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Q.

A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes   [Kerala CEE 2013]

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a

L2

b

L

c

L3

d

L4

answer is A.

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Detailed Solution

We know that,  L=Iω=2πmr2fNow,   ω′=ω/2Hence,  L′=Iω′=mr2ω2=πmr2f∴ LL′=2mπr2fπmr2f⇒L′=L2
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