Question

Work of $3.0\times {10}^{-4}$ joule is required to be done in increasing the size of a soap film from $10cm\times 6cmto10cm\times 11cm$ The surface tension of the film is

• A. $5\times {10}^{-2}N/m$
• B. $3\times {10}^{-2}N/m$
• C. $1.5\times {10}^{-2}N/m$
• D. $1.2\times {10}^{-2}N/m$

Answer 1

Answer: (B)

$3\times {10}^{-2}N/m$

Area increased $=(10\times 11-10\times 6)c{m}^2$

$=50c{m}^2$

Since film has 2 sides
∴ total increased area $50\times 2=100c{m}^2$

work done = surface tension x increase in surface area

$surfacetension=\frac{workdone}{increaseinsurfacearea}$

$\frac{3\times {10}^{-4}}{100\times {10}^{-4}}=0.03N/m=3\times {10}^{-2}N/m$