Wednesday, July 14, 2010

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(n1/3 – 1)

(2) πσD2(n2/3 – 1)

(3) πσD2(n – 1)

(4) πσD2(n4/3 – 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/n1/3

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

E2 = n×4πr2σ = n×4π (R/n1/3) 2σ = 4πR2σn1/3

Or, E2 = πD2σn1/3

The change in energy is E2 E1 = πσD2(n1/3 – 1)

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