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A coin is dropped in a lift. It takes ti...

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)=t_(2)`

B

`t_(1)gtt_(2)`

C

`t_(1)ltt_(2)`

D

`t_(1) lt lt t_(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
When the lift is at rest, `s= (1)/(2) g t_(1)^(2)" " therefore t_(1)^(2)= (2s)/(g)`
But when the lift moves up with an acceleration a, the resultant acceleration `g'=g+a`
and `t_(2)^(2)= (2s)/((g+a))`
`therefore t_(2)^(2) lt t_(1)^(2) " " therefore t_(2) lt t_(1) or t_(1) gt t_(2)`
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