JEE Main 2015PhysicsRotational MotionMediumMCQ

JEE Main 2015 — Rotational Motion Question with Solution

JEE Main 2015 (04 Apr)

Question

From a solid sphere of mass M and radius R, a cube of the maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is:

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Show full solutionCorrect option: D
Correct answer
D4MR293π

Step-by-step explanation

Let a be the length of edge for the cube, with maximum possible volume diagonal length =2R⇒3a=2R⇒a=2R3. As densities of sphere and cube are equal. Let M' be mass of the cube, M43πR3=M'a3⇒M′=3Ma34πR3. Moment of inertia of cube about an axis passing through its center is,  I=M′2a212 =3Ma34πR3×2a212 =Ma58πR3.

also, a=23R⇒I =M×32 R58π×93R3 =4MR293π.

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About this question

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