A coin is dropped in a lift. It takes time `t_(1)` to reach the floor when lift is stationary. It takes time `t_(2)` when lift is moving up with constant acceleration. Then

Options

A. `t_(1) = t_(2)`
B. `t_(1) gt t_(2)`
C. `t_(2) gt t_(1)`
D. `t_(1) gt gt t_(2)`

Answer

B

Explanation

(b) Time `t_(1)` for stationary lift `=sqrt((2h)/(g))`
When lift is moving up with constant acceleration , then `t_(2)=sqrt((2h)/(g+a))`[Relative acceleration =(g+a)]
`therefore t_(1)gt t_(2)`