Question

derive an expression for rise of liquid in a capillary tube and show that the height of the liquid column supported in is inversely proportional to the radius of the tube​

Answer 1

$\huge\red{Answer}$

Capillarity is a physical phenomenon in which liquids flow without the help of

gravity.  Liquids even rise to a height against gravity, through narrow

tubes.

     Capillary action is due to the phenomenon of Surface tension of liquid as well

as adhesive forces between liquid molecules and molecules of the narrow tube.

Surface tension is due to cohesive attraction among liquid molecules.

Derivation:

    When a thin (open or closed at the top) tube is inserted into a liquid in a

container, the liquid inside the tube rises to a height h above the liquid

surface outside.  Let the diameter of the tube be D.  The density of

liquid be ρ.  The surface tension of the liquid be S.

Weight of liquid column acting downwards = m g

   

    W = ρ (πD²/4) h g     --(1)

    The surface on the top liquid inside the capillary tube has a trough (cup) like

shape. Assume the angle of contact with the walls be Ф.  Surface

tension is the contact force per unit length along the circumference of top

surface. This force pulls the liquid vertically upwards.

   

Force upwards = S * πD * CosФ  ----- (2)

=>  

 h =  4 S CosФ / (ρ D g)      

HOPE IT HELPS YOU !!

Answer 2

The expression for rise of liquid in a capillary tube is $h=\frac{2\sigma cos\theta}{r\rho g}$

Explanation:

• Let the height of the liquid rise in the tube be 'h'.
• Let the radius of the capillary tube be 'r'.
• Let the density of the liquid be $\rho$ and the surface tension be $\sigma$.
• The angle between the surface of liquid and the tube be $\theta$ such that $\theta<\frac{\pi}{2}$ .
• now the volume of the liquid is given by, $V=\pi r^2h$
• let the mass of liquid be $m$ then the weight of the liquid rise is,      $W=mg\\W=\rho Vg\\W=\rho (\pi r^2h)g$  
• now the the tension along the y-axis is given by , $T_y=\sigma cos\theta$  
• now given the area of the liquid surface be $A=2\pi r$ we have,                        $T_yA=W\\\sigma cos\theta(2\pi r)=\rho g(\pi r^2h) \\h=\frac{\sigma cos\theta(2\pi r)}{\rho g(\pi r^2)} \\h=\frac{\sigma cos\theta(2 r)}{\rho g( r)}$
• Hence we get final expression for height of liquid as $h=\frac{2\sigma cos\theta}{r\rho g}$  
• From the above expression we clearly get that , $h\propto \frac{1}{r}$.