A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of the another single soap bubble of radius r cm. The value of r is ___________.

Solution

Physics sir · 4 steps

Climb in from the outside air: each soap film you cross adds its own 4T/R to the pressure.

1. Two surfaces each. A soap film has an outer surface and an inner surface. So the excess pressure across one bubble of radius R is 4T/R, not 2T/R.
2. Cross both films. Going from the outside air at P₀ into the space between the bubbles, you cross the 6 cm film. Going from there into the small bubble, you cross the 3 cm film. Add the two jumps.

   Pᵢₙₙₑᵣ = P₀ + 4T/6 + 4T/3 = P₀ + 4T/2

3. Match a single bubble. A single bubble of radius r has internal pressure P₀ + 4T/r. Setting the two equal gives 1/r = 1/6 + 1/3 = 1/2.
4. So. r = 2 cm, so the answer is 2.

The common slip: 3. Counting only the 3 cm film and ignoring the outer 6 cm bubble gives r = 3, not 2. The gas between the bubbles is already above atmospheric pressure.

The idea to keep. For nested bubbles the excess pressures add up, film by film. So the equivalent single bubble has 1/r = 1/r₁ + 1/r₂.

Answer from the NTA final answer key. Solution by PrepWiser.