Q. From a circular ring of mass $M$ and $R$, an arc corresponding to a $90^{0}$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the ring and perpendicular to the plane of ring is $k$ times ${MR}^{2}$. Then the value of $k$ is
A) $\frac{3}{4}$
B) $\frac{7}{8}$
C) $\frac{1}{4}$
D) $1$
Explanation
(a) The moment of inertia of ring $= MR^{2}$
The moment of inertia of removed sector $= \frac{1}{4}MR^{2}$
The moment of inertia of remaining part $= MR^{2} - \frac{1}{4}MR^{2}$
$= \frac{3}{4}\ MR^{2}$
According to question, the moment of inertia of the remaining part $= kMR^{2}$
then, $k = \frac{3}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.