The surface tension of soap solution is 3.5 × 10⁻² Nm⁻¹. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is _________ × 10⁻⁴ J.
(take π=22 / 7)

Solution

Physics sir · 4 steps

A soap bubble has two surfaces, inside and outside, so count the area increase twice.

1. Work equals surface energy. At constant temperature, the work done to stretch a film equals surface tension times the new surface area created: W = T × Δ A.
2. Two surfaces. A soap bubble is a thin film with an inner and an outer surface. Each has area 4π r², so the total area is 2 × 4π r² = 8π r².

   W = T × 8π(r₂² - r₁²)

3. Put in numbers. With r₁ = 0.1 m and r₂ = 0.2 m, r₂² - r₁² = 0.04 - 0.01 = 0.03 m². Then W = 3.5 × 10⁻² × 8 × 22/7 × 0.03 = 0.5 × 10⁻² × 5.28 = 2.64 × 10⁻² J.
4. So. W = 264 × 10⁻⁴ J, so the answer is 264.

The common slip: 132. Counting only one surface of the bubble (4π r²) gives 132, exactly half the correct answer.

The idea to keep. Soap bubble means two surfaces, so use 8π r² for its area. A liquid drop or an air bubble inside a liquid has only one surface, so use 4π r² there.

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