Two drops of same liquid of same radius are falling through air with steady velocity of $2\;m/s$. If the two drops coalesce what would be the terminal velocity?

Given,
Terminal velocity of the each small raindrop $(v) =  2 \; m/s$
Terminal velocity of the big raindrop $(V) = ?$

Let $r$ be the radius of the small rain droplets. Then the terminal velocity of each small raindrop is,

$v = \frac{2\,r^2 (\rho \; - \; \sigma)\,g}{9\; \eta}$ .......... (i)

Let $R$ be the radius of the combined drop. Then the terminal velocity of the large drop is,

$V = \frac{2\,R^2 (\rho \; - \; \sigma)\,g}{9\; \eta}$ .......... (ii)

Since the volume of big raindrop = volume of eight small raindrops. Then

$\frac{4}{3}\; \pi\; R^3 = 2 *\;\frac{4}{3}\; \pi \; r^3$

⇒ $R = 2^{1/3}\;r$ .......... (iii)

Now,
Dividing equation (ii) by (i), we get

$\frac{V}{v} = \frac{R^2}{r^2}$ .......... (iv)

Using equation (iii) in equation (iv), then we get

or, $V = v * \frac{R^2}{r^2} = v * \frac{(2^{1/3} \; r)^2}{r^2} = 2 * 1.58 = 3.17 \;m/s$ 

$\therefore$ Terminal velocity of the big raindrop $(V) = 3.17\;m/s$

Return to Main Menu