A nichrome wire of length l and cross-sectional area A has a resistance R. If you stretch the same wire so that its length becomes 2l (assume the volume remains constant, so the area becomes A/2), what will be the new resistance? Show your reasoning.
Generated by claude-sonnet-4-6 · 2026-06-26 01:10 · grounding rag
Model Answer
Using the formula $R = \rho \dfrac{l}{A}$, the original resistance is:
$$R = \rho \frac{l}{A}$$
When the wire is stretched, the new length = $2l$ and new area = $A/2$ (volume constant).
New resistance:
$$R' = \rho \frac{2l}{A/2} = \rho \frac{2l \times 2}{A} = 4\rho \frac{l}{A} = 4R$$
The new resistance is 4R.
Reasoning: Doubling the length doubles the resistance, and halving the area doubles it again. Both effects multiply, giving $2 \times 2 = 4$ times the original resistance.
Source: Chapter 11, Section 11.5 – Factors on which the resistance of a conductor depends.
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Explanation
- Examiners expect you to state the formula $R = \rho \dfrac{l}{A}$, apply it to both old and new dimensions, and show the algebraic steps clearly.
- The key insight is that stretching changes both length and area simultaneously, so both changes must be accounted for — this is what the volume-constant condition tells you.
- Writing the final answer as 4R (in terms of the original) is sufficient; no numerical value is needed.
- Avoid skipping steps — in a 3-mark question, 1 mark is typically for the formula, 1 for correct substitution, and 1 for the final answer.