A wire of resistance R _ 1 R_ 1 is drawn out so that its length is...

As per the given question, the change in the length is $∆ l = 2l$.

Resistance of a wire is given by, $R = \dfrac{ρl}{A}$

Therefore, ${R}_{1} = ρ\dfrac{{l}_{1}}{{A}_{1}}$

When the length is starched by $2{l}_{1}$, final length will be ${l}_{2} = {l}_{1} + 2{l}_{1} = 3{l}_{1}$. As the volume remains constant ${l}_{2}{A}_{2} = {l}_{1}{A}_{1} ⇒ {A}_{2} = \dfrac{{A}_{1}}{3}$. Therefore,

${R}_{2} = ρ\left(\dfrac{3{l}_{1}}{\dfrac{{A}_{1}}{3}}\right) = 9ρ\dfrac{{l}_{1}}{{A}_{1}}$

$∴ \dfrac{{R}_{2}}{{R}_{1}} = 9$