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Topic: Moment of inertia, radius of gyration, MI of simple objects· Rotational Motion

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 i…

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 'MR\(^2\)'. Then the value of 'K' is :

1. \(\frac{3}{4}\)

2. \(\frac{7}{8}\)

3. \(\frac{1}{4}\)

4. \(\frac{1}{8}\)

1 answer

Best answer

Answer: 1. \(\frac{3}{4}\)

Key idea: moment of inertia of a ring about its axis depends only on mass and radius, since every part is at distance R.

Removing a 90° arc removes a quarter of the mass, so the remaining mass is \(\tfrac34M\).

\(I=\tfrac34MR^2\), so \(K=\tfrac34\). Option 1.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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