Question

Water rises to a height of $10cm$ in a capillary tube and mercury falls to a depth of $3.42cm$ in the same capillary tube. If the density of mercury is $13.6g/c.c$ and the angles of contact for mercury and water are ${135}^o$ and $0^o$ respectively, the ratio of surface tension for water and mercury is

• A. $1:0.15$
• B. $1:3$
• C. $1:6.5$
• D. $1.5:1$

Answer 1

Answer: (C)

$1:6.5$

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')

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

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

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

$\Rightarrow \frac{S_1}{S_2}=\frac{10}{-3.42}\times \frac{\cos {135}^{\circ }}{\cos 0^{\circ }}\times \frac{1}{13.6}$

$\Rightarrow \frac{S_1}{S_2}=\frac{10}{-3.42}\times \frac{-0.707}{1}\times \frac{1}{13.6}$

$\Rightarrow \frac{S_1}{S_2}=0.152\approx \frac{1}{6.5}$