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Hint: The surface tension of the liquid caused by the attraction of the molecules. The surface tension of the liquid shrinks the molecules of the liquid. The liquid resists external force due to the surface tension of it.
The relationship between the work done and the surface tension is given as:
$W=T\times 2\times \Delta A$
Complete step-by-step answer:
Initial dimension of rectangle, ${{l}_{1}}\times {{b}_{1}}=4\ \text{cm}\times \text{2}\ \text{cm}$
Final dimension of rectangle, ${{l}_{2}}\times {{b}_{2}}=5\ \text{cm}\times 4\ \text{cm}$
Work done, W=$3\times {{10}^{-4}}\ \text{J}$
Initial area of the rectangular film is given as:
\[\begin{align}
& \Rightarrow {{A}_{1}}={{l}_{1}}\times {{b}_{1}}=4\ \text{cm}\times 2\ \text{cm} \\
& \Rightarrow {{A}_{1}}=8\ \text{c}{{\text{m}}^{\text{2}}} \\
\end{align}\]
Final area of the rectangular film is given as:
\[\begin{align}
& \Rightarrow {{A}_{2}}={{l}_{2}}\times {{b}_{2}}=5\ \text{cm}\times 4\ \text{cm} \\
& \Rightarrow {{A}_{2}}=20\ \text{c}{{\text{m}}^{\text{2}}} \\
\end{align}\]
Change in area of the rectangular film is given as:
$\begin{align}
& \Rightarrow \Delta A=20\ \text{c}{{\text{m}}^{\text{2}}}-8\ \text{c}{{\text{m}}^{\text{2}}} \\
& \Rightarrow \Delta A=12\ \text{c}{{\text{m}}^{\text{2}}} \\
\end{align}$
The work done required to extend the fluid is given as:
$W=T\times 2\times \Delta A$
Where, T is the surface tension, is the change in area of the rectangular film.
$\begin{align}
& \Rightarrow T=\dfrac{W}{2\Delta A} \\
& \Rightarrow T=\dfrac{3\times {{10}^{-4}}\ \text{J}}{2\times 12\times {{10}^{4}}\ {{\text{m}}^{\text{2}}}} \\
& \Rightarrow T=0.125\ \text{N/m}
\end{align}$
Option C is correct.
Note: The change in area is multiplied by two because there are two surfaces of the rectangular film. The liquid film or bubble is the thin surface of the liquid. The volume inside this film is filled with the air.
The relationship between the work done and the surface tension is given as:
$W=T\times 2\times \Delta A$
Complete step-by-step answer:
Initial dimension of rectangle, ${{l}_{1}}\times {{b}_{1}}=4\ \text{cm}\times \text{2}\ \text{cm}$
Final dimension of rectangle, ${{l}_{2}}\times {{b}_{2}}=5\ \text{cm}\times 4\ \text{cm}$
Work done, W=$3\times {{10}^{-4}}\ \text{J}$
Initial area of the rectangular film is given as:
\[\begin{align}
& \Rightarrow {{A}_{1}}={{l}_{1}}\times {{b}_{1}}=4\ \text{cm}\times 2\ \text{cm} \\
& \Rightarrow {{A}_{1}}=8\ \text{c}{{\text{m}}^{\text{2}}} \\
\end{align}\]
Final area of the rectangular film is given as:
\[\begin{align}
& \Rightarrow {{A}_{2}}={{l}_{2}}\times {{b}_{2}}=5\ \text{cm}\times 4\ \text{cm} \\
& \Rightarrow {{A}_{2}}=20\ \text{c}{{\text{m}}^{\text{2}}} \\
\end{align}\]
Change in area of the rectangular film is given as:
$\begin{align}
& \Rightarrow \Delta A=20\ \text{c}{{\text{m}}^{\text{2}}}-8\ \text{c}{{\text{m}}^{\text{2}}} \\
& \Rightarrow \Delta A=12\ \text{c}{{\text{m}}^{\text{2}}} \\
\end{align}$
The work done required to extend the fluid is given as:
$W=T\times 2\times \Delta A$
Where, T is the surface tension, is the change in area of the rectangular film.
$\begin{align}
& \Rightarrow T=\dfrac{W}{2\Delta A} \\
& \Rightarrow T=\dfrac{3\times {{10}^{-4}}\ \text{J}}{2\times 12\times {{10}^{4}}\ {{\text{m}}^{\text{2}}}} \\
& \Rightarrow T=0.125\ \text{N/m}
\end{align}$
Option C is correct.
Note: The change in area is multiplied by two because there are two surfaces of the rectangular film. The liquid film or bubble is the thin surface of the liquid. The volume inside this film is filled with the air.
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