Question

The end of a capillary tube with radius $r$ is immersed in water. Assuming the tube to be sufficiently long, calculate the heat energy produced when the water rises in the tube.
[Assume angle of contact = $0^{\circ }$ for glass-water system, $\rho =$ density of water, $T=$ surface tension of water]

• A. $\frac{2\pi T^2}{\rho g}$
• B. $\frac{4\pi T^2}{\rho g}$
• C. $0$
• D. $\frac{\pi T^2}{\rho g}$

Answer 1

The correct option is A $\frac{2\pi T^2}{\rho g}$

Under equilibrium,
$2\pi rT\cos 0^{\circ }=\pi r^{2}h\rho g$ [Force due to surface tension = net weight of water risen in the tube]
$h=\frac{2T}{\rho gr}$
Work done by surface tension $=\left(2\pi rT\right)\times h$
$=\frac{4\pi T^2}{\rho g}$
Potential energy of water in the tube :
$V=mgh=\pi r^{2}h\rho g\frac{h}{2}$ [centre of mass of water is at a height $\frac{h}{2}$]
$V=\frac{2\pi T^2}{\rho g}$
Heat Produced = (Work done by surface tension) - (Potential energy of water)
Heat Produced $=\frac{4\pi T^2}{\rho g}-\frac{2\pi T^2}{\rho g}=\frac{2\pi T^2}{\rho g}$