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)

Time taken by coin to reach the floor is given by
$h =\frac{1}{2} g t^{2} (\because u=0)$
$\Rightarrow t =\sqrt{\frac{2 h}{g}}$
In stationary lift,
$t_{1}=\sqrt{\frac{2 h}{g}}$
In upward moving lift with constant acceleration $a$,
$g'=g +a$
$\therefore t_{2}=\sqrt{\frac{2 h}{g +a}}$
Clearly, $g'>g$
Thus, $t_{2} < t_{1}$