Q. From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times '$MR^2$'. Then the value of 'K' is:
A) $\frac{7}{8}$
B) $\frac{1}{4}$
C) $\frac{1}{8}$
D) $\frac{3}{4}$
Explanation
[Solution] The remaining 3/4 of the mass is still at distance R, so $I = \frac{3}{4}MR^2$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.