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