Question

Two equal drops of water are falling through air with a steady velocity $v$. If the drops coalesced what will be the new velocity?

• A. $(2)^{\frac{1}{3}}v$
• B. $(2)^{\frac{3}{2}}v$
• C. $(2)^{\frac{2}{3}}v$
• D. $(2)^{\frac{1}{4}}v$

Answer 1

Answer: (C)

$(2)^{\frac{2}{3}}v$

Answer 2

Let r be the radius of the each drop. The terminal velocity of drop will be given by $v=\frac {2}{9}\frac{ r^2(\rho - \sigma)g}{ \eta}$ ....(i) where $\rho$ is density of drop and a is density of viscous medium of coefficient of viscosity $\eta$ . When two drops each of radius $r$ coalesce to form a new drop, then the radius of coalesced drop will be $R=(2)^{1/3}r$ Hence, new terminal velocity of coalesced drop will be $v'=\frac{2}{9}\left[\frac{(2^{1/3}r)^2(\rho - \sigma)g}{\eta}\right]$ ....(ii) From Eqs. (i) and (ii), we get $\frac{v'}{v}=(2)^{2/3}$ or $v'=(2)^{2/3} v$