Question

The end of a capillary tube with a radius r is immersed in water. Is mechanical energy conserved when the water rises in the tube? The tube is sufficiently long. If not, calculate the energy change.

Answer 1

In the equilibrium position ($\theta =0$ for water and glass)
$2\pi rT=cos\theta =\pi r^{2}hpg$
or $h=\frac{2T}{\rho gr}$
Work done by surface tension =$\left(2\pi rT\right)\times h=\frac{4\pi T^2}{pg}$
The potential energy of water in the tube, U = ($\pi r^{2}hp$)gh/2; it is multiplied by h/2 because the centre of gravity of the water and the capillary tube is at a height h/2
U =$\frac{2\pi T^2}{pg}$
Thus, it is seen that the mechanical energy is not conserved.
Therefore, mechanical energy loss = $\frac{4\pi T^2}{pg}-\frac{2\pi T^2}{pg}=\frac{4\pi T^2}{pg}$
This energy is converted into heat.