Question

Water rises to a height of $10cm$ in a certain capillary tube. An another identical tube when dipped inmercury the level of mercury is depressed by $3.42cm$. Density of mercury is $13.6g/cc$. The angle of contact for water in contact with glass is $0^{}$ and mercury in contact with glass is ${135}^o$. The ratio of surface tension of water to that of Hg is

• A. $1:3$
• B. $1:4$
• C. $1:5.5$
• D. $1:6.5$

Answer 1

The correct option is D $1:6.5$

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')

$\frac{h_1}{h_2}=\frac{2S_1\cos {\theta }_1}{r{\rho }_{1}g}\times \frac{r{\rho }_{2}g}{2S_2\cos {\theta }_2}\Rightarrow \frac{h_1}{h_2}=\frac{S_1\cos {\theta }_1}{{\rho }_1}\times \frac{{\rho }_2}{S_2\cos {\theta }_2}\Rightarrow \frac{S_1}{S_2}=\frac{h_1}{h_2}\times \frac{\cos {\theta }_2}{\cos {\theta }_1}\times \frac{{\rho }_1}{{\rho }_2}\Rightarrow \frac{S_1}{S_2}=\frac{10}{-3.42}\times \frac{\cos {135}^{\circ }}{\cos 0^{\circ }}\times \frac{1}{13.6}\Rightarrow \frac{S_1}{S_2}=\frac{10}{-3.42}\times \frac{-0.707}{1}\times \frac{1}{13.6}\Rightarrow \frac{S_1}{S_2}=0.152\approx \frac{1}{6.5}$