Question

A liquid drop having surface energy $E$ is spread into $512$ droplets of the same size. The final surface energy of the droplets is

• A. $2E$
• B. $4E$
• C. $8E$
• D. $12E$

Answer 1

Answer: (C)

The surface area of the liquid drop is $A=4\pi R^2$
Its surface energy is $E$
When the drop splits in $512$ droplets, the surface area of each droplet is $4\pi r^2$
$\therefore$ Total surface area ${\text{A}}_2$ $=512\times 4\pi r^2$
The volume of the bigger drop is $\frac{4}{3}\pi R^3$ and the volume of small droplets is $512\times \frac{4}{3}\pi r^3$
$\therefore \frac{4}{3}\pi R^3=512\times \frac{4}{3}\pi r^3\Rightarrow r=\frac{R}{8}$
$\therefore A_2=512\times 4\pi r^2=512\times 4\pi {\left(\frac{R}{8}\right)}^2=8A_1$
Surface energy $E=A.T$ ( $T$ is surface tension and $A$ is area)
$\therefore \frac{E_n}{E_0}=\frac{A_2\cdot T}{A_1\cdot T}=\frac{8\cdot A_1}{A_1}=8$
$\therefore E_n=8E$