A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ................ cm.

Solution

Physics sir · 3 steps

Climb out layer by layer: each film adds its own 4T/r, so the two excess pressures add up.

1. One film. A soap film has two surfaces, so the pressure inside a bubble of radius r exceeds the pressure just outside it by 4T/r.
2. Two films stacked. Going from the atmosphere into the big bubble, pressure rises by 4T/6. Crossing the small bubble's film, it rises again by 4T/3. The gas in the small bubble is above atmosphere by the sum.

   4T/r = 4T/3 + 4T/6 ⇒ 1/r = 1/3 + 1/6 = 1/2

3. So. r = (3 × 6)/(3 + 6) = 2 cm, so the equivalent bubble has radius 2 cm.

The common slip: 6. Subtracting the two excess pressures instead of adding them gives 1/r = 1/3 - 1/6, so 6 cm, not 2.

The idea to keep. For bubbles nested inside each other, excess pressures add from outside in. That makes the equivalent radius follow 1/r = 1/r₁ + 1/r₂, like resistors in parallel.

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