Q: Two identical drops of water are falling through air with a steady speed of ‘v’ each. If the drops coalesce to from a single drop, what is the new terminal velocity ?

Sol: From conservation of mass.;

$\large \frac{4}{3} \pi R^3 \rho = 2 \frac{4}{3} \pi r^3 \rho$

$\large R = 2^{1/3} r $

According to Stoke’s Law

$\large v_t \propto r^2 $

$\large \frac{v’}{v} = \frac{R^2}{r^2} = 2^{2/3}$

$\large v’ = 2^{2/3} v $