Problem 80 - University Physics with Modern Physics

Question

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 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread. \(A_{13}\) mm is about (a) \\(\frac{1}{10}\\) \(A_5\) mm; (b) \\(\frac{1}{_4}\\) \(A_5\) mm; (c) \\(\frac{2}{_5}\\) \(A_5\) mm; (d) the same as \(A_5\) mm.

Step-by-Step Solution

Answer

The cross-sectional area of the coating when 13 mm long is closest to (c) \(\frac{2}{5}\) \(A_5\).

Step 1: Understanding the Problem

We need to find the relationship between the cross-sectional area of the coating when the thread is 13 mm long compared to when it is 5 mm long. We know that resistivity is constant and the coating is uniform, which suggests that the total volume of the coating remains the same.

Step 2: Volume Conservation

Since the coating is uniform and the resistivity is constant, the volume of the coating stays the same whether the thread is 5 mm or 13 mm long. The volume, V, of the coating can be given as \(V = A \cdot L\), where \(A\) is the cross-sectional area and \(L\) is the length of the thread.

Step 3: Expressing Cross-Sectional Area

Since the volume remains constant, we can set up the equation \(A_5 \times 5 = A_{13} \times 13\). This allows us to express \(A_{13}\) in terms of \(A_5\): \(A_{13} = \frac{A_5 \times 5}{13}\).

Step 4: Solving for Cross-Sectional Area Ratio

Substitute into the equation: \(A_{13} = \frac{A_5 \times 5}{13}\). Simplifying this, we get \(A_{13} = \left(\frac{5}{13}\right) A_5\).

Step 5: Comparing to the Answer Choices

We need to check which option corresponds to \(\frac{5}{13}\). To approximate, \(\frac{5}{13}\) is approximately \(0.3846\), which is close to \(\frac{2}{5} = 0.4\). Therefore, the correct choice is (c) \(\frac{2}{5}\).

Key Concepts

Cross-Sectional Area

The cross-sectional area is a crucial concept when working with any material that has a defined shape, such as a thread, wire, or rod. In our problem, the cross-sectional area of the coating determines how much conductive material is present along any slice of the thread. To understand how changes in the thread length impact the cross-sectional area, consider a cylinder. Chirally, cross-sectional area is represented as the slice through the object perpendicular to its length.

• When we adjust the length of the thread, the cross-sectional area must also adjust to maintain constant volume if the coating remains uniform.
• The cross-sectional area decreases as the thread elongates because the same volume is distributed over a longer length.

In the exercise, we compare the cross-sectional area at two different lengths: 5 mm and 13 mm. By calculating them, we observe that as the length increases, the cross-sectional area must decrease to conserve volume. This relationship is crucial in applications across physics and engineering where materials change dimensions.

Volume Conservation

Volume conservation plays a critical role in understanding how materials behave when stretched or compressed. With the principle that "volume remains constant," the analysis becomes simpler. In many physical systems, if no material is added or removed, the total volume \( V \) is constant, implying that any change in one measurement (like length) impacts another (like cross-sectional area). This concept can be expressed mathematically as:\[ A_1 \times L_1 = A_2 \times L_2 \]Where:

• \( A \) is the cross-sectional area
• \( L \) is the length of the thread

In our original problem, this principle allows us to deduce the change in cross-sectional area when the thread is stretched from 5 mm to 13 mm. Since the volume stays the same, increasing the length decreases the cross-sectional area proportionally.

Electrical Conductivity

Electrical conductivity indicates how well a material allows the flow of electric current. It is influenced by factors such as the material's resistivity and physical dimensions like cross-sectional area. The formula that defines the relationship between resistivity \( \rho \), length \( L \), cross-sectional area \( A \), and resistance \( R \) is given by:\[ R = \frac{\rho \cdot L}{A} \]In our scenario:

• The resistivity of the coating is constant, meaning any changes to the other variables can impact conductivity.
• A smaller cross-sectional area at a given length increases resistance, thus reducing conductivity.

This is why maintaining uniform resistivity is crucial; the change in cross-sectional area with length affects the overall conductivity of the thread.

Uniform Coating

Uniform coating means the thickness and application of the coating are consistent throughout the surface of the thread. This uniformity is crucial for guaranteeing that the changes in physical properties like resistance or conductivity only result from changes in length or cross-sectional area, and not from variations in the coating itself. Key aspects of the uniform coating include:

• A consistent layer ensures predictable electrical characteristics.
• Uniformity allows the use of simple mathematical relationships to understand changes in the system.

In the problem, the assumption of a uniform coating simplifies calculations because it ensures that the coating's properties do not vary along the length of the thread, allowing us to focus purely on how changes in geometry affect its characteristics.