An object dropped in a stationary lift takes time t 1 to reach the floor. It takes time t 2 when lift is…

When the object is dropped in a stationary lift from height $h,$

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ -h=\left(0\right)t_1+\frac{1}{2}\left(-g\right){t_1}^2 \\ {t}_1=\sqrt{\frac{2h}{g}} \end{gathered}$$

When the object is dropped from a lift which is moving up with an acceleration $a,$then a pseudo force will act on the object since the object is in a non-inertial frame of reference, therefore net acceleration due to gravity and pseudo force on the object will be, $-\left(g+a\right)$in the downward direction.

$$\begin{gathered} \Rightarrow s=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow -h=\left(0\right)t+\frac{1}{2}\left[-\left(g+a\right)\right]{t_2}^2 \\ \Rightarrow t_2=\sqrt{\frac{2h}{g+a}} \end{gathered}$$

Thus, it is clear that $t_1>t_2$