Question

A thin, rectangular film of liquid is extended from $(2\text{cm}\times 3\text{cm})$ to $(4\text{cm}\times 6\text{cm})$. If the work done in the process is $6\times {10}^{-4}\text{J}$, the value of surface tension of the liquid is

• A. $0.133\text{N/m}$
• B. $0.33\text{N/m}$
• C. $0.166\text{N/m}$
• D. $0.12\text{N/m}$

Answer 1

The correct option is C $0.166\text{N/m}$

The thin rectangular film of liquid has two surfaces (Upper and inner surfaces).
Total Initial area $A_i=2\times 2\times 3{\text{cm}}^2=12{\text{cm}}^2$
Total final area $A_f=2\times 4\times 6{\text{cm}}^2=48{\text{cm}}^2$
Increase in area of film,
$\Delta A=(48-12){\text{cm}}^2=36{\text{cm}}^2$
$\Delta A=36\times {10}^{-4}{\text{m}}^2$
Hence, work done during the process,
$W=T\times \Delta A$
$\Rightarrow T=\frac{W}{\Delta A}=\frac{6\times {10}^{-4}}{36\times {10}^{-4}}\text{N/m}$
$\therefore T=0.166\text{N/m}$