Question

The work done in increasing the size of a soap film from 10 × 8 cm to 10 × 11 cm is 3 × 10^-4 J. The surface tension of the film is?

• A. 5 × 10^-2 N/m
• B. 5 × 10^-2 dyne/cm
• C. 0.1 N/m
• D. 0.1 dyne/cm

Answer 1

The correct option is A 5 × 10^-2 N/m


Increase in surface area =$2\times 3\times 10c{m}^2=0.006m^2$ {(a factor of 2 to consider both sides of the surface)}
Work done = $\Delta v=3\times {10}^{-4}J$
$SurfaceTension=\frac{\Delta v}{\Delta A}=\frac{3\times {10}^{-4}}{0.006}\frac{J}{m^2}or\frac{N}{m}$
=0.05$\frac{N}{m}$