Question

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

Answer 1

The net upward force on the top hemisphere of the liquid is just the pressure difference times the area of the equatorial circle :
$F_{upward}=(P_1^{\prime }-P_0)\pi r^2$ where $F_{upward}=$ upward force on the liquid due to air-currents;

$P_1^{\prime }=$ Final pressure; $P_0=$ Initial pressure; $r=$ radius of the drop, assuming it to be spherical.

The surface tension force downward around circle is equal to the surface tension of the circumference, since only one surface contributed to the force :
Thus, $F_{downward}=T\left(2\pi r\right)$ where, $T=$ surface tension.

This gives, $P_1-P_0=\frac{2T}{r}$ which represents the excess pressure inside the liquid drop.