A simple pendulum has a time period `T_(1)`. When its point of suspension is moved vertically upwards according as `y=kt^(2)`, where y is vertical covered and `k=1 ms^(-2)` , its time period becomes `T_(2)` then ` (T_(1)^(2))/(T_(2)^(2))` is (g`=10 ms^(-2))`

Options

A. `(5)/(6)`
B. `(11)/(10)`
C. `(6)/(5)`
D. `(5)/(4)`

Answer

C

Explanation

`T=T_(1)= 2pi sqrt((l)/(g))`
`:.T_(1)^(2)= 4pi^(2)((l)/(g))`
`y= kt^(2)`
`k= 1 ms^(-1)`
`:. Y= t^(2)`
`:. (y_(1))/(y_(2))=(T_(1)^(2))/(T_(2)^(2))=(6)/(5)`