A mass m moves in a circle on a smooth horizontal plane with velocity $v_{0}$ at a radius $R_{0}$ . The mass is attached to a string which passes
through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius $\frac{R_{0}}{2}$
The final value of the kinetic energy is

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$\frac{1}{2}$ ${mv}_{0}^{2}$

${mv}^{2}_{0}$

2 ${mv}_{0}^{2}$

$\frac{1}{4}$ ${mv}_{0}^{2}$

answer is C.

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Detailed Solution

Initial angular momentum $L_{initial} = {mv}_{0} R_{0}$

Initial angular momentum $L_{final} = mv \frac{R_{0}}{2}$

Conservation of angular momentum

$m_{0} v_{0} R_{0} = mv \frac{R_{0}}{2} \Rightarrow v = 2 v_{0}$

$KE = \frac{1}{2} {mv}^{2} = 2 {mv}_{0}^{2}$

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