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JEE Main 202228 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
28 June
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2022 · 28 June · Morning Shift · Physics · Rotational Motion

Match List-I with List-II List-I List-II (A) Moment of inertia of...

Match List-I with List-II

List-I List-II
(A) Moment of inertia of solid sphere
of radius RR about any tangent.
(I) 53MR2\dfrac{5}{3}{MR}^{2}
(B) Moment of inertia of hollow sphere of radius
(R)\left(R\right) about any tangent.
(II) 75MR2\dfrac{7}{5}{MR}^{2}
(C) Moment of inertia of circular ring of radius
(R)\left(R\right) about its diameter.
(III) 14MR2\dfrac{1}{4}{MR}^{2}
(D) Moment of inertia of circular disc of radius
(R)\left(R\right) about any diameter.
(IV) 12MR2\dfrac{1}{2}{MR}^{2}
Match List-I with List-II List-I List-II (A) Moment of inertia of solid sphere of radius R about any tangent. (I)Match List-I with List-II List-I List-II (A) Moment of inertia of solid sphere of radius R about any tangent. (I)

Options

  1. AA−II,B−I,C−IV,D−IIIA - II, B - I, C - IV, D - III
  2. BA−I,B−II,C−IV,D−IIIA - I, B - II, C - IV, D - III
  3. CA−II,B−I,C−III,D−IVA - II, B - I, C - III, D - IV
  4. DA−I,B−II,C−III,D−IVA - I, B - II, C - III, D - IV

Correct answer

A. A−II,B−I,C−IV,D−IIIA - II, B - I, C - IV, D - III

Step-by-step solution

For solid sphere

Match List-I with List-II List-I List-II (A) Moment of inertia of solid sphere of radius R about any tangent. (I)

I0=Icom+MR2{I}_{0} = {I}_{com} + M{R}^{2} (Using parallel axis theorem)

I0=25MR2+MR2{I}_{0} = \dfrac{2}{5}M{R}^{2} + M{R}^{2}

I0=75MR2{I}_{0} = \dfrac{7}{5}M{R}^{2}⇒A→II⇒ A \to II

For hollow sphere

Match List-I with List-II List-I List-II (A) Moment of inertia of solid sphere of radius R about any tangent. (I)

I=Icom+MR2⇒I=53MR2⇒B→II = {I}_{com} + M{R}^{2} ⇒ I = \dfrac{5}{3}M{R}^{2} ⇒ B \to I

For Circular ring

Match List-I with List-II List-I List-II (A) Moment of inertia of solid sphere of radius R about any tangent. (I)

I1+I2=I3{I}_{1} + {I}_{2} = {I}_{3} ( Using perpendicular axis theorem)

By symmetry MOI

About 11 and 22 Axis are same i.e.

I1=I2{I}_{1} = {I}_{2}

∴2I1=I3=MR2(∵Icom=MR2)∴ 2{I}_{1} = {I}_{3} = M{R}^{2}\left(∵ {I}_{com} = M{R}^{2}\right)

I1=MR22⇒C→IV{I}_{1} = \dfrac{M{R}^{2}}{2} ⇒ C \to IV

Similarly in disc

2I1=MR22{∵Icom=MR22}2{I}_{1} = \dfrac{M{R}^{2}}{2}\left\{∵ {I}_{com} = \dfrac{M{R}^{2}}{2}\right\}

I1=MR24⇒D→III{I}_{1} = \dfrac{M{R}^{2}}{4} ⇒ D \to III

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