JEE Main 2014PhysicsGravitationMediumMCQ

JEE Main 2014 — Gravitation Question with Solution

JEE Main 2014 (06 Apr)

Question

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

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Show full solutionCorrect option: D
Correct answer
D 1 2 GM R 1 + 2 2

Step-by-step explanation

F=F1=F2=GM22R2

F 3 = GM 2 4 R 2

∴  FTotal towards the centre.

=2 F+F3

=2GM22R2+GM24R2

= GM 2 4 R 2 2 2 + 1

F T = F CP

GM24R222+1=Mv2R

∴    v = GM 4 R 2 2 + 1 .

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About this question

This is a previous-year question from JEE Main 2014, 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.