JEE Main 2023 · 29 January · morning · Single correct · +4 / −1
Surface tension of a soap bubble is 2.0 × 10⁻² Nm⁻¹. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be:
Take [π=22/7]
- (a) 18 .48 × 10⁻⁴ J
- (b) 5.76 × 10⁻⁴ J
- (c) 0.72 × 10⁻⁴ J
- (d) 9.24 × 10⁻⁴ J
Answer and solution
Answer (a) 18 .48 × 10⁻⁴ J
Solution
Physics sir · 4 steps
A soap bubble is a thin film with two surfaces, so count the area change twice before multiplying by T.
- Two surfaces. The soap film has an inner surface and an outer surface, and both stretch as the bubble grows. The new surface area is therefore 2 × 4π r² for a bubble of radius r.
W = T × Δ A = T × 2 × 4π(r₂² - r₁²)
- Area change. In metres, r₁ = 0.035 and r₂ = 0.07, so r₂² - r₁² = (49 - 12.25) × 10⁻⁴ = 36.75 × 10⁻⁴ m². Then Δ A = 8 × 22/7 × 36.75 × 10⁻⁴ = 924 × 10⁻⁴ m².
- Put in numbers. W = 2.0 × 10⁻² × 924 × 10⁻⁴ = 1848 × 10⁻⁶ J.
- So. W = 18.48 × 10⁻⁴ J, which is option (a).
The idea to keep. Surface energy is T times the new area made. A soap bubble has two free surfaces, so use 8π r². A liquid drop has only one surface, so use 4π r².
Answer from the NTA final answer key. Solution by PrepWiser.