Mechanical Properties of Fluids — Physics STD 11 Science — Question

$\delta\text{R}=$ Small increase in radius of liquid drop due to excess pressure

$\therefore$ W = Force × Displacement.

$\text{W}=\text{P}\times4\pi\text{R}^2\times\delta\text{R}$

$=4\pi(\text{R}+\delta\text{R})^2-4\pi\text{R}^2$

$=8\pi\text{R}(\delta\text{R})(\text{Neglecting }\delta\text{R}^2)$

$=8\pi\text{R}(\delta\text{R})\times\sigma$

$\therefore\text{P}\times4\pi\text{R}^2\times\delta\text{R}=8\pi\text{R}(\delta\text{R})\times\sigma$

$\Rightarrow\text{P}=\frac{2\sigma}{\text{R}}$

$\Rightarrow\text{P}_\text{i}-\text{P}_\text{o}=\frac{2\sigma}{\text{R}}[\because\text{P}=\text{P}_\text{i}-\text{P}_\text{o}]$