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Question

A particles performing uniform circular motion. Its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is


A
L4
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B
2L
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C
4L
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D
L2
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Solution

The correct option is B $$\displaystyle\frac{L}{4}$$
$$\mbox{Angular momentum} \propto \displaystyle\frac{1}{\mbox{Angular frequency}} \propto \mbox{Kinetic Energy}$$
$$\Rightarrow \bar{L} = \displaystyle\frac{K.E.}{\omega}$$
$$\displaystyle\frac{L_1}{L_2}
=
\left(\displaystyle\frac{K.E_1}{\omega_1}\right)\times\displaystyle\frac{\omega_2}{K.E_2}
= 4 \Rightarrow L_2 = \displaystyle\frac{L}{4}$$

Physics

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