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Question

Physics Question on Moment Of Inertia

The diameter of a flywheel is increased by 1%. Increase in its moment of inertia about the central axis is

A

0.01

B

0.005

C

0.02

D

0.04

Answer

0.02

Explanation

Solution

As I=MR2I = MR^2
logI=logM+2logR\therefore \, \log \,I = \log \, M + 2 \, \log \, R
Differentiating, we get dII=0+2dRR\frac{dI}{I} = 0 + 2 \frac{dR}{R}
dII×100=2(dRR)×100\therefore \:\:\frac{dI }{I}\times100 = 2\left(\frac{dR}{R}\right)\times100
=2×1%=2%= 2 \times 1\% = 2\%