Question

derive expression for capillary rise?

Answer 1

Consider the drawing below where s is the surface tension in newton/meter, a is the angle between the surface of the liquid and the wall of the tube and R is the radius of the tube angle "a" depends on the interaction of the molecules of the liquid and the wall ("a" for water and clean glass is 0 deg). the interacting forces are the surface tension and the weight of the liquid suspended above the fluid level of the reservoir into which the tube is inserted. these forces are equal because they are in equilibrium. the total surface tension force is\[Surface ,tension = 2\pi R*s\] The weight of the suspended column is\[wgt . of .column= volume*density*g\] g is the acceleration from gravity. equating these two we get\[V \rho g =2\pi s*R*\cos \left( a \right)\] The cos(a) is introduced because the surface tension is not acting sstaight up against the weight but at an angle a. the volume V is\[vol. = \pi R ^{2}\] substituting we get \[\pi R ^{2}*h*g*\rho =2\pi R*s*\cos(a) \] therefore \[h= 2s*\cos(a)/\rho*g*R\] OK?

Answer 2

Thank you.