A simple pendulum has time period `T_1`. The point of suspension is now moved upward according to the relation `y = Kt^2`. (`K= 1 m s^(-2)`) where y is the vertical displacement. The time period now becomes `T_2`
The ratio of `T_1^2/T_2^2`(Take `g = 10 m s^(-2)`)

Options

A. `(5)/(6)`
B. `(6)/(5)`
C. `1`
D. `(4)/(5)`

Answer

B

Explanation

`T_(1) = 2pi sqrt((l)/(g)) = 2pi sqrt((l)/(10))`
upward accelertion `(d^(2)y)/(dt^(2)) = 2k = 2 xx 1 = 2 m//s^(2)`
`:.` Acceleration `w.r.t.` point of suspension `= 12 m//s^(2)`
`T_(2) = 2pisqrt((l)/(12)) :. (T_(1))/(T_(2)) = sqrt((12)/(10)) :. ((T_(1))/(T_(2)))^(2) = (6)/(5)`
`T_(1) = 2pi sqrt((l)/(g)) = 2pi sqrt((l)/(10))`