Eight spherical rain drops of the same mass and radius are falling down with a terminal speed $5\;cm/s$. If the coalesce to form one big drop, what will be its terminal speed?

Given,
Terminal velocity of the each small raindrop $(v) = 5\;cm/s = 0.05 \; m/s$
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

$\frac{4}{3}\; \pi\; R^3 = 8 *\;\frac{4}{3}\; \pi \; r^3$

⇒ $R = 2\;r$ .......... (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, $V = v * \frac{R^2}{r^2} = v * \frac{(2r)^2}{r^2} = 0.05 * 4 = 0.2 \;m/s$ 

$\therefore$ Terminal velocity of the big raindrop $(V) = 0.2\;m/s$

Return to Main Menu