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Q.

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
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