JEE Main 2022PhysicsRotational MotionMediumMCQ

JEE Main 2022 — Rotational Motion Question with Solution

JEE Main 2022 (26 Jun Shift 2)

Question

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

Choose an option

Show full solutionCorrect option: C
Correct answer
C27

Step-by-step explanation

Under pure rolling condition: v=ωR

Translational K.E=12mv2

Rotational K.E=12×25mR2vR2=15mv2

Total K.E=12mv2+15mv2=710mv2

So, Rotational K·ETotal K·E=15mv2710mv2=27

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About this question

This is a previous-year question from JEE Main 2022, covering the Rotational Motion 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.