Laws Of Motion Question 176

Question: 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

Options:

A) $ t _{1}>t _{2} $

B) $ t _{2}>t _{1} $

C) $ t _{1}=t _{2} $

D) $ t _{1}»t _{2} $

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Answer:

Correct Answer: A

Solution:

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

$ \therefore \ \ t _{1}>t _{2} $



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