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Q.A metal wire of resistance 2Ω is stretched so that its length becomes double the original length. In this condition, the resistance of the wire will be –

(a) 2Ω
(b) 1Ω
(c) 4Ω
(d) 8Ω
Tripura TbseHigher Secondary (+2 Stage) Examination 2024MCQ· 1mImportance★★★★★
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When a wire is stretched, its volume stays constant, so as length doubles, cross-sectional area halves — and resistance depends on both length and area, so it grows by a factor of 4.

Resistance is R=ρLAR = \rho \dfrac{L}{A}. When the wire is stretched, its volume V=ALV = A L remains constant (same amount of metal). If the new length is L′=2LL' = 2L, the new area must be A′=ALL′=AL2L=A2A' = \dfrac{AL}{L'} = \dfrac{AL}{2L} = \dfrac{A}{2}.

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