JEE Main 2022PhysicsWork Power EnergyMediumMCQ

JEE Main 2022 — Work Power Energy Question with Solution

JEE Main 2022 (28 Jun Shift 1)

Question

A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration ac is varying with time t as ac=k2rt2 , where k is a constant. The power delivered to the particle by the force acting on it is -

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Show full solutionCorrect option: C
Correct answer
Cmk2r2t

Step-by-step explanation

ac=k2rt2

Or v2r=k2rt2  Or    v=krt
Therefore, tangential acceleration, at=dvdt=kr

Therefore, tangential force, Ft=mat=mkr

Only tangential force does work.

Power=Ftv=mkrkrt ⇒Power=mk2r2t

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About this question

This is a previous-year question from JEE Main 2022, covering the Work Power Energy 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.