Question

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

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

Answer 1

Answer: (B)

$6\times {10}^{-2}$

$A_1=10\times 6=60{cm}^2=60\times {10}^{-4}{m}^2$

$A_2=10\times 11=110{cm}^2=110\times {10}^{-4}{m}^2$

As the soap film has two free surfaces.

Workdone $=W=T\times 2\Delta A$

$\therefore T=\left\{W/\right(2\cdot \Delta A\left)\right\}$

From given values,

$T=\left[(6\times {10}^{-4})/\{2\times (110-60)\times {10}^{-4}\}\right]$

$=\left(6/100\right)$

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

So option B is correct