Q 11-09-090JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
Eight equal drops of water are falling through air with a steady speed of $10\ \text{cm s}^{-1}$. If the drops coalesce, the new velocity is
Answer: (B) $40\ \text{cm s}^{-1}$
$R^3=8r^3\Rightarrow R=2r$. Terminal speed $\propto r^2$, so $v=4\times10=40\ \text{cm s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics