Q.

A film of water 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 water = 7 .2 x 10-2 N/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-A12×[(10×0.6)-(10×0.5)]×10-4  =2×10-4 m2Work done = T×∆A= 7.2×10-2×2×10-4 = 1.44×10-5 J
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