Question

A certain number of spherical drops of a liquid of radius $$r$$ coalesce to form a single drop of radius $$R$$ and volume $$V$$. If $$T$$ is the surface tension of the liquid, then:
A) Energy = $4VT\left({\displaystyle \frac{1}{r}}-{\displaystyle \frac{1}{R}}\right)$ is released
B) Energy = $3VT\left({\displaystyle \frac{1}{r}}+{\displaystyle \frac{1}{R}}\right)$ is released
C) Energy = $3VT\left({\displaystyle \frac{1}{r}}-{\displaystyle \frac{1}{R}}\right)$ is released
D) Energy is neither released nor absorbed

Answer 1

Hint: In order to find the energy released or absorbed during the process of coalescence, remember the change in energy will be equal to the change in the surface energy. During the process of coalescence, the volume of the total will remain the same.

Complete step by step solution:
Let’s define all the terms given in the question:
Radius of the smaller liquid drops=$$r$$
Radius of the final liquid drop= $$R$$
Volume of the final liquid drop=$$V$$
Surface tension of the liquid=$$T$$
In the question we are asked to find the energy released or absorbed during the process of coalescence
We take $\mathrm{\Delta }E$ as the energy absorbed during the process.
The energy absorbed during the process will be equal to the change in the surface energy. That is,
$\mathrm{\Delta }E=\mathrm{\Delta }SE$
The surface energy is given by,
$SE=A.T$
Where, $A$ is the area.
So the change in the surface energy is given by,
$$\mathrm{\Delta }SE=\mathrm{\Delta }E=4\pi R^{2}T-4\pi r^{2}Tn$$…………………….. (1)
Where, $n$is the number of small drops
We know, during the process of coalescence, volume of the liquid does note change.
That is, $${\displaystyle \frac{4\pi }{3}}{R}^3={\displaystyle \frac{4\pi }{3}}{r}^{3}n=V$$
So the equation (1) will become,
$$\Rightarrow \mathrm{\Delta }E=3S\left({\displaystyle \frac{4\pi }{3}}{R}^3{\displaystyle \frac{1}{R}}-{\displaystyle \frac{4\pi }{3}}{r}^{3}n{\displaystyle \frac{1}{r}}\right)$$
$$\Rightarrow \mathrm{\Delta }E=3T\left({\displaystyle \frac{V}{R}}-{\displaystyle \frac{V}{r}}\right)$$
$$\therefore \mathrm{\Delta }E=3VT\left({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{r}}\right)$$
The energy absorbed= $$3VT\left({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{r}}\right)$$
So the energy released will be the negative of this value, that is,
Energy released= $$-1\left\{3VT\left({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{r}}\right)\right\}$$
Energy released= $$3VT\left({\displaystyle \frac{1}{r}}-{\displaystyle \frac{1}{R}}\right)$$

So the final answer is option (C), Energy = $3VT\left({\displaystyle \frac{1}{r}}-{\displaystyle \frac{1}{R}}\right)$ is released.

Note: The term surface energy can be defined as the work done per unit area given by the force that creates the new surface. The surface energy quantifies the disruption of bonds between the molecules that occurs when a new surface is created
The surface of the liquid counteracts the surface increase through the force by applying a tension force tangentially on the hoop. This tension force is also known as the surface tension.