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

t₁> t₂

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t₂ > t₁

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t₁ = t₂

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t₁ >> t₂

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Solution

The correct option is A
t₁> t₂

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

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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 t1 > t2 For stationary lift t1=√2hg and when the lift is moving up with constant acceleration t1=√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
t₁> t₂

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