A wire of resistance R₁ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is :

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3 : 1

9 : 1

1 : 9

4 : 1

answer is A.

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Detailed Solution

$\begin{array}{l} R = \frac{ρ l}{A} = \frac{ρ l^{2}}{A l} \\ ∵ A l = V o l u m e = constant \\ \Rightarrow R \propto l^{2} \\ \Rightarrow \frac{R}{R_{1}} = {(\frac{1 + 2}{1})}^{2} = \frac{3^{2}}{1} = \frac{9}{1} \\ ⇒ Ratio = 9:1 \end{array}$

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