Thursday, June 14, 2012

Kerala Engineering Entrance 2012 (KEAM Engg. 2012) Questions on Surface Tension

“It doesn't matter how beautiful your theory is; it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong.”
Richard Feynman

Two questions on surface tension were included in the KEAM (Engineering) 2012 question paper. Here are the questions with solution:
(1) If two capillary tubes of radii r1 and r2 in the ratio 1 : 2 are dipped vertically in water, then the ratio of capillary rises in the respective tubes is
(a) 1 : 4
(b) 4 : 1
(c) 1 : 2
(d) 2 : 1
(e) 1 : √2
The capillary rise h due to surface tension is relate to the surface tension S as
            S = hrρg/2 cosθ
where r is the radius of the capillary tube, ρ is the density of the liquid, g is the acceleration due to gravity and θ is the angle of contact.
Evidently h is inversely proportional to r.
Therefore, h1/h2 = r2/r1 = 2 : 1
(2) If the excess pressure inside a soap bubble of radius r1 in air is equal to the excess pressure inside air bubble of radius r2 inside the soap solution, then r1 : r2 is         
(a) 2 : 1
(b) 1 : 2
(c) 1 : 4
(d) √2 : 1
(e) 1 : √2
The excess pressure inside a soap bubble in air is 4S/r where as the excess pressure inside an air bubble in the soap solution is 2S/r where S is the surface tension (of soap solution) and r is the radius of the bubble.
[In the case of the air bubble in the soap solution there is one liquid surfaoe only and that is why the excess pressure is 2S/r and not 4S/r. Remember that in the case of a soap bubble in air there are two liquid surfaces]. 
As given in the question, we have
            4S/r1 = 2S/r2
This gives r1/r2 = 2
Or, r1 : r2 = 2 : 1

No comments:

Post a Comment