Question:
A wire of resistance $R$ is stretched to double its original length. Its new resistance will be
[H]
[width=0.5\linewidth]{107.png}
A wire of resistance $R$ is stretched to double its original length. Its new resistance will be
[H]
[width=0.5\linewidth]{107.png}
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Remember that resistance is directly proportional to the length of the conductor and inversely proportional to the cross-sectional area.
Updated On: May 31, 2026
- $4R$
- $2R$
- $R/2$
- $R/4$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
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 of the wire, and \( A \) is the cross-sectional area of the wire.
Step 2: Meaning
Resistivity (\( \rho \)) is a property of the material that does not change with the dimensions of the conductor. The resistance depends on the length and the cross-sectional area of the conductor.
Step 3: Analysis
When a wire is stretched to double its original length, several changes occur: 1. The length \( L \) of the wire doubles. 2. The volume of the wire remains constant because it is only being reshaped without any loss or gain in material. 3. Since the volume \( V = A \cdot L \), doubling the length while keeping the volume constant means that the cross-sectional area \( A \) must halve. Thus, the new resistance \( R' \) can be calculated as: \[R' = \rho \frac{2L}{A/2} = 4 \rho \frac{L}{A} = 4R\]
Step 4: Conclusion
Doubling the length of a wire and halving its cross-sectional area quadruples the resistance. Final Answer: (A)
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