Question

Eight drops of water each of radius $2\,mm$ are falling through air at a terminal velocity of $8\,cm\,{s^{ - 1}}$. If they coalesce to form a single drop, the terminal velocity of the combined drop will be?

Answer 1

In this question, we shall first calculate the volume of eight drops when they are falling individually and then the volume of the bigger drop which is formed when the eight smaller drops coalesce. On equating the two volumes, we will obtain a relation between the radii of the two drops. Further, we shall derive a relation between the terminal velocity and the radii of the drop and make proper substitutions to get the answer.

Complete step by step answer:
For the body to achieve its terminal or critical velocity ${v_c}$, the viscous force on the object must be equal to the net weight of the body in the fluid.This fluid can be either a gas or a liquid since all fluids cause a hindrance to the motion of an object, better known as drag.

The terminal velocity is mathematically expressed as ${v_c} = \dfrac{2}{9}\dfrac{{({\rho _o} - {\rho _f})g{r^2}}}{\mu }$
where ${\rho _o}$ is the density of the object, ${\rho _f}$ is the density of the fluid, g is the acceleration due to gravity, r is the radius of the object and $\mu$ is the dynamic viscosity of the fluid.
From the above relation we can say that ${v_c} \propto {r^2}$.When the eight drops are falling individually, volume of one drop is given as,
$V = \dfrac{{4\pi {r^3}}}{3}$
where $r$ is the radius of the individual drop.

Volume of eight drops becomes ${V_i} = 8 \times \dfrac{{4\pi {r^3}}}{3}$
where ${V_i}$ represents the total volume of the eight drops when they are falling individually.
Now, when the drops coalesce together to form a bigger drop, the radius of the bigger drop is given as ${r^1}$. The volume of the bigger drop is given by,
${V_f} = \dfrac{{4\pi {r^1}^3}}{3}$
Now since the volume remains conserved throughout, we can say that ${V_i} = {V_f}$
Substituting proper values we have,
$\dfrac{{4\pi {r^1}^3}}{3} = 8 \times \dfrac{{4\pi {r^3}}}{3}$
Cancelling out the common terms we have,
${r^{'3}} = 8{r^3}$
$\Rightarrow {r^1} = 2r$

The radius $r$ of the individual smaller drop is given to be $2\,mm$ which is equal to $0.2\,cm$. Hence, the radius of the bigger drop is $0.4\,mm$. We know that,
${v_c} \propto {r^2}$
So, $\dfrac{{{v_{{c_i}}}}}{{{v_{{c_f}}}}} = \dfrac{{{r^2}}}{{{r^{'2}}}}$
We know that ${v_{{c_i}}} = 8\,cm\,{s^{ - 1}}$
Substituting all the known values, we can say that,
$\dfrac{8}{{{v_{{c_f}}}}} = \dfrac{{{{0.2}^2}}}{{{{0.4}^2}}}$
$\Rightarrow {v_{{c_f}}} = \dfrac{{8 \times {{0.4}^2}}}{{{{0.2}^2}}}$
On further solving this we get,
$\therefore {v_{{c_f}}} = 32\,cm\,{s^{ - 1}}$

Hence, the terminal velocity of the combined drop will be $32\,cm\,{s^{ - 1}}$.

Note: Terminal velocity is the highest velocity acquired by any falling object after which the velocity becomes constant for the rest of the journey. After this, the acceleration also becomes zero and there is no net drag or viscous force acting on the object since the weight becomes equal to the upthrust in such a case. The terminal velocity directly depends on the fluid in which the object is falling.