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 a - Physics MCQ NEET 2021 2026 | ExamTest.live

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From a circular ring of mass M and radius R an arc corresponding to a 9 0^{\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 M R^{2} . Then the value of K is:

Solution & Step-by-step Explanation

The moment of inertia of a whole ring is I = M R^{2} .The removed arc is 9 0^{\circ} out of 36 0^{\circ}, which is 1/4 of the ring.Mass of the remaining part M^{'} = M - \frac{1}{4} M = \frac{3}{4} M .Since the mass distribution for a ring is at constant distance R : I^{'} = M^{'} R^{2} = (\frac{3}{4} M) R^{2} = \frac{3}{4} M R^{2} So, K = 3/4 .

Try it yourself before checking the explanation above.

From a circular ring of mass M and radius R an arc corresponding to a 9 0^{\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 M R^{2} . Then the value of K is:

1/4

1/8

3/4

7/8

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