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. $8.0N{m}^{-1}$
• B. $0.250N{m}^{-1}$
• C. $0.125N{m}^{-1}$
• D. $0.2N{m}^{-1}$

Answer 1

Answer: (C)

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 $A_f=5\times 4=20$ $c{m}^2=20\times {10}^{-4}$ $m^2$

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 $W=S\left(2\Delta A\right)$ where $S$ is the surface tension

$\therefore$ $3\times {10}^{-4}=S\times 2\times 12\times {10}^{-4}$ $⟹S=0.125$ $N{m}^{-1}$