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25.80 If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area \(A\) of the coating compare when the thread is \(13 \mathrm{~mm}\) long versus the starting length of \(5 \mathrm{~mm} ?\) Assume that the resistivity of the coating remains constant and the coat- (b) \(\frac{1}{4} A_{5 \mathrm{~mm}}\) ing is uniform along the thread. \(A_{13 \mathrm{~mm}}\) is about (a) \(\frac{1}{10} A_{5 \mathrm{~mm}}\) (c) \(\frac{2}{5} A_{5 \mathrm{~mm}} ;\) (d) the same as \(A_{5 \mathrm{~mm}}\).

Short Answer

Expert verified
\(A_{13mm} \approx 2/5 A_{5mm}\)

Step by step solution

01

Understand the concept of resistivity

Resistivity is a physical material property that quantifies how strongly a given material opposes the flow of electric current. A low resistivity indicates a material that readily allows the flow of electric current. Here, you're told that the resistivity is constant, meaning it's the same no matter how large the cross-sectional area of the coating becomes.
02

Understand the Role of the Cross-Sectional Area in Conductivity

The amount of conductivity depends on the cross-sectional area of the conductor in this case, the coating. A larger cross-sectional area would mean more paths for the current to travel, resulting in higher conductivity. Since the coating is uniform and the resistivity is constant, we can assume that as the length increases, the cross-sectional area decreases to maintain the same volume. Considering the volume of a cylinder which is \(V = \pi r^2 h\) where \(r\) is the radius (contributing to the cross-sectional area = \(\pi r^2\)) and \(h\) is the height (length of the thread)
03

Compare the Cross-Sectional Area for Different Lengths

Now, compare the cross-sectional area (\(A\)) of the coating when the thread lengths are 5mm and 13mm. If we consider the volume of the 5mm section (which gives us the volume of the coating if we assume uniformity), we can relate that to the volume of the 13mm section. Given that these two volumes are equal (due to the conservation of material), we can set up the following relation: \(\pi r_{5mm}^2 \times 5 = \pi r_{13mm}^2 \times 13\). Simplifying, we find: \(r_{13mm}^2 = (r_{5mm}^2 \times 5)/13\) or \(A_{13mm}= (A_{5mm} \times 5)/13\) due to the area being dependent on the square of the radius.
04

Evaluate the Results

Looking at the given options, we can see that none of the given options exactly matches the result obtained above. However, as \(5/13 = 0.385\) and this is very close to \(2/5 = 0.4\), we can consider \(2/5 A_{{5mm}}\) as a good approximation.

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Key Concepts

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

Conductivity
Conductivity is a fascinating aspect of materials that defines how easily electric current can pass through them. When we talk about conductivity, we're referring to the ability of a material to conduct electrical flow.
  • High conductivity means the material allows current to flow easily.
  • Low conductivity implies the material resists the flow of electricity.
Conductivity is inversely related to resistivity, another important material property. This means if a material has low resistivity, it will have high conductivity, and vice versa. The conductivity of a thread, as mentioned in the exercise, is crucially affected by the cross-sectional area and the material's inherent resistivity. For our specific problem, the conductivity doesn't change as the resistivity remains constant. However, it is intrinsically linked to the cross-sectional area of the coating. This is because the larger the area, the more space there is for the current to navigate through. Think of it like a highway: more lanes mean less traffic congestion, resulting in smoother traffic flow, representing easier passage for the electric current.
Cross-Sectional Area
The cross-sectional area is vital when discussing the flow of electricity in a conductor. It refers to the area of the slice you would see if you were to cut straight through the conductor at any point. For example, this would be a circle when slicing through a round wire.
  • More area implies more room for current to pass through.
  • Less area means less space, which can restrict current flow.
In the problem, the cross-sectional area of the coating made of the aqueous solution directly influences the overall conductivity of the thread. If the thread's length increases, maintaining the same volume, the cross-sectional area must decrease to compensate, assuming uniform coating thickness.The mathematical relationship here shows that as the length increases from 5mm to 13mm, the cross-sectional area reduces according to the volume conservation principle in cylinders: the equation becomes \[A_{13mm} = \frac{A_{5mm} \times 5}{13}\]. This tells us the area reduces as the thread stretches, shrinking the space available for current passage.
Material Property
Material properties such as resistivity and conductivity define how a substance behaves under electric current. These intrinsic properties are consistent for a material and dictate whether it's an excellent conductor or insulator.
  • Resistivity: A measure of how strongly a material opposes the flow of electric current.
  • Conductivity: The ability of a material to allow the flow of electric current.
In our example, the resistivity of the aqueous coating is constant, meaning the way it resists or allows current flow does not change regardless of the coating's shape or size. This assumption simplifies analyses as it means that any variations in conductivity are due only to changes in the geometric factors, like the cross-sectional area, rather than differences in how the material itself conducts electricity. Knowing that the coating's resistivity remains unchanged provides a clear path to understanding how structural changes, such as lengthening the thread, impact the overall conductivity due to changes in cross-sectional area.

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