A particle of mass `m` is fixed to one end of a massless spring of spring constant `k` and natural length `l_(0)`. The system is rotated about the other end of the spring with an angular velocity `omega` ub gravity-free space. The final length of spring is

Options

A. `(k-momega^(2)l_(0))/(momega^(2))`
B. `(momega^(2)l_(0))/(k+momega^(2))`
C. `(momega^(2)l_(0))/(k-momega^(2))`
D. `(k+momega^(2)l_(0))/(momega^(2))`

Answer

C

Explanation

For circular motion
`Kx = momega^(2) (l_(0)+x)`
`rArr (k-momega^(2))x=momega^(2)l_(0)`
`rArr x = (momega^(2)l_(0))/(k-momega^(2))`