Question

Derive an expression for rise of

liquid in a capillary tube (Ascent formula)​

Answer 1

Given:

Given, liquid in a capillary tube.

To find:

We need to derive a expression for the rise of a liquid which is present in a capillary tube.

Solution:

Contemplate of a tube of uniform bore be swaybacked vertically in an exceedingly wet liquid. Since liquid is wet, therefore, the gristle is umblicate.

Let 'r' be the radius of tube, 'R' be the radius of gristle and 'θ' the angle of contact.

In figure (i), X is in atmosphere, Z is at the interface of mercury and atmosphere and Y is on protrusive aspect of gristle.

Therefore, pressure at X and Z is equal and is adequate the gas  pressure.

But, pressure at Y is smaller than that at X and Z by 2T/R.

Therefore, the liquid is not in equilibrium.

Now, so as to realize the equilibrium, the liquid can rise in tube.

Let at equilibrium, liquid rise to height h as shown in figure (ii).

Now in the equilibrium condition,

$& {{P}_{Y}}={{P}_{A}}={{P}_{B}}\text{ }.....\text{(1)} \\ & {{P}_{B}}={{P}_{A}}-\frac{2T}{R}\text{ }.....\text{(2)} \\ & {{P}_{Y}}={{P}_{B}}+\rho gh\text{ }.....\text{(3)}$

From (1), (2) and (3), we have

$\[{{P}_{Y}}={{P}_{B}}-\frac{2T}{R}+\rho gh\]$

$& \Rightarrow \frac{2T}{R}=\rho gh \\ & \Rightarrow h=\frac{2T}{R\rho g}$

But, $\[r=R\cos \theta \]$

$& \Rightarrow R=\frac{r}{\cos \theta } \\ & \therefore h\ \text{=}\frac{2T\cos \theta }{r\rho g}$

This expression is understood as Ascent formula.