A mass `m` moves in a circles on a smooth horizontal plane with velocity `v_(0)` at a radius `R_(0)`. The mass is attached to string which passes through a smooth hole in the plane as shown.
The tension in string is increased gradually and finally `m` moves in a circle of radius `(R_(0))/(2)`. the final value of the kinetic energy is

Options

A. `(a)mv_(0)^(2)`
B. `(b)(1)/(4) mv ._(0)^(2)`
C. `(c)2 mv ._(0)^(2)`
D. `(d)(1)/(2) mv ._(0)^(2)`

Answer

C

Explanation

According to law of conservation of angular momentum
`L_(i)=L_(f)rArr mv_(0)R_(0)=mv'((R_(0))/(2))rArrv'=2v_(0)`
So, final kinetic energy of the particle is
`K_(f)=(1)/(2)mv'^(2)=(1)/(2)m(2v_(0))^(2)=4xx(1)/(2)mv_(0)^(2)=2mv_(0)^(2)`.