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