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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 in the figure. The tension in the string is increased gradually and finally m moves in a circle of radius R0 /2. The final value of the kinetic energy is [CBSE AIPMT 2015]

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a

mv02

b

14mv02

c

2mv02

d

12mv02

answer is C.

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Detailed Solution

According to law of conservation of angular momentum,Li=Lf⇒mv0R0=mv′R02⇒v−′=2v0So, final kinetic energy of the particle,Kf=12mv2=12m2v02=4×12mv02=2mv02

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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 in the figure. The tension in the string is increased gradually and finally m moves in a circle of radius R0 /2. The final value of the kinetic energy is [CBSE AIPMT 2015]