JEE Main 2022PhysicsGravitationMediumMCQ

JEE Main 2022 — Gravitation Question with Solution

JEE Main 2022 (25 Jul Shift 1)

Question

Three identical particle A,B and C of mass 100 kg each are placed in a straight line with AB=BC=13 m. The gravitational force on a fourth particle P of the same mass is F, when placed at a distance 13 m from the particle B on the perpendicular bisector of the line AC. The value of F will be approximately

Choose an option

Show full solutionCorrect option: B
Correct answer
B100G

Step-by-step explanation

Given here, m=100 kg

Gravitational force, FAP=Gm21322,

FBP=Gm2132 and FCP=Gm21322

Net gravitational force, 

Fnet=FBP+FAPcos45°+FCPcos45°

 =G×1041321+12

F≃100G

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About this question

This is a previous-year question from JEE Main 2022, 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.