JEE Main 2019PhysicsGravitationHardMCQ

JEE Main 2019Gravitation Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

Four identical particles of mass M are located at the corners of a square of side a . What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square?

Choose an option

Show full solutionCorrect option: D
Correct answer
D 1.16GMa

Step-by-step explanation


Gravitational force on each mass 'M'
=2GM2a2+GM22a2=2+12GM2a2
This force will provide required centripetal force
2+12GM2a2=Mv2(a/2)
v=1+122GMa=1.16GMa

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About this question

This is a previous-year question from JEE Main 2019, 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.