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,

t_{1} \geq t_{2}

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t_{2} > t_{1}

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t_{1} = t_{2}

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t_{2} >> t_{1}

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Solution

The correct option is D $t_{1} \geq t_{2}$
When lift is stationary the entire distance will be covered by the $c o i n$ .
But when it is in motion the distance is covered by $c o i n + l i f t$ .
So the time taken will be less than the initial time.
Therefore, $t_{1} \geq t_{2}$

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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,

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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,

The correct option is D $t_{1} \geq t_{2}$
When lift is stationary the entire distance will be covered by the $c o i n$ .
But when it is in motion the distance is covered by $c o i n + l i f t$ .
So the time taken will be less than the initial time.
Therefore, $t_{1} \geq t_{2}$