A Conduckor has an inikiah Length L1 and cross Seetionah area, A. Ik is drawn uniformly to a final length L2 =10L1 and anealed. If the conduetor at the new Length is F5se, us hat is the initial value of R
A Conduckor has an inikiah Length L1 and cross Seetionah area, A. Ik is drawn uniformly to a final length L2 =10L1 and anealed. If the conduetor at the new Length is F5se, us hat is the initial value of R
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![### Problem Statement
A conductor has an initial length \( L_1 \) and cross-sectional area \( A \). It is drawn uniformly to a final length \( L_2 = 10L_1 \) and annealed. If the resistance of the conductor at the new length is 75Ω, what is the initial value of \( R \)?
### Explanation
This problem involves understanding the relationship between the length, cross-sectional area, and resistance of a conductor. When the conductor is uniformly drawn out to a longer length, its cross-sectional area decreases. The change in these dimensions affects the electrical resistance of the conductor.
Given:
- Initial length of the conductor: \( L_1 \)
- Final length of the conductor: \( L_2 = 10L_1 \)
- Final resistance of the conductor: 75Ω
To find:
- Initial resistance of the conductor: \( R \)
### Key Concepts
1. **Resistance of a Conductor**:
The resistance \( R \) of a conductor is given by the formula:
\[
R = \rho \frac{L}{A}
\]
where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.
2. **Effect of Drawing on Dimensions**:
When the conductor is drawn out to a new length, its volume remains constant. Therefore:
\[
L_1 \cdot A_1 = L_2 \cdot A_2
\]
Simplifying, since \( L_2 = 10L_1 \):
\[
A_2 = \frac{A_1}{10}
\]
3. **New Resistance**:
With the new length and cross-sectional area, the new resistance \( R_2 \) can be written as:
\[
R_2 = \rho \frac{L_2}{A_2} = \rho \frac{10L_1}{A_1 / 10} = 100 \rho \frac{L_1}{A_1}
\]
Given \( R_2 = 75Ω \), we can solve for the initial resistance \( R_1 \):
\[
R_2 = 100R_1 \Rightarrow 75Ω = 100R_1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5db72156-5f58-48bc-bfda-30e04998aaf5%2F36cc6e44-bb97-4396-bfa0-6ed3ca416861%2Fqvj2ou_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A conductor has an initial length \( L_1 \) and cross-sectional area \( A \). It is drawn uniformly to a final length \( L_2 = 10L_1 \) and annealed. If the resistance of the conductor at the new length is 75Ω, what is the initial value of \( R \)?
### Explanation
This problem involves understanding the relationship between the length, cross-sectional area, and resistance of a conductor. When the conductor is uniformly drawn out to a longer length, its cross-sectional area decreases. The change in these dimensions affects the electrical resistance of the conductor.
Given:
- Initial length of the conductor: \( L_1 \)
- Final length of the conductor: \( L_2 = 10L_1 \)
- Final resistance of the conductor: 75Ω
To find:
- Initial resistance of the conductor: \( R \)
### Key Concepts
1. **Resistance of a Conductor**:
The resistance \( R \) of a conductor is given by the formula:
\[
R = \rho \frac{L}{A}
\]
where \( \rho \) is the resistivity of the material, \( L \) is the length, and \( A \) is the cross-sectional area.
2. **Effect of Drawing on Dimensions**:
When the conductor is drawn out to a new length, its volume remains constant. Therefore:
\[
L_1 \cdot A_1 = L_2 \cdot A_2
\]
Simplifying, since \( L_2 = 10L_1 \):
\[
A_2 = \frac{A_1}{10}
\]
3. **New Resistance**:
With the new length and cross-sectional area, the new resistance \( R_2 \) can be written as:
\[
R_2 = \rho \frac{L_2}{A_2} = \rho \frac{10L_1}{A_1 / 10} = 100 \rho \frac{L_1}{A_1}
\]
Given \( R_2 = 75Ω \), we can solve for the initial resistance \( R_1 \):
\[
R_2 = 100R_1 \Rightarrow 75Ω = 100R_1
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