Between two parallel wires, each of length 10 cm a soap film is trapped. If the separation between them is changed from 2 cm to 4 cm, still being parallel, then the work done for this is [Surface tension of soap solution is 50 dyne/cm)
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a
2×10−4J
b
10−4J
c
4×10−4J
d
5×10−5J
answer is A.
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Detailed Solution
Work done =2TΔA =2×50×(10×2)=100×20=2000erg =2×10−4J