Surface Tension–Questions involving excess of pressure inside a bubble

Here is a question which is popular among question setters:

Two spherical soap bubbles of radii r₁ and r₂ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

(a) (r₁+r₂)/2 (b) √(r₁+r₂) (c) r₁r₂/(r₁+r₂) (d) (r₁+r₂)/√2 (e) √(r₁²+r₂²)

As the temperature is constant, we have P₁V₁+ P₂V₂ =.PV where P₁ and P₂ are the pressures inside the separate bubbles, V₁ and V₂ are their volumes, P is the pressure inside the combined bubble and V is its volume.

Since the bubbles are located in vacuum, the pressure inside the bubble is equal to the excess of pressure 4T/r, where T is the surface tension and ‘r’ is the radius so that we have (4T/r₁)×[(4/3)πr₁³]+ (4T/r₂)×[(4/3)πr₂³] = (4T/R)×[(4/3)πR³] where R is the radius of the combined bubble.

This yields R = √(r₁²+r₂²).

[You can work out this problem by equating the surface energies: 4πr₁²T+4πr₂²T= 4πR²T, from which R = √(r₁²+r₂²)]

Now, consider the following MCQ:

Excess of pressure inside one soap bubble is four times that inside another. Then the ratio of the volume of the first bubble to that of the second is

(a) 1:16 (b) 1:32 (c) 1:64 (d) 1:2 (e) 1:4

If r₁ and r₂ are the radii of the bubbles and T is the surface tension of soap solution, we have (4T/r₁)/(4T/r₂) = 4.

Therefore, r₁/r₂ = ¼. Since the volume is directly proportional to the cube of the radius, the ratio of volumes V₁/V₂ = (r₁/r₂)³ = (¼)³ = 1/64 [Option (d)].

The following MCQ appeared in AIEEE 2004 question paper. This question is popular among question setters and has appeared in other entrance test papers as well:

If two soap bubbles of different radii are connected by a tube,

(a) air flows from the bigger bubble to the smaller bubble till the sizes become equal

(b) air flows from the bigger bubble to the smaller bubble till the sizes are interchanged

(c) air flows from the smaller bubble to the bigger

(d) there is no flow of air

The excess of pressure inside the smaller bubble is greater than that inside the bigger bubble. (Remember, P = 4T/r and hence excess of pressure ‘P’ is inversely proportional to the radius ‘r’). Therefore, air will flow from the smaller bubble to the bigger bubble [Option (c)].