Question

What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature $\left({20}^{\circ }C\right)$ is $4.65\times {10}^{–1}N{m}^{–1}$ . The atmospheric pressure is $1.01\times {10}^5$ Pa. Also give the excess pressure inside the drop.

Answer 1

Radius of the mercury drop, r = 3.00 mm = $3\times {10}^{–3}m$
Surface tension of mercury, $S=4.65\times {10}^{–1}N{m}^{–1}$
Atmospheric pressure, $P_0=1.01\times {10}^5$ Pa
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
$=\frac{2S}{r}+P_0=\frac{2\times 4.65\times {10}^{-1}}{3\times {10}^{-3}}+1.01\times {10}^5=1.031\times {10}^5=1.01\times {10}^{5}Pa$
Excess pressure $=\frac{2S}{r}=\frac{2\times 4.65\times {10}^{-1}}{3\times {10}^{-3}}$
= 310 Pa