question_answer

A coin is dropped in a lift. If 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:

A) \[{{t}_{1}}>{{t}_{2}}\]                    

B) \[{{t}_{2}}>{{t}_{1}}\]

C) \[{{t}_{1}}={{t}_{2}}\]                    

D) \[{{t}_{1}}>>{{t}_{2}}\]

Correct Answer: A

Solution :

From equation of motion, we have                 \[S=ut+\frac{1}{2}g{{t}^{2}}\] where \[u\] is initial velocity, \[t\] is time, \[g\] is acceleration due to gravity. At the time of dropping\[u=0\]. \[\because \]     \[S=\frac{1}{2}gt_{1}^{2}\] \[\Rightarrow \]               \[t_{1}^{2}=\frac{2\,\,s}{g}\] When lift moves up                 \[g=g+a\] \[\therefore \]  \[t_{2}^{2}=\frac{2\,\,s}{g+a}\] \[\Rightarrow \]               \[t_{2}^{2}<t_{1}^{2}\] \[i.e.,\] \[{{t}_{2}}<{{t}_{1}}\]                     or            \[{{t}_{1}}>{{t}_{2}}\]