Question

In a drop of radius r breaks up into 27 small drops then how much will be the change in the surface energy. The surface tension of the liquid is T.

Answer 1

Hint:If a single drop breaks up into 27 small drops which means that the volume of one single drop is divided into 27 individual drops. So, the total volume of the drop remains constant. Also, the drop is in the shape of a sphere, so the volume of the drop is equal to the volume of the sphere.

Complete step by step answer:
Given: The radius of a single drop $=r$
This single drop breaks up into 27 small drops.
Let the radius of the small drops be $R$ and the surface tension of the liquid be $T$.
Then, according to the question,
Volume of one single drop $=27\times$ Volume of small drops
As we know, the volume of a drop is same as the volume of a sphere
So, the volume of a drop having radius $r$ is $={\displaystyle \frac{4}{3}}\pi r^3$
And similarly, the volume of the small drop having radius $R$ is $={\displaystyle \frac{4}{3}}\pi R^3$
Then, putting these values into the expression we get,
${\displaystyle \frac{4}{3}}\pi r^3=27\times \left({\displaystyle \frac{4}{3}}\pi R^3\right)$
Solving this we get,
$r^3=27\times R^3$
Taking cube root of both sides we get,
$r=3\times R$
Or,
$R={\displaystyle \frac{r}{3}}$
Now, the change in the surface energy of the drop $\left(\mathrm{\Delta }U\right)$ is given by the following formula,
$\mathrm{\Delta }U=T\left(\mathrm{\Delta }A\right)$
Where, $\mathrm{\Delta }A$ is the change in the surface area of the liquid drop or (The final surface area of the small drops – The initial surface area of a one big drop)
So, $\mathrm{\Delta }A=27\left(4\pi R^2\right)-4\pi r^2$
Putting this value of $\mathrm{\Delta }A$ in the surface energy formula$\left(\mathrm{\Delta }U\right)$ we get,
$\mathrm{\Delta }U=T\left[27\left(4\pi R^2\right)-4\pi r^2\right]$
Putting $R={\displaystyle \frac{r}{3}}$ in the expression we get,
$\begin{array}{c}\mathrm{\Delta }U=4\pi T\left[27{\left({\displaystyle \frac{r}{3}}\right)}^2-r^2\right]\\ =4\pi r^{2}T\left(27\times {\displaystyle \frac{1}{9}}-1\right)\\ =4\pi r^{2}T\left(3-1\right)\\ =8\pi r^{2}T\end{array}$
Therefore, the change in the surface energy of the drop is $8\pi r^{2}T$.

Note: If we divide one big liquid drop into a number of small drops having equal volume and surface area then the change in the surface energy is the work per unit surface area of the tension force that develops the new surface of the drop. The unit of this energy is the same as that of work done.