Question

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6cm{s}^{-1}$. If they coalesce to form one big drop, what will be the terminal speed of bigger drop? (Neglect the buoyancy of the air)

• A. $1.5cm{s}^{-1}$
• B. $6cm{s}^{-1}$
• C. $24cm{s}^{-1}$
• D. $32cm{s}^{-1}$

Answer 1

Answer: (C)

Let now radius of big drop is $R$.
Then $\frac{4}{3}\pi R^3=\frac{4}{3}\times \pi r^3.8$
$R=2r$
where $r$ is radius of small drops.
Now, terminal velocity of drop in liquid.
$v_e=\frac{2}{9}\times \frac{r^2}{\eta }\rho -\sigma g$
where $r$ is coefficient of viscosity and $\rho$ is density of drop a is density of liquid.
Terminal speed drop is $6cm{s}^{-1}$
$\therefore 6=\frac{2}{9}\times \frac{r^2}{\eta }\rho -\sigma g$ .....(i)
Let terminal velocity becomes $v^{\prime }$ after coalesce, then
$v^{\prime }=\frac{2}{9}\frac{R^2}{\eta }\rho -\sigma g$ ....(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{6}{v^{\prime }}=\frac{\frac{2}{9}\frac{r^2}{\eta }\rho -\sigma g}{\frac{2}{9}\frac{R^2}{\eta }\rho -\sigma g}$
Or $\frac{6}{v^{\prime }}=\frac{r^2}{2r^2}$
Or $v^{\prime }=24cm{s}^{-1}$