Question

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

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

Step-by-step solution

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 = \(0.1 m\times 0.11 m = 0.011 m^2\)

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 = \(0.011 m^2 - 0.006 m^2 = 0.005 m^2\)

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). \(3 \times 10^{-4}\,J = T \times 0.005\,m^2\)

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).

Key concepts

Work Done in Physics

In physics, work done is an important concept that describes the energy required to perform a task. When dealing with surface tension, work is often related to changes in the surface area of a material, like a soap film. The basic formula for work done is:
\[ W = T \times \Delta A \]
where

• \(W\) represents work done, measured in Joules (J),
• \(T\) is the surface tension, measured in Newtons per meter (N/m), and
• \(\Delta A\) is the change in surface area, measured in square meters (m²).

This formula indicates that the work done is proportional to the change in surface area and the surface tension of the material. In our problem, when a soap film changes size, work is done to increase its surface area, reflecting the energy used in this process.

Change in Surface Area

The change in surface area of a material refers to the difference between the final and initial surface areas. For our soap film, we begin by determining the initial and final areas.
The initial dimensions of the soap film are 10 cm by 6 cm, which translates to 0.1 m by 0.06 m after converting from centimeters to meters. The initial surface area, \(A_1\), is therefore:
\[ A_1 = 0.1 \, \text{m} \times 0.06 \, \text{m} = 0.006 \, \text{m}^2 \]
For the final dimensions, 10 cm by 11 cm converts to 0.1 m by 0.11 m, resulting in a final surface area, \(A_2\):
\[ A_2 = 0.1 \, \text{m} \times 0.11 \, \text{m} = 0.011 \, \text{m}^2 \]
The change in surface area, \(\Delta A\), is the difference between \(A_2\) and \(A_1\):
\[ \Delta A = A_2 - A_1 = 0.011 \, \text{m}^2 - 0.006 \, \text{m}^2 = 0.005 \, \text{m}^2 \]
Understanding how to calculate this change is crucial for solving problems related to work done and surface tension.

Units Conversion

Correct units are crucial in physics as they ensure accuracy and consistency. In this exercise, measurements are first given in centimeters, and it’s vital to convert them to meters because the standard unit for length in the International System of Units (SI) is the meter.
When converting, remember:

• 1 centimeter equals 0.01 meters.

For example, to convert 10 cm to meters, multiply by 0.01 to obtain 0.1 meters.
In addition to length, this concept also applies to other quantities like work and tension. Here, work is presented in Joules, and surface tension is in Newtons per meter (N/m). Consistent units allow us to relate these different quantities using formulas, ensuring calculations are correct and meaningful.

Physics Problem Solving

Approaching physics problems systematically is key to finding accurate solutions. Here’s a step-by-step approach for our exercise:

• Identify what is given: Work done \(W = 3 \times 10^{-4} \) J, and initial and final dimensions.
• Convert all units to standard SI units to maintain consistency.
• Calculate the change in surface area \(\Delta A\).
• Use the work done formula \(W = T \times \Delta A\) to find the unknown, in this case, the surface tension \(T\).
• Solve for \(T\) using known values for \(W\) and \(\Delta A\).

By following each logical step, complex problems become more manageable. This method not only applies to this exercise but can be generalized to other physics problems, promoting structured and effective problem solving.