Question

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. $\frac{2}{ \sqrt 15} R$
• B. $\frac{2}{ \sqrt 5} R$
• C. $\frac{3}{\sqrt15} R$
• D. $\frac{\sqrt 3}{\sqrt15} R$

Answer 1

Answer: (A)

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

Answer 2

$\frac{2}{5} MR^2 = \frac{ 1}{2} Mr^2 + Mr^2$
or $\, \, \, \, \, \, \, \, \, \, \frac{2}{5} MR^2 \frac{3}{2}Mr^2$
$\therefore \, \, \, \, \, \, \, \, r= \frac{2}{ \sqrt {15}} R$