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)

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 $