Question

If work done in increasing the size of rectangular soap film with dimensions $8cm\times 3.75cmto10cm\times 6cm$ is $2\times 10–4J$. The surface tension of film in newton per meter is

• A. $6.6\times {10}^{-2}$
• B. $165\times {10}^{-2}$
• C. $8.25\times {10}^{-2}$
• D. $3.3\times {10}^{-2}$

Answer 1

The correct option is D

$3.3\times {10}^{-2}$

Step1: Given data

The initial area of the film is $A=8cm\times 3.75cm$

The final area of the film is $A\text{'}=10cm\times 6cm$

The work done in obtaining this increase in the area of the soap film is $W=2\times {10}^{-4}J$

Step2: Surface tension

1. The propensity of fluid surfaces to shrink to the smallest feasible surface area is known as surface tension.
2. The difference in interactions between the molecules of the fluid and the molecules of the flask, beaker, or storage wall can be used to calculate the surface tension.

Step3: Formula used

The surface energy of a soap film is given by

$SE=S∆A[whereS=surfacetensionofthefilm,∆A=changeintheareaofsoap]$

Step4: Relation for the surface energy of the soap film to obtain the surface tension of the film.

The given work done will be equal to the surface energy of the soap

$SE=W=2\times {10}^{-4}J$

Change in the area will be $∆A=2(A\text{'}-A)$

Substituting the values we get

$$\begin{gathered} ∆A=2\left[\right(10\times 6)-(8\times 3.75\left)\right] \\ ∆A=60c{m}^2 \end{gathered}$$

Substituting the value of $SE=W=2\times {10}^{-4}J$ and $∆A=60c{m}^2=60\times {10}^{-4}m$ in the formula

$$\begin{gathered} SE=S∆A \\ 2\times {10}^{-4}=S(60\times {10}^{-4}) \\ S=\frac{2\times {10}^{-4}}{60\times {10}^{-4}} \\ S=3.3\times {10}^{-2}N/m \end{gathered}$$

Hence, option D is the correct answer.