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ExamsJEE MainPhysics

A soap bubble of radius 2.4 * 10⁻⁴ m sits in air inside a cylinder initially at pressure 10⁵ N/m². The air is compressed isothermally until the bubble's radius is halved. The soap film has surface tension 0.08 N/m. The final air pressure in the cylinder is found to be 8.08 * 10ⁿ N/m². Find n.

  1. 5
  2. 3
  3. 4
  4. 2

Correct answer: 5

Solution

The inside pressure equals cylinder pressure plus excess pressure 4T/r. As the bubble's radius halves isothermally, its volume drops by a factor of 8, so by Boyle's law the inside pressure increases 8-fold. The new excess pressure also doubles (4T/(r/2) = 8T/r). Solving gives the cylinder pressure of order 10⁵, so n = 5.

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