question_answer

 Acapillary tube of radius r is dipped vertically in a liquid of density d, surface tension T and angle of contact 0 then the pressure difference just below the two surfaces, one in the beaker and the other in the capillary tube , is :

A) \[\frac{2T}{r}\]                                 

B) \[\frac{T}{r\cos \theta }\]

C) \[\frac{2T\cos \theta }{r}\]                         

D) \[\frac{T\cos \theta }{r}\]

Correct Answer: C

Solution :

When a capillary tube is dipped in water, the water meniscus inside the tube is concave. The pressure just below the meniscus is less than  the pressure just above it by \[\frac{2T}{R},\] where T is surface tension, R is radius of curvature of the meniscus. The pressure on the surface of water is atmospheric pressure P. The pressure just below the plane surface of water outside the tube is also P, but that just below the meniscus  inside the tube is \[P-\frac{2T}{R}.\]Pressure at all points in the same level of water must be same. Therefore, to make up the deficiency of pressure \[\frac{2T}{R}\] below the meniscus, water begins to flow from outside the tube. The rising of water in the capillary stops at a certain height h. In this position the pressure of the water column of height h becomes equal to \[\frac{2T}{R}.\] \[P=h\rho g=\frac{2T}{R}\]where \[\rho \]is density g is gravity. If r is the radius of tube and \[\theta \]is angle of contact then \[R=r/\cos \theta \] \[\therefore \] \[P=h\rho g=\frac{2T\cos \theta }{r}\]is correct.