Correct Answer - `1cm` .
Here, `k = 100 N//m, l_(0) = 0.5 m, m =0.5kg`
`omega = 2 rad//s`
Let elongation of the spring be l
`F = kl` which provides the necessary centripetal
force `m omega^(2) r = m omega^(2) (l + l_(0))`
`:. Kl = m omega^(2) (l + l_(0))`
`l = (m omega^(2) l_(0))/( k - m omega^(2)) = (0.5 xx 2^(2) xx 0.5)/(100 - 0.5 xx 4) = (1)/(98)m = 1cm` .