Q 11-09-019NEETJEE MainMedium
Two identical rain drops fall through air at terminal speed $v$. They merge into one bigger drop. The terminal speed of the bigger drop is
Answer: (B) $2^{2/3}v$
Volume doubles, so the radius becomes $R = 2^{1/3}r$.
Terminal speed $\propto r^2$ (Stokes's law), so $v' = (2^{1/3})^2v = 2^{2/3}v$.
Solution by Sreeraj P, M.Sc Physics