NEET 2021 Physics Ring with Removed Sector MCQ Question
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 ‘MR²’. Then the value of ‘K’ is:
3/4
7/8
1/4
1/8
Correct Answer
Detailed Explanation
The moment of inertia of the full ring is MR². Removing a 90° sector means removing 1/4 of the ring, so the remaining moment of inertia is (3/4)MR². Therefore, K = 7/8.
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