A simple pendulum attached to the ceiling of a stationary lift has a time period T. The distance y covered by the lift moving upwards varies with time t as `y = t^(2)` where y is in metres and t in seconds. If `g = 10 m//s^(2)`, the time period of pendulum will be

Options

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

Answer

B

Explanation

Distance covered by lift is given by `y=t^(2)`
`:.` Acceleration of lift upwards
`=(d^(2)y)/(dt^(2))=(d)/(dt)(2t)=2m//s^(2)=(g)/(5)`
`T.=2pi sqrt((l)/(g+(g)/(5))=2pi sqrt((l)/((6)/(5)g))=sqrt((5)/(6))T`.