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Questions 3-23 · Q23

Q.A rectangular wire frame of size 2 cm × 2 cm, is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm × 3 cm, calculate the work done in the process. The surface tension of soap film is 3×10⁻² N/m. [Ans. 3×10⁻⁵ J]

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Step 1. A soap film on a wire frame has two surfaces, so its effective area is twice the frame's own area. Initial frame area =2×2=4= 2×2 = 4 cm² =4×10−4= 4×10^{-4} m², so initial two-sided film area =8×10−4= 8×10^{-4} m². Final frame area =3×3=9= 3×3 = 9 cm² =9×10−4= 9×10^{-4} m², so final two-sided film area =18×10−4= 18×10^{-4} m². …

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