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(C) The volume of the disc is $V = \pi R^2 t = \pi R^2 (\frac{R}{6}) = \frac{\pi R^3}{6}$.Let the radius of the sphere be $r$. The volume of the spher

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$\frac{I}{5}$

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(C) The volume of the disc is $V = \pi R^2 t = \pi R^2 (\frac{R}{6}) = \frac{\pi R^3}{6}$.Let the radius of the sphere be $r$. The volume of the sphere is $V = \frac{4}{3} \pi r^3$.Since the volume remains constant,$\frac{\pi R^3}{6} = \frac{4}{3} \pi r^3$,which gives $r^3 = \frac{R^3}{8}$,so $r = \frac{R}{2}$.The moment of inertia of the disc is $I = \frac{1}{2} M R^2$.The moment of inertia of the sphere is $I' = \frac{2}{5} M r^2$.Substituting $r = \frac{R}{2}$,we get $I' = \frac{2}{5} M (\frac{R}{2})^2 = \frac{2}{5} M \frac{R^2}{4} = \frac{1}{10} M R^2$.Comparing with $I = \frac{1}{2} M R^2$,we have $M R^2 = 2I$.Therefore,$I' = \frac{1}{10} (2I) = \frac{I}{5}$.

$A$ circular disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. It is melted and recasted into a solid sphere. The moment of inertia of the sphere about its diameter as axis of rotation is

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