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)