Here is a question which is popular among question setters:

**Two spherical soap bubbles of radii r _{1} and r_{2} in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to **

**(a) (r _{1}+r_{2})/2 (b) √(r_{1}+r_{2}) (c) r_{1}r_{2}/(r_{1}+r_{2}) (d) (r_{1}+r_{2})/√2 (e) √(r_{1}^{2}+r_{2}^{2}) **

As the temperature is constant, we have P_{1}V_{1}+ P_{2}V_{2} =.PV where P_{1} and P_{2 }are the pressures inside the separate bubbles, V_{1} and V_{2 }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/r_{1})×[(4/3)πr_{1}^{3}]+ (4T/r_{2})×[(4/3)πr_{2}^{3}] = (4T/R)×[(4/3)πR^{3}] where R is the radius of the combined bubble.

This yields R = √(r_{1}^{2}+r_{2}^{2}).** **

[You can work out this problem by equating the surface energies: 4πr_{1}^{2}T+4πr_{2}^{2}T= 4πR^{2}T, from which R = √(r_{1}^{2}+r_{2}^{2})]

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 r_{1} and r_{2} are the radii of the bubbles and T is the surface tension of soap solution, we have (4T/r_{1})/(4T/r_{2}) = 4.

Therefore, r_{1}/r_{2} = ¼. Since the volume is directly proportional to the cube of the radius, the ratio of volumes V_{1}/V_{2} = (r_{1}/r_{2})^{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**

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)].

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