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} $
Show Answer
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} $
