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

Options

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

Answer

A

Explanation

Time taken by the coin to reach the floor in the stationary lift is
`h = (1)/(2) g t _(1)^(2) u _(1) = 0`
`t_(1)^(2) = (2h)/(g) :. t_(1) = sqrt((2h)/(g)) " ". . . (1)`
When the lift is moving up with constant acceleration a ,
g' = g +a
`h = (1)/(2) (g +a) t_(2)^(2)`
`t_(2)^(2) = (2h)/(g +a) :. t_(2) = sqrt((2h)/(g +a))`
`:. t_(2) lt t_(1)`