Q 11-04-131JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
An object of mass $5\ \text{kg}$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10\ \text{N}$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to: [Use $g = 10\ \text{m s}^{-2}$]
Answer: (B) $\sqrt2 : \sqrt3$
Retarding acceleration from air resistance $= \dfrac{10}{5} = 2\ \text{m s}^{-2}$.
Going up: $a_u = g + 2 = 12\ \text{m s}^{-2}$. Coming down: $a_d = g - 2 = 8\ \text{m s}^{-2}$.
The same height $h$ is covered each way from/to rest: $h = \frac12 a t^2$, so $t\propto\dfrac1{\sqrt a}$.
$$\frac{t_u}{t_d} = \sqrt{\frac{a_d}{a_u}} = \sqrt{\frac{8}{12}} = \frac{\sqrt2}{\sqrt3}$$
Solution by Sreeraj P, M.Sc Physics