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 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
Updated On: Jun 14, 2022
- $\frac{2}{ \sqrt 15} R $
- $\frac{2}{ \sqrt 5} R $
- $\frac{3}{\sqrt15} R $
- $\frac{\sqrt 3}{\sqrt15} R $
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The Correct Option is A
Solution and Explanation
$ \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 $
or $ \, \, \, \, \, \, \, \, \, \, \frac{2}{5} MR^2 \frac{3}{2}Mr^2$
$\therefore \, \, \, \, \, \, \, \, r= \frac{2}{ \sqrt {15}} R $
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