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The moment of inertia of a uniform circular disc of radius $R$ and mass $M$ about an axis passing from the edge of the disc and normal to the disc is:
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Verified Answer
The correct answer is:
$\frac{3}{2} M R^2$
M.I. of disc about its normal $=\frac{1}{2} M R^2$
M.I. about its one edge $=M R^2+$ $\frac{M R^2}{2}$
$$
\therefore \quad \text { M.I. }=\frac{3}{2} M R^2
$$
M.I. about its one edge $=M R^2+$ $\frac{M R^2}{2}$
$$
\therefore \quad \text { M.I. }=\frac{3}{2} M R^2
$$
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