Question

A rectangular film of liquid is extended from $(4cm\times 2cm)$ to $(5cm\times 4cm)$. If the work done is $3\times {10}^{-4}J$, the value of the surface tension of the liquid is

• A. $8.0N/m$
• B. $0.250N/m$
• C. $0.125N/m$
• D. $0.2N/m$

Answer 1

The correct option is C $0.125N/m$

Initial area of the film
$A_i=4\times 2=8$ $c{m}^2=8\times {10}^{-4}$ $m^2$

Final area of the film
$A_f=5\times 4=20c{m}^2=20\times {10}^{-4}{m}^2$

Thus change in area
$\Delta A=(20-8)\times {10}^{-4}=12\times {10}^{-4}{m}^2$

Work done $W=3\times {10}^{-4}J$

Using for soap films $W=S\left(2\Delta A\right)$
$2\Delta A$ because both side area will be taken (top and bottom)
where $S$ is the surface tension
$\therefore$ $3\times {10}^{-4}=S\times 2\times 12\times {10}^{-4}$
$⟹S=0.125$ $N/m$