A wire of resistance R 1 $R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

Question

Aicharya · Accepted Answer

As per the given question, the change in the length is $∆ l = 2 l$ .
Resistance of a wire is given by, $R = \frac{ρ l}{A}$
Therefore, $R_{1} = ρ \frac{l_{1}}{A_{1}}$
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 $l_{2} A_{2} = l_{1} A_{1} \Rightarrow A_{2} = \frac{A_{1}}{3}$ . Therefore,
$R_{2} = ρ (\frac{3 l_{1}}{\frac{A_{1}}{3}}) = 9 ρ \frac{l_{1}}{A_{1}}$
$∴ \frac{R_{2}}{R_{1}} = 9$

A wire of resistance R 1 $R_{1}$ is drawn out so that its length is increased by twice of its original…

Solution