Question

What is the formula of capillarity?

Answer 1

Hint: When a tube open on both the ends of very small radius is immersed in water the water level in the tube rises where as if we dip the same tube in liquids which does not wet the surface of the walls e.g. mercury the level is suppressed. The phenomenon of rise and fall of liquid in a capillary tube in comparison to the surrounding is called capillarity. The measure of this fall or rise in level of liquids is given by capillarity formula.

Formula used:
$h={\displaystyle \frac{2\sigma \cos \theta }{r\rho g}}$

Complete answer:
Consider a capillary tube of radius ‘r’ dipped in a liquid of surface tension $\sigma$ and density $\rho$. Suppose the liquid wets the sides of the tube. Then its meniscus will be concave. The shape of the meniscus of water will be nearly spherical if the capillary tube is of sufficiently narrow bore. As the pressure is greater on the concave side of a liquid surface, so excess of pressure at point just above the meniscus compared to point just below the meniscus is,
 $p={\displaystyle \frac{2\sigma \cos \theta }{r}}$
Where ‘r’ is the radius of the capillary tube, $\theta$ is the angle of contact. Due to this excess pressure p, the liquid rises in the capillary tube of height ‘h’ when the hydrostatic pressure exerted by the liquid column becomes equal to excess pressure p. Therefore, at equilibrium we have
$\begin{array}{rl}& p=h\rho g={\displaystyle \frac{2\sigma \cos \theta }{r}}\\ & \therefore h={\displaystyle \frac{2\sigma \cos \theta }{\rho gr}}\end{array}$
Hence the above expression is the formula for capillarity.

Note: If we take into account the volume of the liquid contained in the meniscus then, the above expression obtained does not hold valid. In fact the same expression gets reduced by a factor of ${\displaystyle \frac{r}{3}}$ . However the factor ${\displaystyle \frac{r}{3}}$ can be neglected if the tube has a very small radius.