question_answer

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

A) t1 > t2                                    

B) t2 > t1

C) t1 = t2                                    

D) t1 >> t2

Correct Answer: A

Solution :

Key Idea: When lift is moving up, resultant acceleration increases. From equation of motion, we have\[s=ut+\frac{1}{2}g{{t}^{2}}\]where u is initial velocity, t is time and g is acceleration due to gravity. At the time of dropping u = 0. \[\because \]         \[s=\frac{1}{2}gt_{1}^{2}\] \[\Rightarrow \]      \[t_{1}^{2}=\frac{2s}{g}\] When lift moves up               \[g'=g+a\] \[\therefore \]          \[t_{2}^{2}=\frac{2s}{g+a}\] \[\Rightarrow \]        \[t_{2}^{2}<t_{1}^{2}\]         i.e., \[{{t}_{2}}<{{t}_{1}}\]or \[{{t}_{1}}<{{t}_{2}}\]