Question

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

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

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 $
$R = 2 r$
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 $6 \, cm \, s^{-1}$
$ \therefore \, 6 =\frac{2}{9} \times \frac{r^2}{\eta} \rho - \sigma \, g $ .....(i)
Let terminal velocity becomes $v'$ after coalesce, then
$v' = \frac{2}{9} \frac{R^2}{\eta} \rho - \sigma \, \, g$ ....(ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{6}{v'} = \frac{\frac{2}{9} \frac{r^{2}}{\eta} \rho-\sigma g}{\frac{2}{9} \frac{R^{2}}{\eta} \rho -\sigma g}$
Or $\frac{6}{v'} = \frac{r^{2}}{2r^{2}} $
Or $v' = 24 \, cm \, s^{-1}$