The lower end of a capillary tube of radius r is placed vertically in water. Then with the rise of water in the capillary, heat evolved is

The correct option/answer is + π r²h²dg/2J.

Let us find the heat evolved during the rise of water in a capillary tube of radius _$r$.
When a capillary tube is dipped in water, water rises to a height _$h$.
The mass of the water column of height _$h$ and radius _$r$ is:
$m=\mathrm{Volume}\times \mathrm{density}=\pi r^{2}hd$
where _$d$ is the density of water.

The center of mass of this risen water column lies at a height of _{$\frac{h}{2}$} from the free surface.
Therefore, the gain in potential energy (_{$U_g$}) of the water column is:
$U_g=mg\left(\frac{h}{2}\right)=\left(\pi r^{2}hd\right)g\left(\frac{h}{2}\right)=\frac{\pi r^2{h}^{2}dg}{2}$

The work done by the force of surface tension (_$W$) in raising the water column is:
$W=\mathrm{Force}\times \mathrm{distance}=(2\pi rTcos\theta )\times h$
For a capillary rise, we know the relation:
$h=\frac{2Tcos\theta }{rdg}\Rightarrow 2\pi rTcos\theta =\pi r^{2}hdg$
Substituting this back into the work formula:
$W=\left(\pi r^{2}hdg\right)\times h=\pi r^2{h}^{2}dg$

The energy that is lost as heat (_$H$ in mechanical units) is the difference between the work done by surface tension and the gain in potential energy:
$W-U_g=\pi r^2{h}^{2}dg-\frac{\pi r^2{h}^{2}dg}{2}=\frac{\pi r^2{h}^{2}dg}{2}$

To convert this energy into heat units (calories/Joules), we divide by the mechanical equivalent of heat (_$J$):
$\mathrm{Heat\; Evolved}=+\frac{\pi r^2{h}^{2}dg}{2J}$