The time period of a mass loaded spring is 'T'. If `20%` of its turns are cut and removed and same mass is attached to the remaining spring then it oscillates with a period of

Options

A. `(5T)/(4)`
B. `(sqrt(5)T)/(2)`
C. `(4T)/(5)`
D. `(2T)/(sqrt(5))`

Answer

D

Explanation

For a spring, `k xx l` = constant
`k xx 100 = k^(1) xx (100 - 200) rArr k^(1) = (5)/(4) k`
`(T^(1))/(T) = sqrt((k)/(k_(1))) = sqrt((4)/(5)) rArr T^(1) = (2)/(sqrt(5))T`