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Chapter 7: Problem 993

The work done increasing the size of a soap film from $10 \mathrm{~cm} \times 6 \mathrm{~cm}\( to \)10 \mathrm{~cm} \times 11 \mathrm{~cm}\( is \)3 \times 10^{-4}$ Joule. The surface tension of the film is (A) \(1.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (B) \(3.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (C) \(6.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\) (D) \(11.0 \times 10^{-2} \mathrm{~N} / \mathrm{m}\)

Short Answer

Expert verified
The surface tension of the soap film is \(6.0 \times 10^{-2}\) N/m, which corresponds to option (C).

Step by step solution

01

Calculate the initial and final surface areas of the soap film

The initial dimensions of the soap film are 10 cm x 6 cm, and the final dimensions are 10 cm x 11 cm. We must convert these dimensions to meters and then calculate the initial surface area A1 and the final surface area A2 of the soap film. Initial dimensions: \(10 cm \times 6 cm = 0.1 m \times 0.06 m\) Final dimensions: \(10 cm \times 11 cm = 0.1 m \times 0.11 m\) A1 = Length × Width = \(0.1 m \times 0.06 m = 0.006 m^2\) A2 = Length × Width = \(0.1 m\times 0.11 m = 0.011 m^2\)
02

Calculate the change in surface area

Now, we will find the difference between the final surface area (A2) and the initial surface area (A1). This will give us the change in surface area ∆A. ∆A = A2 - A1 = \(0.011 m^2 - 0.006 m^2 = 0.005 m^2\)
03

Substitute the values into the work done formula

We are given that the work done (W) is equal to \(3 \times 10^{-4}\) Joule. Now using the formula W = T × ∆A, we will substitute the values for W and ∆A to find the surface tension (T). \(3 \times 10^{-4}\,J = T \times 0.005\,m^2\)
04

Solve for surface tension (T)

Now, we will solve the equation from the previous step to find the surface tension (T). T = \(\frac{3 \times 10^{-4}\,J}{0.005\,m^2}\) T = \(6 \times 10^{-2}\,\frac{N}{m}\) The surface tension of the soap film is \(6.0 \times 10^{-2}\) N/m, which corresponds to option (C).

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