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Q.A wire of resistance 1000 Ω1000\ \Omega and length ll is increased to twice its original length. Calculate its new resistance.

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2019Subjective· 1mImportance★★★★★
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Stretching a wire at constant volume doubles its length and halves its area, so resistance — which depends on l/Al/A — goes up by a factor of 4.

Step 1 — Volume conservation

V=A l=const  ⟹  A′=All′=Al2l=A2V = A\,l = \text{const} \implies A' = \frac{Al}{l'} = \frac{Al}{2l} = \frac{A}{2}

Step 2 — New resistance

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