Question

Water rises to a height of $20\text{cm}$ in a capillary tube and mercury falls to a depth of $3.5\text{cm}$ in the same capillary tube. If the density of mercury is $13.6{\text{g cm}}^{-3}$ and its angle of contact is ${135}^{\circ }$ and density of water is $1{\text{g cm}}^{-3}$ and its angle of contact is $0^{\circ }$, then the ratio of surface tensions of the two liquids is approximately equal to

• A. 1 : 14
• B. 5 : 17
• C. 1 : 5
• D. 5 : 27

Answer 1

The correct option is B 5 : 17

The height $h$ through which a liquid will rise in a capillary tube of radius $r$ is given by
$h=\frac{2\text{cos}\theta }{r\rho g}$
whose S is the surface tension, $\rho$ is the density of the liquid and $\theta$ is the angle of contact.
For identical capillary tubes, radius remains same. Hence we can say that
$\frac{h_1}{h_2}=\frac{2S_1\text{cos}{\theta }_1}{r{\rho }_{1}g}\times \frac{r{\rho }_{2}g}{2S_2\text{cos}{\theta }_2}$
$\Rightarrow \frac{h_1}{h_2}=\frac{S_1\text{cos}{\theta }_1}{{\rho }_1}\times \frac{{\rho }_2}{S_2\text{cos}{\theta }_2}$
$\Rightarrow \frac{S_1}{S_2}=\frac{h_1}{h_2}\times \frac{\text{cos}{\theta }_2}{\text{cos}{\theta }_1}\times \frac{{\rho }_1}{{\rho }_2}.......\left(1\right)$
Given,
Rise in capillary tube containing water $\left(h_1\right)=20\text{cm}$
Fall in capillary tube containing mercury $\left(h_2\right)=3.5\text{cm}$
Density of water $\left({\rho }_1\right)=1{\text{g/cm}}^3$
Density of mercury $\left({\rho }_2\right)=13.6{\text{g/cm}}^3$
Angle of contact of water with capillary tube $\left({\theta }_1\right)=0^{\circ }$
Angle of contact of mercury with capillary tube $\left({\theta }_2\right)={135}^{\circ }$
Thus from $\left(1\right)$,
$\frac{S_1}{S_2}=\frac{20}{-3.5}\times \frac{\text{cos}{135}^{\circ }}{\text{cos}{0}^{\circ }}\times \frac{1}{13.6}$
$\Rightarrow \frac{S_1}{S_2}=0.29$.
$\Rightarrow \frac{S_1}{S_2}\approx 5:17$
Thus, option (b) is the correct answer.