Problem 43
Question
A water film is made between two straight parallel wires of length \(10 \mathrm{~cm}\) separated by \(5 \mathrm{~mm}\) from each other. If the distance between the wires is increased by \(2 \mathrm{~mm}\). How much work will be done? Surface tension for water is 72 dyne \(\mathrm{cm}^{-1}\). (a) \(288 \mathrm{erg}\) (b) \(72 \mathrm{erg}\) (c) \(144 \mathrm{crg}\) (d) \(216 \mathrm{crg}\)
Step-by-Step Solution
Verified Answer
288 erg (option a).
1Step 1: Understanding the Problem
We have a water film created between two parallel wires, each 10 cm long and initially separated by 5 mm. The wires are then moved to increase the separation by 2 mm. We need to calculate the work done, given the surface tension of water as 72 dyne/cm.
2Step 2: Calculate Initial and Final Area of Water Film
Initially, the separation between the wires is 5 mm, so the area of each side of the film is \( A_{\text{initial}} = 10 \times 0.5 = 5 \text{ cm}^2 \). When the separation increases by 2 mm, the final separation becomes 7 mm, thus the final area is \( A_{\text{final}} = 10 \times 0.7 = 7 \text{ cm}^2 \). Since the film has two sides, we need to double these areas.
3Step 3: Calculate the Change in Surface Area
The change in the surface area due to the movement of the wires is calculated as \( \Delta A = 2 \times (A_{\text{final}} - A_{\text{initial}}) = 2 \times (7 - 5) = 4 \text{ cm}^2 \).
4Step 4: Calculate the Work Done
Work is done against the surface tension, which is given by the formula \( W = \gamma \times \Delta A \), where \( \gamma = 72 \text{ dyne/cm} \) is the surface tension.\[ W = 72 \times 4 = 288 \text{ dyne cm} = 288 \text{ erg} \]
5Step 5: Select the Correct Answer
The work done to increase the distance between the wires is 288 erg, which corresponds to option (a) 288 erg.
Key Concepts
Work Done in PhysicsSurface Area ChangeParallel Wires in Physics
Work Done in Physics
In physics, work is defined as the energy transferred to or from an object via the application of force along a displacement. It is calculated by the formula:
The work done can be likened to the force the liquid's surface tension applies over a given distance. This results in an increase in the film's area. For the given exercise, the work done is calculated using the formula:
- \( W = F \times d \), where \( F \) is force and \( d \) is displacement.
The work done can be likened to the force the liquid's surface tension applies over a given distance. This results in an increase in the film's area. For the given exercise, the work done is calculated using the formula:
- \( W = \gamma \times \Delta A \)
Surface Area Change
The concept of surface area change is critical in problems involving thin films. Initially, in the exercise, the water film has a certain surface area based on the separation between two wires. The increase in separation leads to an increase in the surface area of the water film.
This change in area is essential to determine the work done to increase the film's surface, by utilizing the difference \( \Delta A \) in initial and final surface areas.
- Initially, we calculate the surface area with the formula: \( A = \, \text{length} \times \text{width} \).
- With two sides considered, we double the calculated area.
This change in area is essential to determine the work done to increase the film's surface, by utilizing the difference \( \Delta A \) in initial and final surface areas.
Parallel Wires in Physics
Parallel wires are often used within physics to demonstrate concepts related to forces, current flow, and tension within different contexts. In this particular situation, they serve to suspend a water film.
- When placed parallel, they create a frame—holding the water film in the gap between them.
- The distance between these wires is crucial as it directly affects the film's surface area.
Other exercises in this chapter
Problem 41
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