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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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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