A piece of wire has resistance of 1.25 2. The wire is made of a material with resistivity 8.2x10-7 Qm. What is the dimension of the wire? O Length = 10 m; diameter = 1.45 mm Length = 10 m; diameter = 2.89 mm O Length 15 m; diameter = 3.14 mm O Length 15 m; diameter = 1.11 cm
A piece of wire has resistance of 1.25 2. The wire is made of a material with resistivity 8.2x10-7 Qm. What is the dimension of the wire? O Length = 10 m; diameter = 1.45 mm Length = 10 m; diameter = 2.89 mm O Length 15 m; diameter = 3.14 mm O Length 15 m; diameter = 1.11 cm
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Chapter1: Units, Trigonometry. And Vectors
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![### Determining the Dimensions of a Wire Given Resistance and Resistivity
**Question:**
A piece of wire has a resistance of 1.25 Ω. The wire is made of a material with resistivity \(8.2 \times 10^{-7} \, \Omega \cdot \text{m}\). What is the dimension of the wire?
**Options:**
1. Length = 10 m; diameter = 1.45 mm
2. Length = 10 m; diameter = 2.89 mm
3. Length = 15 m; diameter = 3.14 mm
4. Length = 15 m; diameter = 1.11 cm
---
To solve this problem, we'll use the formula for resistance in terms of resistivity, length, and cross-sectional area:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) (rho) is the resistivity,
- \( L \) is the length,
- \( A \) is the cross-sectional area.
The cross-sectional area \( A \) for a wire with a circular cross-section can be expressed as:
\[ A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4} \]
Plugging this into the resistance formula gives:
\[ R = \frac{4 \rho L}{\pi d^2} \]
Given \( R = 1.25 \, \Omega \) and \( \rho = 8.2 \times 10^{-7} \, \Omega \cdot \text{m} \), we can calculate the dimensions \( L \) and \( d \) for each option to verify which one satisfies the resistance formula.
**Calculations:**
- For each option, compute the value of \(\frac{4 \rho L}{\pi d^2}\) and check if it equals the given resistance \(1.25 \, \Omega\).
Finally, select the option where the calculated resistance is closest to 1.25 Ω.
This mathematical analysis helps us understand the relationship between resistance and physical dimensions, crucial for applications in electronics and material science.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff89c5e82-2527-49f4-96f0-d8c4ad99581b%2F4cd3bd9c-303a-454c-bca2-fa1e8b7510fd%2F091i6q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Dimensions of a Wire Given Resistance and Resistivity
**Question:**
A piece of wire has a resistance of 1.25 Ω. The wire is made of a material with resistivity \(8.2 \times 10^{-7} \, \Omega \cdot \text{m}\). What is the dimension of the wire?
**Options:**
1. Length = 10 m; diameter = 1.45 mm
2. Length = 10 m; diameter = 2.89 mm
3. Length = 15 m; diameter = 3.14 mm
4. Length = 15 m; diameter = 1.11 cm
---
To solve this problem, we'll use the formula for resistance in terms of resistivity, length, and cross-sectional area:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) (rho) is the resistivity,
- \( L \) is the length,
- \( A \) is the cross-sectional area.
The cross-sectional area \( A \) for a wire with a circular cross-section can be expressed as:
\[ A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4} \]
Plugging this into the resistance formula gives:
\[ R = \frac{4 \rho L}{\pi d^2} \]
Given \( R = 1.25 \, \Omega \) and \( \rho = 8.2 \times 10^{-7} \, \Omega \cdot \text{m} \), we can calculate the dimensions \( L \) and \( d \) for each option to verify which one satisfies the resistance formula.
**Calculations:**
- For each option, compute the value of \(\frac{4 \rho L}{\pi d^2}\) and check if it equals the given resistance \(1.25 \, \Omega\).
Finally, select the option where the calculated resistance is closest to 1.25 Ω.
This mathematical analysis helps us understand the relationship between resistance and physical dimensions, crucial for applications in electronics and material science.
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