JEE Main 2021 — Rotational Motion Question with Solution
From: JEE Main 2021 (Online) 18th March Morning Shift
Question
A thin circular ring of mass M and radius r is rotating about its axis with an angular speed . Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become :
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
net = 0, so angular momentum is conserved
By angular momentum conservation
Iii = Iff
(MR2) = (MR2 + 2mR2)f
f =
By angular momentum conservation
Iii = Iff
(MR2) = (MR2 + 2mR2)f
f =
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This is a previous-year question from JEE Main 2021, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.