Question 232: Properties of Matter

Question

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

Accepted Answer

When a spherical drop falls through a viscous medium (like air), it reaches a constant speed called terminal velocity $\left(\mathrm{v}_{\mathrm{t}}ight)$ when the net force on it is zero.
The forces involved are :
Weight ( W ) is acting downwards, $\mathrm{W}=\mathrm{mg}=$ Volume $	imes ho 	imes \mathrm{g}$.
Buoyant force ( $\mathrm{F}_{\mathrm{b}}$ ) is acting upwards.
Viscous drag force ( $\mathrm{F}_{\mathrm{v}}$ ) is acting upwards
According to Stokes' Law $\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{r}$ v.
At equilibrium, the net force must be zero.
$$ \mathrm{W}-\mathrm{F}_{\mathrm{b}}=\mathrm{F}_{\mathrm{v}} $$
This leads to the standard formula for terminal velocity :
$$ v_t=\frac{2 r^2 g(ho-\sigma)}{9 \eta} $$
Where :
$\mathrm{r}=$ Radius of the drop
$ho=$ Density of the drop (water)
$\sigma=$ Density of the medium (air)
$\eta=$ Coefficient of viscosity
$\mathrm{g}=$ Acceleration due to gravity
Since $\mathrm{g}, ho, \sigma$, and $\eta$ are constants for both the small and big drops, the terminal velocity is directly proportional to the square of the radius.
$$ v_t \propto r^2 $$
When 64 small drops combine to form one big drop, the total volume remains conserved.
Let r be the radius of a small drop.
Let R be the radius of the big drop.
Let $\mathrm{n}=64$ be the number of drops.
Volume of Big Drop $=\mathrm{n} 	imes$ (Volume of Small Drop)
$$\Rightarrow $$ $\frac{4}{3} \pi \mathrm{R}^3=\mathrm{n} 	imes \frac{4}{3} \pi \mathrm{r}^3$
$$\Rightarrow $$ $\mathrm{R}^3=\mathrm{n} \cdot \mathrm{r}^3$
$$\Rightarrow $$ $$ \mathrm{R}=\mathrm{n}^{1 / 3} \cdot \mathrm{r} $$
Substituting $\mathrm{n}=64$ :
$$ \mathrm{R}=64^{1 / 3} \cdot \mathrm{r} $$
$$\Rightarrow $$ $$ \mathrm{R}=4 \mathrm{r} $$
So, the ratio of the terminal velocities for big and small drop using the fact that terminal velocity is proportional to square of radius,
$$ \frac{\mathrm{v}_{	ext {big }}}{\mathrm{v}_{	ext {small }}}=\frac{\mathrm{R}^2}{\mathrm{r}^2}=\left(\frac{\mathrm{R}}{\mathrm{r}}ight)^2=\left(\frac{4 \mathrm{r}}{4}ight)^2=16 $$
Given $\mathrm{v}_{	ext {small }}=10 \mathrm{~cm} / \mathrm{s}$.
$$ v_{	ext {big }}=16 	imes v_{	ext {small }} $$
$$\Rightarrow $$ $v_{\mathrm{big}}=16 	imes 10 \mathrm{~cm} / \mathrm{s}$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{big}}=160 \mathrm{~cm} / \mathrm{s}$

Properties of Matter

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

Integer / Numeric Answer Type