Rotational Motion

JEE Main level, single correct, medium

63 of 280

Question 63 of 280

Moment of inertia (M.I.)\left(M.I.\right) of four bodies, having same mass and radius, are reported as;
I1=M.I.{I}_{1}=M.I. of thin circular ring about its diameter,
I2=M.I.{I}_{2}=M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I3=M.I.{I}_{3}=M.I. of solid cylinder about its axis and
I4=M.I.{I}_{4}=M.I. of solid sphere about its diameter.
Then:

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Answer: (C) I1=I2=I3>I4{I}_{1}={I}_{2}={I}_{3}\gt {I}_{4}

Ring I1=MR22{I}_{1}=\dfrac{M{R}^{2}}{2} about diameter
Disc I2=MR22{I}_{2}=\dfrac{M{R}^{2}}{2}
Solid cylinder I3=MR22{I}_{3}=\dfrac{M{R}^{2}}{2}
Solid sphere I4=25MR2{I}_{4}=\dfrac{2}{5}M{R}^{2}
I1=I2=I3>I4{I}_{1}={I}_{2}={I}_{3}\gt {I}_{4}

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