Question 8: Properties of Matter

Question

A spherical liquid drop of radius $R$ acquires the terminal velocity $v_1$ when falls through a gas of viscosity $\eta$. Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity $v_2$ falling through the same gas. The ratio of terminal velocities $v_1 / v_2$ is $\_\_\_\_$ .

Accepted Answer

When a spherical object falls through a viscous fluid, it eventually reaches a constant speed called terminal velocity $\left(\mathrm{v}_{\mathrm{t}}ight)$. This occurs when the downward force of gravity is balanced by the upward buoyant force and the viscous drag force.
The force of gravity acting on the sphere is $F_g=m g=\frac{4}{3} \pi r^3 ho g$, where $ho$ is the density of the liquid.
Using Archimedean's principle of buoyant force $\mathrm{F}_{\mathrm{b}}=\frac{4}{3} \pi \mathrm{r}^3 \sigma \mathrm{~g}$, where $\sigma$ is the density of the gas.
Using Stokes' Law the viscous drag is $\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{r} \mathrm{v}_{\mathrm{t}}$
At terminal velocity,
$$ \mathrm{F}_{\mathrm{v}}=\mathrm{F}_{\mathrm{g}}-\mathrm{F}_{\mathrm{b}} $$
$$\Rightarrow $$ $6 \pi \eta r v_t=\frac{4}{3} \pi r^3(ho-\sigma) g$
$$\Rightarrow $$ $\mathrm{v}_{\mathrm{t}}=\frac{2 \mathrm{r}^2(ho-\sigma) \mathrm{g}}{9 \eta}$
For the same liquid and gas, terminal velocity is proportional to the square of the radius :
$$ \mathrm{v} \propto \mathrm{r}^2 $$
The large drop of radius $R$ is broken into 64 identical droplets of radius $r$. Since the total volume remains constant :
$$ V_R=64 	imes V_r $$
$$\Rightarrow $$ $\frac{4}{3} \pi R^3=64 	imes \frac{4}{3} \pi r^3$
$$\Rightarrow $$ $\mathrm{R}^3=64 \mathrm{r}^3$
$$\Rightarrow $$ $$ R=4 r \Rightarrow r=\frac{R}{4} $$
Let $\mathrm{v}_1$ be the terminal velocity of the large drop and $\mathrm{v}_2$ be the terminal velocity of a small droplet. Using the relationship $\mathrm{v} \propto \mathrm{r}^2$ :
$$ \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{R}^2}{\mathrm{r}^2} $$
Substituting $\mathrm{r}=\frac{\mathrm{R}}{4}$ :
$$ \frac{\mathrm{v}_1}{\mathrm{v}_2}=\frac{\mathrm{R}^2}{\left(\frac{\mathrm{R}}{4}ight)^2} $$
$$\Rightarrow $$ $\frac{v_1}{v_2}=\frac{R^2}{R^2 / 16}$
$$\Rightarrow $$ $$ \frac{v_1}{v_2}=16 $$
Therefore, the ratio of terminal velocities $\frac{v_1}{v_2}$ is 16 .
Hence, the correct option is (D).

Answer

4

Answer

0.25

Answer

32

Answer

16

Properties of Matter

A spherical liquid drop of radius R acquires the terminal velocity v_1 when falls through a gas of viscosity \eta. Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity v_2 falling through the same gas. The ratio of terminal velocities v_1 / v_2 is \_\_\_\_ .

Choose the correct answer: