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A water film spans the gap between two parallel straight wires, each 10 cm long and initially 0.5 cm apart. The wires are pulled apart by an additional 1 mm while staying parallel. How much work must be done? (Surface tension of water = 7.2 × 10⁻² N/m.)
- 7.22 × 10⁻⁶ J
- 1.44 × 10⁻⁵ J
- 2.88 × 10⁻⁵ J
- 5.76 × 10⁻⁵ J
Correct answer: 1.44 × 10⁻⁵ J
Solution
A free liquid film has two air-water surfaces. Work done = T * ΔA_total = T * 2 * (L * Δx). With L = 0.10 m, Δx = 1 mm = 0.001 m, T = 7.2e-2 N/m: W = 7.2e-2 * 2 * (0.10*0.001) = 7.2e-2 * 2e-4 = 1.44e-5 J.
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