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Chapter 12: Problem 38

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 increased by \(1 \mathrm{~mm}\), while still maintaining their parallelism, how much work will have to be done? (surface tension of water is \(\left.7.2 \times 10^{-2} \mathrm{Nm}^{-1}\right)\) (a) \(\left.7.22 \times 10^{-6}\right]\) (b) \(1.44 \times 10^{-5} \mathrm{~J}\) (c) \(\left.2.88 \times 10^{-8}\right]\) (d) \(5.76 \times 10^{-5} \mathrm{~J}\)

Short Answer

Expert verified
The work done is \(1.44 \times 10^{-5} \text{ J}\). Option (b) is correct.

Step by step solution

01

Understand the Problem

We need to calculate the work done to increase the separation between two parallel wires in a water film. The work done depends on the change in surface area and the surface tension of water.
02

Define Initial and Final Conditions

Initially, the wires are separated by \(0.2\, \text{cm}\), and each wire is \(10\, \text{cm}\) long. The separation is increased by \(0.1\, \text{cm}\), so the final separation is \(0.3\, \text{cm}\).
03

Calculate Initial Surface Area

The initial surface area of the water film is calculated using the formula: \(2 \times \text{length of wires} \times \text{initial separation}\). So, \[ A_\text{initial} = 2 \times 10\, \text{cm} \times 0.2\, \text{cm} = 4\, \text{cm}^2.\]
04

Calculate Final Surface Area

The final surface area, after increasing the separation, is \(2 \times \text{length of wires} \times \text{final separation}\). Thus, \[ A_\text{final} = 2 \times 10\, \text{cm} \times 0.3\, \text{cm} = 6\, \text{cm}^2.\]
05

Calculate Change in Surface Area

The change in the surface area is \( \Delta A = A_\text{final} - A_\text{initial} = 6\, \text{cm}^2 - 4\, \text{cm}^2 = 2\, \text{cm}^2.\)
06

Convert Units for Surface Area

Convert the change from \(\text{cm}^2\) to \(\text{m}^2\) by using the conversion \(1 \text{ cm}^2 = 10^{-4} \text{ m}^2\). Hence, \(\Delta A = 2 \times 10^{-4}\, \text{m}^2\).
07

Calculate Work Done using Surface Tension

The work done \(W\) is given by \(W = \text{surface tension} \times \Delta A\). Therefore, \[ W = 7.2 \times 10^{-2} \text{ N/m} \times 2 \times 10^{-4} \text{ m}^2 = 1.44 \times 10^{-5} \text{ J}.\]
08

Select the Correct Answer Option

The calculated work done matches the option \(b\). So, the correct answer is \( b) 1.44 \times 10^{-5} \text{ J} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Tension
Surface tension is a fascinating property of liquids that causes the surface layer to behave like a stretched elastic membrane. It arises due to unbalanced molecular cohesive forces at the surface, pulling the molecules downward and sideways. These forces create a tendency to minimize the surface area, which is why droplets form spherical shapes. The unit of surface tension is Newton per meter (N/m).
Understanding surface tension helps explain the incredible abilities of small insects to "walk" on water, the creation of soap bubbles, and even plays a crucial role in biological processes.
The surface tension of water is notably high due to hydrogen bonding, approximately 0.072 N/m at room temperature. This high surface tension is what requires additional force (or work) when trying to change the shape or size of a water surface, like in the problem at hand.
Change in Surface Area
Changing the surface area of a liquid film involves either expanding or contracting the boundary area over which the liquid molecules are spread. In physics, when you change the surface area of a liquid, you are essentially doing work against the surface tension.
Mathematically, the work done is directly proportional to the change in surface area multiplied by the surface tension.
For example, increasing the separation of parallel wires in a water film increases the surface area, requiring work to be done against the surface tension.
  • Initially, the surface area is determined by the product of the length of the wires and their initial separation.
  • Similarly, when the separation is increased, the new or final surface area is calculated. The change in surface area \( \Delta A \) is the final surface area minus the initial surface area.
This change determines the amount of work needed to achieve the increased surface area.
Parallel Wires Separation
The separation between parallel wires is an important factor in determining the work done when changing the surface area of a liquid film between them. When you have a water film between two wires, increasing their separation requires stretching the film and consequently altering its surface area.
This process requires a force to overcome the surface tension, effectively doing work.
  • Initially, a specified separation creates a certain area of contact between the water film and wires.
  • By increasing the separation, you expand the surface area the film spans, thus requiring an increase in energy input to maintain the film.
In practical terms, the initial and final separation amounts are critical inputs for calculating both the change in surface area and the consequent work needed to move the wires apart. Understanding these principles is crucial in fields ranging from material science to mechanical engineering.

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Most popular questions from this chapter

A body floats in a liquid contained in a beaker. If the whole system falls under gravity, then the upthrust on the body due to liquid is IUP SEE 2009] (a) equal to the weight of the body in air (b) equal to the weight of the body in liquid (c) zero (d) equal to the weight of the immersed part of the body

The rate of flow of liquid through a capillary tube of radius \(r\) is \(V\), when the pressure difference across the two ends of the capillary is \(p\). If pressure is increased by \(3 p\) and radius is reduced to \(r / 2\), then the rate of flow becomes (a) \(\sqrt{V / 9}\) (b) \(3 V / 8\) (c) \(V / 4\) (d) \(V / 3\)

A rain drop of radius \(1.5 \mathrm{~mm}\), experiences a drag force \(F=\left(2 \times 10^{-5} \mathrm{v}\right) \mathrm{N}\), while falling through air from a height \(2 \mathrm{~km}\), with a velocity \(\bar{v} .\) The terminal velocity of the rain drop will be nearly (use \(g=10 \mathrm{~ms}^{-2}\) ) (a) \(200 \mathrm{~ms}^{-1}\) (b) \(80 \mathrm{~ms}^{-1}\) (c) \(7 \mathrm{~ms}^{-1}\) (d) \(3 \mathrm{~ms}^{-1}\)

An aeroplane of \(\operatorname{mass} 3 \times 10^{4} \mathrm{~kg}\) and total wing area of \(120 \mathrm{~m}^{2}\) is in a level flight at some height. The difference in pressure between the upper and lower surfaces of its wings in kilo pascal is \(\left(g=10 \mathrm{~ms}^{-2}\right.\) ) (a) \(2.5\) (b) \(5.0\) (c) \(10.0\) (d) \(12.5\)

A fluid flows through a horizental pipe having two different cross-sections of area \(A\) and \(2 A\). If the pressure at the thin cross-section is \(p\) and fluid velocity is \(v\), the velocity and pressure at the thicker cross-section is (take the density of fluid as \(\rho\) ) (a) \(\frac{v}{2}, p+\frac{1}{2} \rho v^{2}\) (b) \(\frac{v}{4}, p+\frac{3}{8} \rho v^{2}\) (c) \(\frac{v}{2}, p+\frac{3}{8} \rho v^{2}\) (d) \(v_{2} p+\frac{3}{4} \rho v^{2}\)
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