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A wire has an initial resistance of 4Ω, calculated using R = ρℓ/A. If the wire is stretched such that its volume remains unchanged, what will be the resistance of the stretched wire?

1. The resistance stays unchanged
2. The resistance changes to 8Ω
3. The resistance changes to 16Ω
4. The resistance changes to 32Ω

Correct answer: The resistance changes to 16Ω

Solution

When a wire is stretched, its length increases and its cross-sectional area decreases, but the volume remains constant. Resistance is proportional to the square of the length when volume is constant. Since the length doubles, the resistance becomes 4 times the original, i.e., 4Ω × 4 = 16Ω.

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