Question

Water rises up in a glass capillary up to a height of $9.0cm$, while mercury falls down by $3.4cm$ in the same capillary. Assume angles of contact for water glass and mercury glass $0^0$ and ${135}^0$ respectively. Determine the ratio of surface tension of mercury and water ($cos{135}^0=-0.71$)

Answer 1

The height $h$ through which a liquid will rise in a capillary tube of radius $r$ is given by

$h=\frac{2S\cos \theta }{r\rho g}$ where $S$ is the surface tension,

$\rho$ is the density of the liquid and $\theta$ is the angle of contact.

For identical capillary tubes and two liquids ( water '1' & mercury '2')

$\frac{h_1}{h_2}=\frac{2S_1\cos {\theta }_1}{r{\rho }_{1}g}\times \frac{r{\rho }_{2}g}{2S_2\cos {\theta }_2}$

$\Rightarrow \frac{h_1}{h_2}=\frac{S_1\cos {\theta }_1}{{\rho }_1}\times \frac{{\rho }_2}{S_2\cos {\theta }_2}$

$\Rightarrow \frac{S_1}{S_2}=\frac{h_1}{h_2}\times \frac{\cos {\theta }_2}{\cos {\theta }_1}\times \frac{{\rho }_1}{{\rho }_2}$

$\Rightarrow \frac{S_1}{S_2}=\frac{9}{-3.4}\times \frac{\cos {135}^{\circ }}{\cos 0^{\circ }}\times \frac{1}{13.6}$

$\Rightarrow \frac{S_1}{S_2}=\frac{9}{-3.4}\times \frac{-0.71}{1}\times \frac{1}{13.6}$

$\Rightarrow \frac{S_1}{S_2}=0.1382\approx \frac{1}{7.23}$

$\Rightarrow \frac{S_2}{S_1}=7.23$