Question
Derive an expression for the excess of pressure inside a liquid drop.

Answer

Consider a liquid drop of radius R and $\sigma$ the surface tension of liquid.

Excess pressure inside the liquid drop, $P = P_i - P_0$ ($\therefore$ liquid drop has only one free surface)$\delta\text{R}=$ Small increase in radius of liquid drop due to excess pressure
$\therefore$ W = Force × Displacement.
W = (Excess pressure × Area) × Increase in radius$\text{W}=\text{P}\times4\pi\text{R}^2\times\delta\text{R}$
Increase in surface area of liquid drop = Final surface area - Initail surface area$=4\pi(\text{R}+\delta\text{R})^2-4\pi\text{R}^2$
$=8\pi\text{R}(\delta\text{R})(\text{Neglecting }\delta\text{R}^2)$
Increase P.E. = Increase in surface area × Surface tension$=8\pi\text{R}(\delta\text{R})\times\sigma$
Since the drop is in equilibrium.$\therefore\text{P}\times4\pi\text{R}^2\times\delta\text{R}=8\pi\text{R}(\delta\text{R})\times\sigma$
$\Rightarrow\text{P}=\frac{2\sigma}{\text{R}}$
$\Rightarrow\text{P}_\text{i}-\text{P}_\text{o}=\frac{2\sigma}{\text{R}}[\because\text{P}=\text{P}_\text{i}-\text{P}_\text{o}]$

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