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Q.
A liquid drop having surface energy $E$ is spread into $512$ droplets of the same size. The final surface energy of the droplets is
NTA AbhyasNTA Abhyas 2022
Solution:
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 $\textit{A}_{2}$ $=512\times 4\pi r^{2}$
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)
$\therefore \frac{E_{n}}{E_{0}}=\frac{A_{2} \cdot T}{A_{1} \cdot T}=\frac{8 \cdot A_{1}}{A_{1}}=8$
$\therefore \, E_{n}=8E$