Question

One end of a capillary tube with a radius $r$ is immersed in water. Is mechanical energy conserved when the water rises in the tube? If not, calculate the energy change.
Consider water to be perfectly wetting in nature. The tube is sufficiently long.

• A. Yes, mechanical energy is concerved
• B. No, Loss in mechanical energy $=\frac{4\pi T^2}{\rho g}$
• C. No, Loss in mechanical energy $=\frac{2\pi T^2}{\rho g}$
• D. No, Loss in mechanical energy $=\frac{\pi T^2}{8\rho g}$

Answer 1

The correct option is C No, Loss in mechanical energy $=\frac{2\pi T^2}{\rho g}$

For the vertical equilibrium of liquid column in capillary tube,
$2\pi rT\cos \theta =\pi r^{2}h\rho g...\left(i\right)$
$\Rightarrow h=\frac{2T\cos \theta }{\rho gr}$
For water( perfectly wetting nature), $\theta =0^{\circ }$
$\therefore h=\frac{2T}{\rho gr}$
Since force due to surface tension acting in vertically upward direction is,
$F=2\pi rT\cos \theta =2\pi rT$
where $2\pi r=$ perimeter of meniscus;
Work done by surface tension when liquid rises through height $h$,
$W=\left(2\pi rT\right)\times h=\frac{4\pi T^2}{\rho g}...\left(ii\right)$
The potential energy of water in the tube,
$U=mg{h}^{\prime }$
$U=\left(\pi r^{2}h\rho \right)\times g\times \frac{h}{2}...\left(iii\right)$
It is multiplied by $h^{\prime }=\frac{h}{2}$ because the whole mass of liquid in the tube is concentrated at the centre of mass at a height of $\frac{h}{2}$.
Substituting in Eq.$\left(iii\right)$ in terms of $T$,
$U=\frac{2\pi T^2}{\rho g}$
Consider $U=0$ at the initial level of water, the increase in $PE$ of water in the tube is $\Delta U=U-0=U$

Since all the work done by the force due to surface tension has not been converted into gain in $PE$, mechanical energy is not conserved.
$W\ne \Delta U$
Hence, loss in mechanical energy :
$\Delta E=W-\Delta U$
$\Rightarrow \Delta E=\frac{4\pi T^2}{\rho g}-\frac{2\pi T^2}{\rho g}=\frac{2\pi T^2}{\rho g}$
The loss in mechanical energy is dissipated as heat.