Three spherical raindrops of equal size are falling vertically through air with a terminal velocity of $0.15\;m/s$. What would be the terminal velocity if these three drops were to coalesce to form a large spherical drop?

Given,
Terminal velocity of the each small raindrop $(v) =  0.15 \; m/s$
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

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

⇒ $R = 3^{1/3}\;r$ .......... (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, $V = v * \frac{R^2}{r^2} = v * \frac{(3^{1/3} \; r)^2}{r^2} = 0.15 * 2.08 = 0.312 \;m/s$ 

$\therefore$ Terminal velocity of the big raindrop $(V) = 0.312\;m/s$

Return to Main Menu