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ExamsJEE MainPhysics

A soap bubble of radius R has a thin film of thickness d (with d much less than R). It collapses to form a single spherical liquid drop. Find the ratio of the excess pressure inside the resulting drop to the excess pressure inside the original bubble.

  1. (1/2)*(R/(3d))^(1/3)
  2. (R/(3d))^(1/3)
  3. (1/2)*(3d/R)^(1/3)
  4. 2*(R/(3d))^(1/3)

Correct answer: (1/2)*(R/(3d))^(1/3)

Solution

The soap film volume (surface area times thickness) is conserved when it becomes a full drop. From that, find the drop radius r, then take the ratio of 2T/r (drop) to 4T/R (bubble, which has two surfaces).

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