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JEE PHYSICS MCQ: MECHANICAL PROPERTIES OF FLUIDS

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A film of water is formed between two straight parallel wires of length 20 cm20 \mathrm{~cm} with separation of 0.5 cm0.5 \mathrm{~cm}. The work done to increase the separation by 0.1 cm0.1 \mathrm{~cm} is:
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Correct option: B
28×10−6 J28 \times 10^{-6} \mathrm{~J}
Option B (28×10−6 J28 \times 10^{-6} \mathrm{~J}) is correct because it is the work done to increase the separation, calculated using the surface tension and the change in surface area.

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