The surface tension of a soap solution is 0.03 N·m⁻¹. A soap bubble is expanded so that its diameter increases from 2 cm to 6 cm. If the work required is απ×10⁻⁴ J, what is α?
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Question

The surface tension of a soap solution is 0.03 N·m⁻¹. A soap bubble is expanded so that its diameter increases from 2 cm to 6 cm. If the work required is απ×10⁻⁴ J, what is α?

G1 Physics · Accepted Answer

A soap bubble has two liquid surfaces. Therefore its surface energy is U = 2T(4πR²) = 8πTR². Initial radius: R₁ = 1 cm = 1×10⁻² m. Final radius: R₂ = 3 cm = 3×10⁻² m. The work required equals the increase in surface energy: W = 8πT(R₂²-R₁²). Substitute T = 0.03 N·m⁻¹: W = 8π(0.03)[(3×10⁻²)²-(1×10⁻²)²] = 8π(0.03)(8×10⁻⁴) = 1.92π×10⁻⁴ J. Therefore, α = 1.92. Correct option: C.