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Physics Question on mechanical properties of fluid

Consider a soap film on a rectangular frame of wire of area 4×4 cm24 \times 4\, cm ^{2}. If the area of the soap film is increased to 4×5 cm24 \times 5 \,cm ^{2}, the work done in the process will be (The surface tension of the soap film is 3×10−2 N/m.3 \times 10^{-2} \,N / m . )

A

12×10−6J12\times 10^{-6}J

B

24×10−6J24\times 10^{-6}J

C

60×10−6J60\times 10^{-6}J

D

96×10−6J96\times 10^{-6}J

Answer

24×10−6J24\times 10^{-6}J

Explanation

Solution

Work done, W=2T×AW=2 T \times A
W=6×10−2×[4×5−4×4]×10−4W=6 \times 10^{-2} \times[4 \times 5-4 \times 4] \times 10^{-4}
=6×10−6×4=6 \times 10^{-6} \times 4
=24×10−6J=24 \times 10^{-6} J
(Film has two free surfaces).