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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. ThenA. `t_(1) = t_(2)`B. `t_(1) gt t_(2)`C. `t_(2) gt t_(1)`D. `t_(1) gt gt t_(2)` |
Answer» Correct Answer - B (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)` |
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