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NEET 2021 · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from NEET 2021

Q139NEET 2021Medium

Moment of Inertia & Radius of Gyration

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 center of the ring and perpendicular to the plane of the ring is 'K' times 'MR2'. Then the value of 'K' is:

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Answer: (b)

  1. Removing a 90∘90^\circ arc removes 14\frac{1}{4} of the mass; remaining mass 3M4\frac{3M}{4}.
  2. All remaining mass is still at distance RR: I=34MR2I = \frac{3}{4}MR^2.
  3. So K=34K = \frac{3}{4}, answer (b).
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