JEE Main 2018PhysicsWork Power EnergyMediumMCQ

JEE Main 2018 — Work Power Energy Question with Solution

JEE Main 2018 (08 Apr)

Question

A particle is moving in a circular path of radius a under the action of an attractive potential U=-k2r2. Its total energy is:

Choose an option

Show full solutionCorrect option: D
Correct answer
DZero

Step-by-step explanation

U=-K2r2⇒F=-dUdr=2K2r3=Kr3⇒mv2r=Kr3⇒v2=Kmr2
⇒12mv2=K2r2⇒E=K2r2-K2r2=0.

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About this question

This is a previous-year question from JEE Main 2018, 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.