Q 11-06-012NEETNEET 2021Top questionEasy
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 :
Answer: (B) $\dfrac{3}{4}$
The remaining $270°$ arc has mass $\dfrac{3}{4}M$, and all of it is at distance $R$ from the axis:
$$I = \frac{3}{4}MR^2 \;\Rightarrow\; K = \frac{3}{4}$$
Solution by Sreeraj P, M.Sc Physics