Q. From a circular disc of radius $R$ and mass $9\ M$, a small disc of mass $M$ and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
A) $\frac{40}{9}MR^{2}$
B) $MR^{2}$
C) $4\ MR^{2}$
D) $\frac{4}{9}MR^{2}$
Explanation
(a) $I = I_{1} - I_{2} = \frac{9MR^{2}}{2} - \frac{MR^{2}}{18}$
$= \frac{81MR^{2} - MR^{2}}{18} = \frac{40MR^{2}}{9}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.