A solid sphere of radius R has moment of inertia I about its geometrical axis. It is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc) is also equal to I, then the value of r is equal to

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$\frac{2}{\sqrt{5}} R$

$\frac{2}{\sqrt{15}} R$

$\frac{\sqrt{3}}{\sqrt{15}} R$

$\frac{3}{\sqrt{15}} R$

answer is A.

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Detailed Solution

$I = \frac{2}{5} M R^{2} = (M r^{2} + \frac{1}{2} M r^{2})$

$\Rightarrow \frac{2}{5} R^{2} = \frac{3}{2} r^{2} \Rightarrow r = \frac{2 R}{\sqrt{15}}$

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