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From a circular ring of mass $M$ and radius $R$ an arc corresponding to a $90^\circ$ 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
$\frac{3}{4}$
Detailed Solution
Removing a $90^\circ$ arc leaves $\frac{3}{4}$ of the ring: $M_{remain} = \frac{3}{4}M$
All of the remaining mass is at distance R from the axis:
$I = M_{remain}R^2 = \frac{3}{4}MR^2 \Rightarrow K = \frac{3}{4}$