What is the excess pressure inside a drop of mercury of radius $3.00 ext{ mm}$ at room temperature?

Question

What is the excess pressure inside a drop of mercury of radius $3.00	ext{ mm}$ at room temperature? Surface tension of mercury at that temperature is $4.65 	imes 10^{-1}	ext{ N/m}$.

Professor PC Thomas Classes · Answer

A liquid drop has only one liquid-air interface. Excess pressure $\Delta P = \frac{2S}{r} = \frac{2 	imes 4.65 	imes 10^{-1}}{3.0 	imes 10^{-3}} = \frac{9.3 	imes 10^{-1}}{3 	imes 10^{-3}} = 3.1 	imes 10^2	ext{ Pa} = 310	ext{ Pa}$.