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JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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A ring of mass 10 kg and diameter 0.4 m is rotated about its axis. If it makes 2100 revolutions per minute, then its angular momentum will be
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Correct option: B
88  kg  ×m2/s88\;kg\; \times {m^2}/s
Option B (88  kg  ×m2/s88\;kg\; \times {m^2}/s) is correct because it is the result of calculating the angular momentum using the formula L=IωL = I\omega with the given parameters.

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