Question

Two drops of the same radius are falling through the air with a terminal velocity of $10m/s$. If these two drops coalesce, then the terminal velocity of the new drop will be?

Answer 1

Step 1: Given that:
Two drops having the same radius ($r$), are falling through the air with a terminal velocity $v_1=10m/s$.

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

Let the terminal velocity for smaller and bigger drops are $v_1$ and $v_2$ respectively.
Step 2: Formula used:
Let $R$ is the radius of the bigger drop, as the volume remains the same, so we can write the volume of the big drop is equal to the volume of two small drops.

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

So,

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

Now the terminal velocity($v$) of a drop is directly proportional to the square of the radius($r$), so

$\begin{array}{rcl}\frac{v_1}{v_2}& =& \frac{r^2}{R^2}\\ & =& \frac{r^2}{2^{{\displaystyle \frac{2}{3}}}{r}^2}\\ & =& \frac{1}{2^{{\displaystyle \frac{2}{3}}}}\\ & & \end{array}$

So,
$\begin{array}{rcl}{v}_2& =& 2^{\frac{2}{3}}{v}_1\\ & =& 4^{\frac{1}{3}}\times 10\\ & =& 10\times {\left(4\right)}^{\frac{1}{3}}\\ & =& 15.87m/s\end{array}$

Hence, the terminal velocity of the new drop will be$15.87m/s$.