Q: 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 it’s 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

(a) $\large \frac{2}{\sqrt{15}}R$

(b) $\large \frac{2}{\sqrt{5}}R$

(c) $\large \frac{3}{\sqrt{15}}R$

(d) $\large \frac{\sqrt{3}}{\sqrt{15}}R$

Ans: (a)

Sol: According to question;

$\large \frac{2}{5}MR^2 = \frac{1}{2}Mr^2 + M r^2$ ; (By applying parallel axes theorem)

$\large \frac{2}{5}MR^2 = \frac{3}{2}Mr^2 $

$\large r = \frac{2}{\sqrt{15}}R$