Question

The terminal velocity of a tiny droplet is $V$. $N$ number of such identical droplets combine together forming a bigger drop. Find the terminal velocity of the bigger drop.

Answer 1

Hint: In order to find the required solution of the given question we will apply the concept of terminal velocity. Firstly, we will equate the formula for terminal velocity of both smaller and bigger droplets, and further solve it to find the answer.

Formula used:
 $V=\dfrac{2}{9}{{r}^{2}}=\dfrac{\left( \rho -\sigma \right)g}{\eta }$
$V=\dfrac{4}{3}\pi {{r}^{3}}$

Complete step-by-step answer:
The maximum constant velocity acquired by a body while falling through a viscous fluid is called its terminal velocity.
Mathematically terminal velocity is given as,
$V=\dfrac{2}{9}{{r}^{2}}=\dfrac{\left( \rho -\sigma \right)g}{\eta }$
Where $\rho $ and $\sigma $ are mass densities of sphere and fluid respectively. We can infer from the equation that the terminal velocity depends on the square of the radius of the sphere and is inversely proportional to the viscosity of the medium. Viscosity is used to measure a medium’s internal resistance against the flow.
$\Rightarrow \dfrac{V}{{{r}^{2}}}=\dfrac{2}{9}\left[ \dfrac{\left( \rho -\sigma \right)g}{\eta } \right]$
According to the question terminal velocity for the smaller drop is given by,
$\Rightarrow \dfrac{V}{{{r}^{2}}}=\dfrac{2g}{9\eta }\left( \rho -\sigma \right)$ -- (1)
Similarly, terminal velocity for the bigger drop is given by,
$\Rightarrow \dfrac{{{V}^{'}}}{{{R}^{2}}}=\dfrac{2g}{9\eta }\left( \rho -\sigma \right)$ -- (2)
Dividing equation (1) by (2),
$\Rightarrow \dfrac{V}{{{V}^{'}}}=\dfrac{{{r}^{2}}}{{{R}^{2}}}$
$\Rightarrow \dfrac{V}{{{V}^{'}}}=V{{\left( \dfrac{R}{r} \right)}^{2}}$ -- (3)
If $N$ number of such identical droplets combine together forming bigger drop, then
Volume of one big drop = Volume of $N$droplets
Since, volume of the sphere is given by,
 $V=\dfrac{4}{3}\pi {{r}^{3}}$
$\dfrac{4}{3}\pi {{R}^{3}}=N\left( \dfrac{4}{3}\pi {{r}^{3}} \right)$
$\Rightarrow {{R}^{3}}=N\left( {{r}^{3}} \right)$
$\Rightarrow {{R}^{3}}={{N}^{1/3}}\left( r \right)$
Thus, terminal velocity of bigger drop can be given by,
Terminal velocity of bigger drop \[={{\left( \dfrac{R}{r} \right)}^{2}}\times V\]
From equation (1)
$={{\left( \dfrac{{{N}^{1/3}}}{\dfrac{r}{3}} \right)}^{2}}\times V={{\left( {{N}^{1/3}} \right)}^{2}}\times V$
$={{N}^{2/3}}\times V$ from equation (2)
Hence, the terminal velocity of the bigger drop is ${{N}^{2/3}}V$.

Note: Stokes law is used to calculate the viscosity of the fluid. It is a mathematical equation that expresses the settling velocities of small spherical particles in a fluid medium. Mathematically, it is represented as,
$F\propto {{\eta }^{a}}{{r}^{b}}{{v}^{c}}$
$\Rightarrow F=k{{\eta }^{a}}{{r}^{b}}{{v}^{c}}$ ; here $k$ is the constant of proportionality.