Question

Eight raindrops each of radius R fall through air terminal velocity 6 cm / s square .what is the terminal velocity of the bigger drop formed by coalescing these drops together ???????????????

Answer 1

Answer on Question# 36960 – Physics - Mechanics

Eight raindrops each of radius R fall through air terminal velocity 6 cm / s. what is the terminal velocity of the bigger drop formed by coalescing these drops together?

Solution

Let $r$ be the radius of the bigger drop formed. The volume of the big drop is equal to the total volume of the 8 small drops, that is,

$$\frac{4}{3} \pi r^3 = 8 \cdot \frac{4}{3} \pi R^3 \rightarrow r = 2R.$$

By the Stoke’s law, the terminal velocity of the drop of radius $R$ is

$$v_R = \frac{2R^2 (\rho - \rho_0) g}{9 \eta}.$$

The terminal velocity of the drop of radius $r = 2R$ is

$$v_r = \frac{2(2R)^2 (\rho - \rho_0) g}{9 \eta} = 4 \cdot \frac{2R^2 (\rho - \rho_0) g}{9 \eta} = 4 * v_R = 4 * 6 \frac{\mathrm{cm}}{\mathrm{s}} = 24 \frac{\mathrm{cm}}{\mathrm{s}}.$$

Answer: $24 \frac{\mathrm{cm}}{\mathrm{s}}$.