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

EAMCET 2010 Questions (MCQ) on Surface Tension

Today we will discuss two questions on surface tension which appeared in EAMCET 2010 question papers.

The following multiple choice question appeared in EAMCET 2010 (Engineering) question paper:

The excess pressure inside a spherical soap bubble of radius 1 cm is balance by a column of oil (Sp. gravity 0.8) 2 mm high. The surface tension of oil is

(1) 3.92 N/m

(2) 0.0392 N/m

(3) 0. 392 N/m

(4) 0.00392 N/m

The excess pressure ∆P insie a spherical bubble of radius r is given by

∆P = 4 T/r where T is the surface tension.

The pressure p exerted by a liquid column of height h is given by

p = hρg where ρ is the density of the liquid.

Therefore we have

4 T/r = hρg from which T = rhρg/4

Substituting for known values, T = (1×10^–2×2×10^–3×0.8×10³×10)/4 = 0.004 N/m [Option (4)].

[If you substitute g = 9.8 ms^–2 instead of 10 ms^–2 (as we did above for convenience), you will get the answer as 0.00392 N/m].

Here is the question which appeared in EAMCET 2010 (Agriculture and Medicine) question paper:

A spherical liquid drop of diameter D breaks up to n identical spherical drops. If the surface tension of the drop is ‘σ’, the change in energy in this process is

(1) πσD²(n^1/3 – 1)

(2) πσD²(n^2/3 – 1)

(3) πσD²(n – 1)

(4) πσD²(n^4/3 – 1)

The surface energy (E₁) of a drop of radius R is given by

E₁ = 4πR²σ

Or E₁ = πD²σ where D is the diameter of the drop.

When a drop of radius R breaks into n identical droplets, the radius r of each droplet is given by (on equating the volumes)

(4/3) πR³ = n×(4/3) πr³

Therefore, r = R/n^1/3

The total surface energy (E₂) of all the n droplets is given by

E₂ = n×4πr²σ = n×4π (R/n^1/3) ²σ = 4πR²σn^1/3

Or, E₂ = πD²σn^1/3

The change in energy is E₂ – E₁ = πσD²(n^1/3 – 1)