Question

What is the potential energy of the soap film formed on a frame of area $4 \times 10^{-3} m^{2}$ ? If the area of the film is reduced by half, the change in its potential energy will be (Surface tension of the soap solution is $10^{-3} N / m$ )

• A. $32 \times 10^{-5} J, 16 \times 10^{-5} J$
• B. $16 \times 10^{-5} J, 8 \times 10^{-5} J$
• C. $48 \times 10^{-5} J, 12 \times 10^{-5} J$
• D. $36 \times 10^{-5} J ,2 \times 10^{-5} J$

Answer 1

Answer: (A)

Potential energy of the film
$=2 T A$
$=2 \times 40 \times 10^{-3} \times 4 \times 10^{-3}$
$=32 \times 10^{-5} J$
Change in potential energy
$=2 T \Delta A$
$=2 \times 40 \times 10^{-3}\left[4 \times 10^{-3}-2 \times 10^{-3}\right] $
$=16 \times 10^{-5} J$