Q.A small spherical drop of mercury of radius $2.0\ \text{mm}$ is formed in air. Taking the surface tension of mercury as $0.465\ \text{N/m}$, calculate the excess pressure inside the drop over the pressure of the surrounding air.

Given: $r = 2.0\ \text{mm} = 2.0\times10^{-3}\ \text{m}$, $T = 0.465\ \text{N/m}$.

A liquid drop has only one free surface, so: …

Use $\Delta P = 2T/r$, the excess-pressure formula for a single free surface, since a liquid drop ( …

• Using the soap-bubble formula $4T/r$ for a plain liquid drop, which has only a single surface, not two. …