Q86 / 180 · NEET 2019 (Odisha) · Physics
A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time $t_1$ if the elevator is at rest and in time $t_2$ if the elevator is moving uniformly. The which of the following option is correct?
- A$t_1 < t_2$ or $t_1>t_2$ depending upon whether the lift is going up or down
- B$t_1 < t_2$
- C$t_1>t_2$
- Correct answerD$t_1=t_2$
Solution
Let $\mathrm{h}$ be the height through which the coin is dropped. Then, according to the equation of motion, it is given as
$$
\begin{aligned}
\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \quad[\because \mathrm{u}=0] \\
\Rightarrow \quad \mathrm{t} \propto \frac{1}{\sqrt{\mathrm{g}}}
\end{aligned}
$$
As the elevator is moving uniformly i.e. its velocity is constant, so the acceleration is zero.
$\therefore$ Relative acceleration of the lift when it is either moving upward or downward is given as, $\mathrm{g}^{\prime}=\mathrm{g} \pm \mathrm{a}=\mathrm{g} \pm 0=\mathrm{g}$
Hence, the time for the coin to reach the floor will remains same i.e. $t_1=t_2$.
Hence, the correct answer is option: 'D'.