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Question: The work done in increasing the size of a soap film from \(10{\text{cm}} \times {\text{6cm}}\) to \(...

The work done in increasing the size of a soap film from 10cm×6cm10{\text{cm}} \times {\text{6cm}} to 10cm×11cm10{\text{cm}} \times 11{\text{cm}} is 3×10−4J3 \times {10^{ - 4}}{\text{J}} . Find the surface tension of the film.
A) 1×10−2Nm−11 \times {10^{ - 2}}{\text{N}}{{\text{m}}^{ - 1}}
B) 6×10−2Nm−16 \times {10^{ - 2}}{\text{N}}{{\text{m}}^{ - 1}}
C) 3×10−2Nm−13 \times {10^{ - 2}}{\text{N}}{{\text{m}}^{ - 1}}
D) 1⋅5×10−2Nm−11 \cdot 5 \times {10^{ - 2}}{\text{N}}{{\text{m}}^{ - 1}}

Explanation

Solution

To increase the size of the soap film work must be done. This work done will be equal to the surface energy of the film. The surface energy of the soap film depends on the surface tension of the film and the change in the area of the film.

Formula used:
The surface energy of a soap film is given by, SE=SΔASE = S\Delta A where SS is the surface tension of the film, and ΔA\Delta A is the change in the area of the soap film.

Complete step by step answer:
Step 1: List the parameters known from the question.
The initial area of the film is A=10cm×6cmA = 10{\text{cm}} \times {\text{6cm}} and the final area of the film is A′=10cm×11cmA' = 10{\text{cm}} \times 11{\text{cm}} .
The work done in obtaining this increase in the area of the soap film is W=3×10−4JW = 3 \times {10^{ - 4}}{\text{J}} .
This work done will be equal to the surface energy of the soap film i.e., SE=W=3×10−4JSE = W = 3 \times {10^{ - 4}}{\text{J}} .
Step 2: Express the relation for the surface energy of the soap film to obtain the surface tension of the film.
The surface energy of a soap film is given by, SE=SΔASE = S\Delta A --------- (1)
where SS is the surface tension of the face of the film and ΔA\Delta A is the change in the area of the soap film.
The change in area will be ΔA=2(A′−A)\Delta A = 2\left( {A' - A} \right)
Substituting for A=10cm×6cmA = 10{\text{cm}} \times {\text{6cm}} and A′=10cm×11cmA' = 10{\text{cm}} \times 11{\text{cm}} in the above relation we get,
⇒ΔA=2[(10×11)−(10×6)]=100cm2\Rightarrow \Delta A = 2\left[ {\left( {10 \times 11} \right) - \left( {10 \times {\text{6}}} \right)} \right] = 100{\text{c}}{{\text{m}}^2}
Substituting for SE=3×10−4JSE = 3 \times {10^{ - 4}}{\text{J}} and ΔA=10−2m2\Delta A = {10^{ - 2}}{{\text{m}}^2} in equation (1) we get,
⇒3×10−4=S×10−2\Rightarrow 3 \times {10^{ - 4}} = S \times {10^{ - 2}}
⇒S=3×10−410−2=3×10−2Nm−1\Rightarrow S = \dfrac{{3 \times {{10}^{ - 4}}}}{{{{10}^{ - 2}}}} = 3 \times {10^{ - 2}}{\text{N}}{{\text{m}}^{ - 1}}

∴\therefore The surface tension of the film is S=3×10−2Nm−1S = 3 \times {10^{ - 2}}{\text{N}}{{\text{m}}^{ - 1}}. Thus the correct option is (C).

Note:
A soap film has two sides. So while calculating the change in surface area of the film, we multiply the change by 2 as the change in area is for both the faces of the film. Also, while substituting values of physical quantities in any equation make sure that all the quantities are expressed in their respective S.I. units. If not, then the necessary conversion of units must be done. Here, the change in the area ΔA=100cm2\Delta A = 100{\text{c}}{{\text{m}}^2} is converted to ΔA=10−2m2\Delta A = {10^{ - 2}}{{\text{m}}^2} while substituting in equation (1).