Question

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 costant acceleration. Then
A. `t_(1)gtt_(2)`
B. `t_(2)gtt_(1)`
C. `t_(1)=t_(2)`
D. `t_(1)ge t_(2)`

Answer 1

Correct Answer - A
For stationary lift `t_(1) = sqrt((2h)/g)` and when the lift is moveing up with constant acceleration `t_(2)=sqrt((2h)/(g+a))`