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Question

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 m moves in a
circle of radius R0/2. Then, the final value of the kinetic energy of particle is

A
12mv02
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B
mv02
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C
14mv02
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D
2mv02
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Solution

The correct option is D 2mv02
Since no external torque is acting on the system, thus according to law of conservation of angular momentum,
Initial angular momentum = Final angular momentum
mv0R0=mvrmv0R0=mvR02 (New radius, r=R02)v=2v0New kinetic energy = 12mv2=2mv20

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