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}×0.8×10^{3}×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) π*σD*^{2}(*n*^{1/3} – 1)

(2) π*σD*^{2}(*n*^{2/3} – 1)

(3) π*σD*^{2}(*n* – 1)

(4) π*σD*^{2}(*n*^{4/3} – 1)

The surface energy (*E*_{1}) of a drop of radius *R* is given by

*E*_{1} =_{ }4π*R*^{2}*σ*

Or *E*_{1 }= π*D*^{2}*σ* 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) π*R*^{3} = *n*×(4/3) π*r*^{3}

Therefore, *r = R/n*^{1/3}

The total surface energy (*E*_{2}) of all the *n* droplets is given by

*E*_{2 }= *n*×4π*r*^{2}*σ* = *n*×4π* *(*R/n*^{1/3})^{ }^{2}*σ = *4π*R*^{2}*σn*^{1/3}

Or, *E*_{2 }= π*D*^{2}*σn*^{1/3}

The change in energy is *E*_{2 }–* E*_{1} = π*σD*^{2}(*n*^{1/3} – 1)