Answer to Question #138572 in Electric Circuits for Gary

Question

A clean glass capillary tube, of internal diameter 0.04 cm, is held vertically with its lower end below the surface of clean water in a beaker, and with 10 cm of the tube above the surface. To what height will the water rise in the tube? What will happen if the tube is now depressed until only 5 cm of its length is above the surface? The surface tension of water is 7.2 x 10-2 Nm-1.

Answer 1

Let $d$ and $r$ represent the diameter and radius respectively.

$r=\frac{d}{2}=\frac{0.04cm}{2}=0.02cm=2×10^{-4}m\\$

The height $h$ through which a liquid will rise in a capillary tube of radius $r$ is given by $h=\frac{2Scos\theta}{r\rho\textsf{g}}$

Where $S$ is the surface tension, $\rho$ is the density of the liquid and $\theta$ is the angle of contact.

$h=\frac{2×7.2×10^{-2}Nm^{-1}×cos(0)}{2×10^{-4}m×10³kgm^{-3}×9.81ms^{-2}}\\ h=0.073m\\ h=7.3cm$