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Q.

A film of water is formed between two straight parallel wires of length 10cm 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 water =  7.2×10−2N/m)

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a

7.22×10−6 Joule

b

1.44×10−5 Joule

c

2.88×10−5 Joule

d

5.76×10−5 Joule

answer is B.

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Detailed Solution

Increment in area of soap film =  A2−A1 =2×[(10×0.6)−(10×0.5)]×10−4=2×10−4m2Work done =  T×ΔA=7.2×10−2×2×10−4=1.44×10−5J
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