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Question

A particle performs uniform circular motion with of an angular momentum L. If the frequency of partide'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 A L4
Given - Angular momentum is (L) * Frequency of particle's motion (0) is doubled (20) and kinetic energy kEkE2 (halved)
we have L=mvr and kE=12mv2 foequency v=v2πv From the above relations we get L=kEπ0 H kineticenergy is halved and frequency is doubled Lf=kE4π0=L4LL4 Henu, option (d) is correct.

2007082_1475443_ans_f7ae801daa7147d596a3dbceca6189a1.PNG

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