Question

Six small raindrops each of radius 1.5 mm, come down with a terminal velocity of $6cm{s}^{–1}.$ They coalesce to form a bigger drop. What is the terminal velocity (nearest integer in $m{s}^{–1})$ of the bigger drop?

Answer 1

Radius of each small drop r = 1.5 mm
= 0.15 cm
Terminal velocity for small drops $v_t=6cm{s}^{-1}$
Let the density of water $=\rho$
and the density of air $={\rho }^{\prime }$
Then $v_t=\frac{2r^2}{9\eta }(\rho -{\rho }^{\prime })g$
$6cm{s}^{-1}=\frac{2\times (0.15{)}^2}{9\eta }(\rho -{\rho }^{\prime })g$ ...(i)
When the six drops combine, let the radius of the bigger drop be R.
Then volume of bigger drop = 6 × (volume of a small drop)
$\Rightarrow \frac{4}{3}\pi R^3=6\times \frac{4}{3}\pi r^3$
$\Rightarrow R^3=6r^3$
$\Rightarrow R=(6{)}^{\frac{1}{3}}r=(6{)}^{\frac{1}{3}}\left(0.15\right)cm$
Let the terminal velocity for this drop be $V_T$
Then $V_T=\frac{2\times 6^{\frac{2}{3}}(0.15{)}^2}{9\eta }(\rho -{\rho }^{\prime })g$ ...(ii)
Dividing equation (ii) by equation (i),
We get $\frac{V_T}{6}=(6{)}^{\frac{2}{3}}$
$V_T=6^{\frac{5}{3}}m{s}^{-1}$
$V_T=19.81m{s}^{-1}$