From a circular ring of mass M and radius R,a | vedclass

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(A) The moment of inertia of a complete circular ring of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its pla

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$\frac{3}{4}$

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(A) The moment of inertia of a complete circular ring of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I = MR^{2}$.Since the mass is uniformly distributed along the circumference,the mass of the remaining part of the ring after removing a $90^{\circ}$ sector (which is $1/4$ of the total circumference) is $M' = M - \frac{1}{4}M = \frac{3}{4}M$.The moment of inertia of a point mass $m$ at a distance $R$ from the axis is $I = mR^{2}$. Since every point on the ring is at the same distance $R$ from the central axis,the moment of inertia of any part of the ring with mass $M'$ is simply $I' = M'R^{2}$.Substituting $M' = \frac{3}{4}M$,we get $I' = \frac{3}{4}MR^{2}$.Comparing this with $I' = KMR^{2}$,we find $K = \frac{3}{4}$.

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