Question

A liquid of density $\rho$ and surface tension rises to a height h in capillary tube of diameter d. Angle of constant between the tube and liquid is zero. The weight of the liquid in the capillary tube is :

Answer 1

Ф = Angle of contact = 0 deg.

S = surface tension

d/2 = radius of the tube

ρ = density of liquid

h = height of the column inside tube

The force of surface tension pulling the liquid upwards in the tube

= S $\times$ π d cosФ = π d S

The force pulling the liquid downwards = weight = W = $\frac{\pi d^2}{4}$$\times$h $\times$ρ $\times$g

So: π d S = W = $\frac{\pi }{4}$ $\times d^{2}h\rho g$

height of liquid $=h=\frac{4S}{\left[d\rho g\right]}$

Weight W= πd²/4 $\times$h g ρ or W= π d S