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

Solution:

As $I = MR^2$
$\therefore \, \log \,I = \log \, M + 2 \, \log \, R$
Differentiating, we get $\frac{dI}{I} = 0 + 2 \frac{dR}{R}$
$ \therefore \:\:\frac{dI }{I}\times100 = 2\left(\frac{dR}{R}\right)\times100$
$= 2 \times 1\% = 2\% $

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