Q. One quarter of the disc of mass $m$ is removed. If $r$ be the radius of the disc, the new moment of inertia is
A) $\frac{3}{2}mr^{2}$
B) $\frac{mr^{2}}{2}$
C) $\frac{3}{8}mr^{2}$
D) None of these
Explanation
(c) Moment of inertia of whole disc about an axis through centre of disc and perpendicular to its plane is $I = \frac{1}{2}mr^{2}$
As one quarter of disc is removed, new mass,
$m^{'} = \frac{3}{4}m$
$\therefore I^{'} = \frac{1}{2}\left( \frac{3}{4}m \right)r^{2} = \frac{3}{8}mr^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.