A particle performs uniform circular motion with an angular momentum L. If the angular frequency of the particle is doubled, and kinetic energy is halved, its angular momentum becomes:

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answer is D.

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Detailed Solution

L=Iω=12Iω212ω=2Kω, where K=12Iω2 is the kinetic energy. If ω is doubled and K is halved, the value of L becomes one-fourth. Hence the correct choice is (4).

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