JEE Main 2018PhysicsRotational MotionHardMCQ

JEE Main 2018Rotational Motion Question with Solution

JEE Main 2018 (08 Apr)

Question

From a uniform circular disc of radius R and mass 9 M, a small disc of radius R3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B4MR2

Step-by-step explanation

By parallel axis theorem,

Mass of smaller disc = 9M π R 2 π ( R 3 ) 2 =M

MI of smaller disc =MR322+M2R32

MI=MR218+4MR29

MI=MR22

MI=MIBigger-MISmaller

=9MR22-MR22

=4MR2

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About this question

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