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Derive an expression for capillary rise for a liquid having a concave meniscus.
Answer
388.8k+ views
Hint: Capillary rise is defined as a rise in a liquid above the level of zero pressure due to a net upward force produced by the attraction of the water molecules to a solid surface. Eg- glass and soil. This is applicable where the adhesion of the liquid to the solid is greater than the cohesion of the liquid to itself.
Complete step by step solution:
Apart from capillary rise, there is another term which is known as capillary depression. The depression of the meniscus of a liquid contained in a tube where the liquid does not wet the walls of the container. The meniscus is shaped convex upward, and this results in a depression of the meniscus.
The reason why the capillary action takes place is that when the adhesion to the walls is stronger than the cohesive forces between the liquid molecules.
Now, to derive the expression for capillary rise for a liquid having concave meniscus:
Let us consider a capillary tube of radius $ r $ which is partially immersed in a wetting liquid of density $ \rho $ . Let the capillary rise be $ h $ and also let $ \theta $ be the angle of contact at the edge of contact of the concave meniscus and glass figure. If $ R $ is the radius of curvature of the meniscus then, $ r = R\cos \theta $
Now, by analysing capillary action with the help of Laplace's law for a spherical membrane, surface tension $ T $ is the tangential force per unit length acting along the contact line. Also remember that it is directed into the liquid which makes an angle with the capillary wall. So, the gauge pressure within the liquid at a depth $ h $ , i.e., at the level of the free liquid surface open to the atmosphere, is
$ p - {P_o} = \rho gh......(1) $
Now, by applying Laplace’s law, we get,
$ P - {P_o} = \dfrac{{2T}}{R}.......(2) $
On comparing the above two equations, we get,
$ h\rho g = \dfrac{{2T}}{R} $
As $ r = R\cos \theta $ , so we can write the above equation,
$ h = \dfrac{{2T\cos \theta }}{{r\rho g}} $
$ h = \dfrac{{2T\cos \theta }}{{r\rho g}} $
In this case, the meniscus is convex and is obtuse. Then, $ cos \theta $ is negative but so is $ h $ , indicating a fall or depression of the liquid in the capillary. $ T $ is positive in both cases.
This is the required expression.
From this expression, we can conclude that the narrower the tube, the greater is the capillary rise.
Note:
The action of capillary rise can be seen in our day to day life. The supply of water to the leaves at the top of even a tall tree is through the capillary rise, further, cotton dresses are preferred in summers because they have fine pores that act as capillaries for sweat.
Complete step by step solution:
Apart from capillary rise, there is another term which is known as capillary depression. The depression of the meniscus of a liquid contained in a tube where the liquid does not wet the walls of the container. The meniscus is shaped convex upward, and this results in a depression of the meniscus.
The reason why the capillary action takes place is that when the adhesion to the walls is stronger than the cohesive forces between the liquid molecules.
Now, to derive the expression for capillary rise for a liquid having concave meniscus:
Let us consider a capillary tube of radius $ r $ which is partially immersed in a wetting liquid of density $ \rho $ . Let the capillary rise be $ h $ and also let $ \theta $ be the angle of contact at the edge of contact of the concave meniscus and glass figure. If $ R $ is the radius of curvature of the meniscus then, $ r = R\cos \theta $
Now, by analysing capillary action with the help of Laplace's law for a spherical membrane, surface tension $ T $ is the tangential force per unit length acting along the contact line. Also remember that it is directed into the liquid which makes an angle with the capillary wall. So, the gauge pressure within the liquid at a depth $ h $ , i.e., at the level of the free liquid surface open to the atmosphere, is
$ p - {P_o} = \rho gh......(1) $
Now, by applying Laplace’s law, we get,
$ P - {P_o} = \dfrac{{2T}}{R}.......(2) $
On comparing the above two equations, we get,
$ h\rho g = \dfrac{{2T}}{R} $
As $ r = R\cos \theta $ , so we can write the above equation,
$ h = \dfrac{{2T\cos \theta }}{{r\rho g}} $
$ h = \dfrac{{2T\cos \theta }}{{r\rho g}} $
In this case, the meniscus is convex and is obtuse. Then, $ cos \theta $ is negative but so is $ h $ , indicating a fall or depression of the liquid in the capillary. $ T $ is positive in both cases.
This is the required expression.
From this expression, we can conclude that the narrower the tube, the greater is the capillary rise.
Note:
The action of capillary rise can be seen in our day to day life. The supply of water to the leaves at the top of even a tall tree is through the capillary rise, further, cotton dresses are preferred in summers because they have fine pores that act as capillaries for sweat.
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