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Q.

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
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