A mass `m` moves in a circles on a smooth horizontal plane with velocity `v_(0)` at a radius `R_(0)`. The mass is atteched 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 cricle of radius `(R_(0))/(2)`. the final value of the kinetic energy is

Options

A. `mv_(0)^(2)`
B. `1/4mv_(0)^(2)`
C. `2mv_(0)^(2)`
D. `1/2mv_(0)^(2)`

Answer

c

Explanation

(c )Conserving angular momemtum
`L_(i)=L_(f)`
`implies mv_(0)R_(0)=mv'((R_0)/(2))`
`impliesv" "=2v_(0)`
So,final kinetic energy of the particle is
`K_(f)=(1)/(2)mv'^(2)=(1)/(2)m(2v_(0))^(2)`
`4(1)/(2)mv'^(2)=2mv_(0)^(2)`