JEE Main 2015PhysicsWork Power EnergyMediumMCQ

JEE Main 2015 — Work Power Energy Question with Solution

JEE Main 2015 (11 Apr Online)

Question

A particle is moving in a circle of radius r under the action of a force F=αr2 which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy=0 for r=0):

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Show full solutionCorrect option: A
Correct answer
A56αr3

Step-by-step explanation

Force F=αr2 will act as centripetal force

⇒αr2=mv2r⇒mv2=αr3

∴ Kinetic energy K.E. =12mv2

⇒K.E.=αr32

Also dU=-F→.dr→

∆U=-∫F→.dr→

=-∫αr2-dr

=α∫0rr2dr

U-0=αr33

⇒U=αr33

So total mechanical energy

E=U+K

=αr33+αr32

E=5αr36

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About this question

This is a previous-year question from JEE Main 2015, 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.