Consider a soap film on a rectangular frame of wire of area $3 \times 3 \,cm^2$. If the area of the soap film is increased to $5 \times 5 \,cm^2$, the work done in the process will be (surface tension of soap solution is $2.5 \times 10^{-2} \,N/m$).
- A$9 \times 10^{-6} \,J$
- B$16 \times 10^{-6} \,J$
- C$40 \times 10^{-6} \,J$
- D$80 \times 10^{-6} \,J$
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