Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of 6cms−1. If they coalesce to form one big drop, what will be its terminal speed?

Question:
Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of 6cms⁻¹. If they coalesce to form one big drop, what will be its terminal speed?

Solution:

When eight equal drops coalesce, volume of water remains same and a bigger drop is formed. Lets say r is radius of drop before these combine and R is radius after these combine.
We have

\[8 \times \frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3}\]

\[ \to r = 2R\]

and we know that terminal velocity V_T​∝ radius², so we have

\[\frac{{{V_T}'}}{{{V_T}}} = \frac{{{R^2}}}{{{r^2}}}\]

\[ \to {V_T}' = {\left( {\frac{R}{r}} \right)^2} \times {V_T} = {(2)^2} \times 6 = 24c{m^{ - 1}}\]

Thus, the terminal speed is 24cms^-1