Easy MCQ +4 / -1 PYQ · JEE Mains 2022

Match List-I with List-II

List-I List-II
(A) Moment of inertia of solid sphere of radius R about any tangent. (I) ${5 \over 3}M{R^2}$
(B) Moment of inertia of hollow sphere of radius (R) about any tangent. (II) ${7 \over 5}M{R^2}$
(C) Moment of inertia of circular ring of radius (R) about its diameter. (III) ${1 \over 4}M{R^2}$
(D) Moment of inertia of circular disc of radius (R) about any diameter. (IV) ${1 \over 2}M{R^2}$

Choose the correct answer from the options given below :

  1. A A - II, B - I, C - IV, D - III Correct answer
  2. B A - I, B - II, C - IV, D - III
  3. C A - II, B - I, C - III, D - IV
  4. D A - I, B - II, C - III, D - IV

Solution

<p>(A) Moment of inertia of solid sphere of radius R about a tangent $= {2 \over 5}M{R^2} + M{R^2} = {7 \over 5}M{R^2}$</p> <p>$\Rightarrow$ A $-$ (II)</p> <p>(B) Moment of inertia of hollow sphere of radius R about a tangent $= {2 \over 3}M{R^2} + M{R^2} = {5 \over 3}M{R^2}$</p> <p>$\Rightarrow$ B $-$ (I)</p> <p>(C) Moment of inertia of circular ring of radius (R) about its diameter = ${{\left( {M{R^2}} \right)} \over 2}$</p> <p>$\Rightarrow$ C $-$ (IV)</p> <p>(D) Moment of inertia of circular disc of radius (R) about any diameter</p> <p>$= {{M{R^2}/2} \over 2} = {{M{R^2}} \over 4}$</p> <p>$\Rightarrow$ D $-$ (III)</p>

About this question

Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass

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