Question

A liquid of density $\rho$ and surface tension $\sigma$ rises to a height $h$ in a capillary tube of diameter $d$. The potential energy of the liquid in the capillary tube is:
[Assume liquid to be perfectly wetting in nature]

• A. $h\rho g$
• B. $\frac{2\pi {\sigma }^2}{\rho g}$
• C. $2\pi {\sigma }^2\rho g$
• D. $\frac{2\pi {\sigma }^2}{\rho gh}$

Answer 1

The correct option is B $\frac{2\pi {\sigma }^2}{\rho g}$

The mass of the liquid in the capillary tube can be considered to be concentrated at the centre of mass of the liquid column of height $h$ i.e $\text{COM}$ is at a height $\frac{h}{2}$.

The potential energy of the liquid in the capillary tube is,
$U=mg\times \frac{h}{2}$
$U=\left(\pi r^{2}h\rho \right)\times g\times \left(\frac{h}{2}\right)$
$U=\frac{\pi (rhg{)}^2\rho }{2g}....\left(1\right)$
The equation for capillary rise,
$h=\frac{2\sigma \cos \theta }{\rho rg}$
For perfectly wetting liquid, $\cos \theta =1$
$\Rightarrow \sigma =\frac{h\rho rg}{2}$
$\Rightarrow rhg=\frac{2\sigma }{\rho }....\left(2\right)$

Substituting Eq.$\left(2\right)$ in Eq.$\left(1\right)$, the $PE$ of a liquid in the capillary tube can be written as,
$U=\frac{\pi }{2g}{\left(\frac{2\sigma }{\rho }\right)}^2\rho$
$\therefore U=\frac{2\pi {\sigma }^2}{\rho g}$
Hence, the correct choice is $\left(b\right)$