Question

A capillary tube of radius $r$ is dipped vertically in a liquid of density $d$, surface tension $T$ and angle of contact $\theta$ then the pressure difference just below the two surfaces, one in the beaker and the other in the capillary tube , is :

• A. $ \frac{2T}{r} $
• B. $ \frac{T}{r\cos \theta } $
• C. $ \frac{2T\cos \theta }{r} $
• D. $ \frac{T\cos \theta }{r} $

Answer 1

Answer: (C)

When a capillary tube is dipped in water, the water meniscus inside the tube is concave. The pressure just below the meniscus is less than the pressure just above it by $\frac{2 T}{R}$, where $T$ is

surface tension, $R$ is radius of curvature of the meniscus. The pressure on the surface of water is atmospheric pressure $P$. The pressure just below the plane surface of water outside the tube is also $P$, but that just below the meniscus inside the tube is $P-\frac{2 T}{R}$. Pressure at all points

in the same level of water must be same. Therefore, to make up the deficiency of pressure $\frac{2 T}{R}$ below the meniscus, water begins to flow from outside the tube. The rising of water in the capillary stops at a certain height $h$. In this position the pressure of the water column of height $h$ becomes equal to $\frac{2 T}{R}$.
$P=h \rho g=\frac{2 T}{R}$
where $\rho$ is density
$g$ is gravity. If $r$ is the radius of tube and $\theta$ is angle of contact then
$R=r / \cos \theta$
$\therefore P=h \rho g=\frac{2 T \cos \theta}{r}$ is correct.