radius R . The gravitational force on ' m ' due to M is . A spherical part of radius is removed from the big sphere as shown in the figure and the gravitational force on m due to remaining part of is found to be . The value of ratio is

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Gravitation Question 12 English

This question includes a diagram. The text above accompanies the figure.

Choose an option

Show full solutionCorrect option: B
Correct answer
B12 : 11

Step-by-step explanation

JEE Main 2025 (Online) 22nd January Morning Shift Physics - Gravitation Question 12 English Explanation

Given mass of whole sphere =

= mass of removed sphere

Due to whole sphere,

(using Newton's gravitational law)

=Force due to whole sphere Force due to removed sphere

So,

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About this question

This is a previous-year question from JEE Main 2025, covering the Gravitation chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.