Sixty four rain drops of radius 1 mm each falling down w-ith a terminal velocity of 10 cm/s coalesce to fonn a bigger drop. The terminal velocity of bigger drop is_____cm/s

We need to find the terminal velocity of a bigger drop formed by the coalescence of 64 small raindrops.

Number of small drops: $$n = 64$$

Radius of each small drop: $$r = 1$$ mm

Terminal velocity of each small drop: $$v = 10$$ cm/s

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

$$R^3 = n \times r^3 = 64 r^3$$

$$R = 4r$$

Terminal velocity is proportional to the square of the radius:

$$v_t \propto r^2$$

(This comes from Stokes' law: $$v_t = \frac{2r^2(\rho - \sigma)g}{9\eta}$$)

$$\frac{V}{v} = \frac{R^2}{r^2} = \frac{(4r)^2}{r^2} = 16$$

$$V = 16 \times v = 16 \times 10 = 160 \text{ cm/s}$$

Therefore, the terminal velocity of the bigger drop is 160 cm/s.