A film of soap solution is formed between two straight parallel wires of length 10 cm each separated by 0.5 cm . If their separation is increased by 1 mm while still maintaining their parallelism. How

Question

A film of soap solution is formed between two straight parallel wires of length 10 cm each separated by 0.5 cm . If their separation is increased by 1 mm while still maintaining their parallelism. How much work will have to be done? (surface tension of solution = 65 × 1 0 − 2 N / m )

TestHub · Accepted Answer

The increase in surface area of the film is, Δ A = A 2 ​ − A 1 ​ A 1 ​ = 2 × l × b = 2 × 10 × 1 0 − 2 × 0.5 × 1 0 − 2 A 2 ​ = 2 × l × ( b + 1 ) = 2 × 10 × 1 0 − 2 × ( 0.5 + 0.1 ) × 1 0 − 2 A 2 ​ − A 1 ​ = [ 2 × 10 × 1 0 − 2 × ( 0.5 + 0.1 ) × 1 0 − 2 ] − [ 2 × 10 × 1 0 − 2 × 0.5 × 1 0 − 2 ] = 2 × 1 0 − 4 m 2 ∴ Work done = Increase in surface energy = TdA = ( 65 × 1 0 − 2 ) × ( 2 × 1 0 − 4 ) = 1.3 × 1 0 − 4 = 13 × 1 0 − 5 J

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