A particle of mass m is tied to one end of a light string and the other end of the string passes through a small hole at the centre of the table as shown. The particle is rotating along a circle of radius R on the smooth table with speed V0. Now the free end of the string is slowly pulled downward by a distance R/2. Then the final tension in the string is
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a
mv022R
b
2mv02R
c
4mv02R
d
8mv02R
answer is D.
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Detailed Solution
No external torque is acting on the particle about the axis of rotation.Conserving angular momentum, mv.R2=mv0R⇒v=2v0∴Final tension, T=m.v2R/2=m.(2v0)2R/2⇒T=8mv02R