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(C) The height to which water rises in a capillary tube is given by $h = \frac{2T}{\rho g r}$.The potential energy of the water column of mass $m$ is

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$\frac{2 \pi T^2}{\rho g}$

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(C) The height to which water rises in a capillary tube is given by $h = \frac{2T}{\rho g r}$.The potential energy of the water column of mass $m$ is $U = \frac{mgh}{2}$.Since $m = \pi r^2 h \rho$, we have $U = \frac{(\pi r^2 h \rho) g h}{2} = \frac{\pi r^2 \rho g h^2}{2}$.Substituting $h = \frac{2T}{\rho g r}$, we get $U = \frac{\pi r^2 \rho g}{2} \left( \frac{2T}{\rho g r} \right)^2 = \frac{2 \pi T^2}{\rho g}$.The work done by the surface tension force is $W = (2 \pi r T) h = 2 \pi r T \left( \frac{2T}{\rho g r} \right) = \frac{4 \pi T^2}{\rho g}$.From the principle of conservation of energy, the heat evolved $Q$ is the difference between the work done and the potential energy gained:$Q = W - U = \frac{4 \pi T^2}{\rho g} - \frac{2 \pi T^2}{\rho g} = \frac{2 \pi T^2}{\rho g}$.

The heat evolved for the rise of water when one end of the capillary tube of radius $r$ is immersed vertically into water is (Assume surface tension $= T$ and density of water to be $\rho$)

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