A particle performs uniform circular motion with an angular momentum L. If the frequency of particle's motion is doubled and its kinetic energy halved, the angular momentum becomes
A
2L
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B
4L
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C
L2
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D
L4
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Solution
The correct option is DL4 Kinetic energy in rotation is K=12Iω2=12(I×ω)×ω ⇒K=12Lω ⇒L=2Kω
So, L2L1=K2K1×ω1ω2=12×12 ⇒L2L1=14 L2=L4