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

• A. $t_1 > t_2$
• B. $t_2 > t_1$
• C. $t_1 = t_2$
• D. $t_1 >> t_2$

Answer 1

Answer: (A)

$t_1 > t_2$

Answer 2

From equation of motion, we have $\, \, \, \, \, \, \, \, s = ut + \frac{1}{2}gt^2$ where u is initial velocity, t is time, g is acceleration due to gravity. At the time of dropping u = 0 $\therefore \, \, \, \, \, \, \, \, \, s = \frac{1}{2} gt^2_1 \Rightarrow t^2_1 = \frac{2s}{g}$ When lift moves up $\, \, \, \, \, \, \, \,$ g' = g + a $\therefore \, \, \, \, \, \, \, \, \, \, \, t^2_2 = \frac{2s}{g + a} \Rightarrow t^2_2 < t^2_1$ $ie, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, t_2 < t_1$