Question

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

Answer 1

Step 1: Given parameters

Consider two equal drops of water, having the same radius $r$, falling through the air with a steady velocity $v$.

Now when they coalesce, a bigger drop is formed but the volume of water remains the same.

Step 2: Formula used

Let $R$ is the radius of the bigger drop, as the volume remains the same, so we can write

$2\times \frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}{\mathrm{\pi R}}^3$

$\Rightarrow R=2^{\frac{1}{3}}r......(1)$.

And we know that the terminal velocity is directly propotional to the square of the radius, i.e.,

$V_T\propto radiu{s}^2$, so we have,

$\frac{{V^{\text{'}}}_T}{V_T}=\frac{R^2}{r^2}\dots \dots \left(2\right)$

where ${V^{\text{'}}}_T$= terminal velocity of the bigger drop and $V_T$= terminal velocity of the smaller drop.

Step 3: Calculate the new velocity of the coalesced drop

Now substituting equation (1) in equation (2), we get,

$\begin{array}{rcl}\frac{{V^{\text{'}}}_T}{V_T}& =& \frac{{\left(2^{{\displaystyle \frac{1}{3}}}r\right)}^2}{r^2}\\ {V^{\text{'}}}_T& =& 2^{\frac{2}{3}}{V}_T\mathrm{m}/\mathrm{s}\\ {{\mathrm{V}}^{\text{'}}}_{\mathrm{T}}& =& 2^{\frac{2}{3}}v\left(\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{s}$}\right.\right)\end{array}$

So the new velocity of the bigger drop is $2^{\frac{2}{3}}$times the steady velocity $v$.