Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 ~cm / s coalesce to form a bigger drop. The terminal velocity of bigger drop is \_\_\_\_ cm / s .

Question

Quantrex Academy · Accepted Answer

Correct answer: 160

At terminal velocity, gravitational force equals Stokes drag:

mg = 6\pi\eta rv

For a sphere with mass

m = \frac{4}{3}\pi r^3 \rho

, this gives

v = \frac{2r^2\rho g}{9\eta}

Therefore,

v \propto r^2

When 64 drops coalesce:

64 \times \frac{4}{3}\pi r_1^3 = \frac{4}{3}\pi r_2^3

, so

r_2 = 4r_1

New terminal velocity:

\frac{v_2}{v_1} = \left(\frac{r_2}{r_1}\right)^2 = 16

v_2 = 16 \times 10 = 160

cm/s

Sixty four rain drops of radius 1 mm each falling down with a...

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