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Question

Physics Question on Moment Of Inertia

Match List-I with List-II.

| List - I| | List - II
---|---|---|---
A. | Moment of inertia of solid sphere of Radius R about any tangent | i.| 53MR2\frac{5}{3}MR^2
B.| Moment of inertia of hollow sphere of radius (R) about any tangent| ii.| 75MR2\frac{7}{5}MR^2
C.| Moment of inertia of circular ring of radius (R) about its diameter| iii.| 14MR2\frac{1}{4}MR^2
D.| Moment of inertia of circular disc of radius (R) about any diameter| iv.| 12MR2\frac{1}{2}MR^2

Choose the correct answer from the options given below.

A

A-II, B-I, C-IV, D-III

B

A-I, B-II, C-IV, D-III

C

A-II, B-I, C-III, D-IV

D

A-I, B-II, C-III, D-IV

Answer

A-II, B-I, C-IV, D-III

Explanation

Solution

The correct answer is (A) : A-II, B-I, C-IV, D-III
(A) Moment of inertia of solid sphere of radius R about a tangent
=25MR2+MR2=75MR2= \frac{2}{5} MR² + MR² = \frac{7}{5} MR²
⇒ A – (II)
(B) Moment of inertia of hollow sphere of radius R about a tangent
=23MR2+MR2=53MR2= \frac{2}{3} MR² + MR² = \frac{5}{3} MR²
⇒ B – (I)
(C) Moment of inertia of circular ring of radius (R) about its diameter
= (MR2)2\frac{(MR²) }{ 2}
⇒ C – (IV)
(D) Moment of inertia of circular ring of radius (R) about any diameter
= MR222=MR24\frac{\frac{MR²}{2} }{2} = \frac{MR²}{4}
⇒ D – (III)