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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Solution

The correct option is A >
For stationary lift $t_{1} = \sqrt{\frac{2 h}{g}}$
and when the lift is moving up with constant acceleration $t_{2} = \sqrt{\frac{2 h}{g + a}} ∴ t_{1} > t_{2}$

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A coin is dropped in a lift. It takes time t1 to reach the floor when lift is stationary. It takes time t2 when lift is moving up with constant acceleration. Then

The correct option is A > For stationary lift t1=√2hg and when the lift is moving up with constant acceleration t2=√2hg+a ∴ t1>t2

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

The correct option is A >
For stationary lift $t_{1} = \sqrt{\frac{2 h}{g}}$
and when the lift is moving up with constant acceleration $t_{2} = \sqrt{\frac{2 h}{g + a}} ∴ t_{1} > t_{2}$