JEE Main 2022PhysicsRotational MotionMediumNumerical

JEE Main 2022 — Rotational Motion Question with Solution

JEE Main 2022 (25 Jun Shift 2)

Question

Moment of Inertia (M.I.) of four bodies having same mass M and radius 2R are as follows
I1= M.I. of solid sphere about its diameter
I2= M.I. of solid cylinder about its axis
I3= M.I. of solid circular disc about its diameter
I4= M.I. of thin circular ring about its diameter
If 2I2+I3+I4=xI1 then the value of x will be _____ .

Enter your answer

Show full solutionCorrect answer: 5
Correct answer
5

Step-by-step explanation

∵ 2I2+I3+I4=xI1

Now

For solid sphere about its diameter I1=25MR2,

For solid cylinder about its axis I2=12MR2,

For circular disc about its diameter I3=14MR2

And for circular ring about its diameter I4=12MR2

⇒2MR22+MR24+MR22=x×25MR2

⇒234+12=x×25

⇒2=x×25

⇒x=5

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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.