Two soap bubbles of radius 2 cm and 4 cm , respectively, are in contact with each other. The radius of curvature of the common surface, in cm , is _________.

Solution

Physics sir · 4 steps

The common film has its own excess pressure: the pressure difference between the two bubbles.

1. Pressure in each bubble. A soap bubble has two surfaces, so the excess pressure inside is 4T/r. The smaller bubble (r₁ = 2 cm) has the higher pressure, P₀ + 4T/r₁. The larger one (r₂ = 4 cm) has P₀ + 4T/r₂.
2. The common film. The shared film is also a soap film with two surfaces. It has the smaller bubble on one side and the larger bubble on the other. So the pressure difference across it must be 4T/r, where r is its radius of curvature.

   4T/r = 4T/r₁ - 4T/r₂ ⇒ r = (r₁ r₂)/(r₂ - r₁)

3. Put in numbers. r = (2 × 4)/(4 - 2) = 8/2 = 4 cm. The film bulges into the larger bubble, because the smaller bubble pushes harder.
4. So. The radius of curvature of the common surface is 4 cm.

The common slip: 1.33. Adding the reciprocals, 1/r = 1/2 + 1/4, gives r = 4/3 ≈ 1.33 cm. The film feels the difference of the two pressures, not their sum.

The idea to keep. For any curved soap film, the pressure difference across it is 4T/r. Equate that to the difference of the two bubble pressures and the radius drops out. Here T cancels completely.

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