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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 constant acceleration. Then
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The correct answer is:
$t_1\gtt_2$
For stationary lift $\left.t_1=\sqrt{\frac{2 h}{g}} \right\rvert\,$ and when the lift is moving up with constant acceleration $t_2=\sqrt{\frac{2 h}{g+a}} \quad \therefore t_1\gtt_2$
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