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JEE Main 202124 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
24 February
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2021 · 24 February · Morning Shift · Physics · Rotational Motion

Moment of inertia (M . I .) of four bodies, having same mass and...

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:

Options

  1. AI1+I2=I3+52I4{I}_{1} + {I}_{2} = {I}_{3} + \dfrac{5}{2}{I}_{4}.
  2. BI1+I3<I2+I4{I}_{1} + {I}_{3} < {I}_{2} + {I}_{4}
  3. CI1=I2=I3>I4{I}_{1} = {I}_{2} = {I}_{3} > {I}_{4}
  4. DI1=I2=I3<I4{I}_{1} = {I}_{2} = {I}_{3} < {I}_{4}

Correct answer

C. I1=I2=I3>I4{I}_{1} = {I}_{2} = {I}_{3} > {I}_{4}

Step-by-step solution

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} > {I}_{4}

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