Problem 36
Question
The work done in increasing the size of a rectangular soap film with dimensions \(8 \mathrm{~cm} \times 3.75 \mathrm{~cm}\) to \(10 \mathrm{~cm} \times 6 \mathrm{~cm}\) is \(2 \times 10^{-4} \mathrm{~J}\). The surface tension of the film in \(\mathrm{Nm}^{-1}\) is (a) \(1.65 \times 10^{-2}\) (b) \(3.3 \times 10^{-2}\) (c) \(6.6 \times 10^{-2}\) (d) \(8.25 \times 10^{-2}\)
Step-by-Step Solution
Verified Answer
The surface tension is (b) \(3.3 \times 10^{-2} \text{ Nm}^{-1}\).
1Step 1: Calculate Initial Area of the Soap Film
The initial dimensions of the soap film are given as \(8 \text{ cm} \times 3.75 \text{ cm}\). Calculate the area by multiplying the length and width: \[ A_1 = 8 \times 3.75 = 30 \text{ cm}^2 \]. Since the film has two surfaces (front and back), the effective initial area is \(2 \times 30 = 60 \text{ cm}^2\).
2Step 2: Calculate Final Area of the Soap Film
The final dimensions of the soap film are given as \(10 \text{ cm} \times 6 \text{ cm}\). Calculate the area by multiplying the length and width: \[ A_2 = 10 \times 6 = 60 \text{ cm}^2 \]. Taking both surfaces into account, the effective final area is \(2 \times 60 = 120 \text{ cm}^2\).
3Step 3: Calculate Change in Area
The change in area of the soap film can be calculated by subtracting the initial area from the final area: \[ \Delta A = A_2 - A_1 = 120 \text{ cm}^2 - 60 \text{ cm}^2 = 60 \text{ cm}^2 \].
4Step 4: Convert Area Change to Square Meters
Since surface tension is typically given in \(\text{Nm}^{-1}\), convert the area from square centimeters to square meters: \[ 60 \text{ cm}^2 = 60 \times 10^{-4} \text{ m}^2 = 0.006 \text{ m}^2 \].
5Step 5: Use Work-Energy Principle to Find Surface Tension
The work done is related to surface tension and change in area by the equation \( W = \gamma \times \Delta A \). Rearrange to solve for \(\gamma\): \[ \gamma = \frac{W}{\Delta A} = \frac{2 \times 10^{-4} \text{ J}}{0.006 \text{ m}^2} = 3.33 \times 10^{-2} \text{ Nm}^{-1} \].
6Step 6: Select Closest Answer Choice
From the given options, the closest value to the calculated surface tension \(3.33 \times 10^{-2} \text{ Nm}^{-1}\) is option (b) \(3.3 \times 10^{-2}\).
Key Concepts
work-energy principlesoap filmchange in area
work-energy principle
The work-energy principle is a fundamental concept in physics that relates the work done on an object to the change in its energy. This principle states that the work done by external forces on an object is equal to the change in kinetic energy of that object. However, in the context of surface tension and soap films, it helps us understand how the work done is used to change the area of the film.
When you stretch a soap film, you're essentially doing work to overcome the forces of surface tension, allowing the film's area to increase. Surface tension acts like a "skin" on the surface of the film, and as you increase the film's area, you're stretching this "skin." Using the work-energy principle, the relationship between the work done (\( W \)) and surface tension (\( \gamma \)) can be expressed as:
When you stretch a soap film, you're essentially doing work to overcome the forces of surface tension, allowing the film's area to increase. Surface tension acts like a "skin" on the surface of the film, and as you increase the film's area, you're stretching this "skin." Using the work-energy principle, the relationship between the work done (\( W \)) and surface tension (\( \gamma \)) can be expressed as:
- \( W = \gamma \times \Delta A \) - where \( \Delta A \) is the change in area.
soap film
A soap film is a thin layer of liquid surrounded by air. It forms the colorful bubbles that we often see when playing with soap. Scientifically, a soap film behaves as if it is trying to minimize its surface area due to surface tension, a physical phenomenon where the surface of a liquid acts like an elastic skin.
Soap films can demonstrate various optical effects, such as iridescence, due to the interference of light reflecting from the film's multiple surfaces. As it tries to reduce its energy, a soap film naturally seeks to minimize its surface area, assuming shapes like spheres (in bubbles) when unconfined.
Importantly, when working with soap films, remember that the film has two surfaces – the one facing you and the one facing away from you. Thus, when calculating properties like area, you need to take both surfaces into account, effectively doubling the calculated single-layer area.
Soap films can demonstrate various optical effects, such as iridescence, due to the interference of light reflecting from the film's multiple surfaces. As it tries to reduce its energy, a soap film naturally seeks to minimize its surface area, assuming shapes like spheres (in bubbles) when unconfined.
Importantly, when working with soap films, remember that the film has two surfaces – the one facing you and the one facing away from you. Thus, when calculating properties like area, you need to take both surfaces into account, effectively doubling the calculated single-layer area.
change in area
The concept of change in area is straightforward in the context of the problem, yet it is crucial for understanding how surface tension works. Initially, you start with a piece of soap film of certain dimensions. By applying work, the dimensions of the film change, leading to a new total area.
To calculate the change in area (\( \Delta A \)):
To calculate the change in area (\( \Delta A \)):
- First determine the initial area (\( A_1 \)) of the film based on its starting dimensions and multiply by two, accounting for both surfaces of the film.
- Next, calculate the final area (\( A_2 \)) under the new dimensions, also remembering to double it for both sides.
- The change in area is then \( \Delta A = A_2 - A_1 \).
Other exercises in this chapter
Problem 34
What is the excess pressure inside a bubble of soap solution of radius \(5.00 \mathrm{~mm}\), given that the surface tension of soap solution at the temperature
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A frame made of a metallic wire enclosing a surface area \(A\) is covered with a soap film. If the area of the frame of metallic wire is reduced by \(50 \%\), t
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A mercury drop of radius \(1 \mathrm{~cm}\) is broken into 106 droplets of equal size. The work done is \(\left(S=35 \times 10^{-1} \mathrm{Nm}^{-1}\right)\) (a
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A film of water is found between two straight parallel wires of length \(10 \mathrm{~cm}\) each separated by \(0.2 \mathrm{~cm}\). If their separation is increa
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