Question

Work of $6.0\times {10}^{-4}$ joule is required to be done in increasing the size of a soap film from $10cm\times 6cm$ to $10cm\times 11cm$. The surface tension of the film is:

• A. $5\times {10}^{-2}N/m$
• B. $6\times {10}^{-2}N/m$
• C. $1.5\times {10}^{-2}N/m$
• D. $1.2\times {10}^{-2}N/m$

Answer 1

Answer: (B)

$6\times {10}^{-2}N/m$

$\Delta A=2[10\times 11-10\times 6]$ film has 2 surfaces

$=2\times 50c{m}^2$

$=2\times 50\times {10}^{-4}{m}^2$

$\Delta \bigcup =\Delta AT$

$6\times {10}^{-4}=2\times 50\times {10}^{-4}\times T$

$T=6\times {10}^{-2}N/m$