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Q.

A mass m moves in a circle on a smooth horizontal plane with velocity v₀ at a radius R₀. 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 it moves in a circle of radius R₀/2. The final value of the kinetic energy is

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a

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

b

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

c

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

d

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

answer is D.

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Detailed Solution

According to law of conservation of angular momentum L_i = L_f

${mv}_{0} R_{0} = mv (\frac{R_{0}}{2}) \Rightarrow v^{'} = 2 v_{0}$

Final K.E of the particle is

$K_{f} = \frac{1}{2} {mv}^{2} = \frac{1}{2} m {(2 v_{0})}^{2} = 2 {mv}_{0}^{2}$

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A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius R0. 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 it moves in a circle of radius R0/2. The final value of the kinetic energy is