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If a wire of resistance $1 Ω$ is stretched to double its length, then resistance will be

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1/4 Ω

2 Ω

4 Ω

1/2 Ω

answer is D.

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Detailed Solution

R = 1 Ω

$l_{1} = 2 l$

$area = A/2$

$R = \frac{ρ l}{A}$

${new resistance R}^{|} = ρ (\frac{2 l}{A/2})$

$R^{|} = 4 \frac{ρ l}{A} \Rightarrow R^{|} = 4 R$

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