A metal wire of resistance $6 Ω$ is stretched so that its length is increased to twice of original length. Calculate its new resistance.

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Solution

Step 1 Given data:
Resistance $(R_{1}) = 6 Ω$
Length $(l_{1}) = l$
Length $(l_{2}) = 2 l_{1} = 2 l$
Step 2 Formula used:
Resistance $R_{1} = ρ \frac{l_{1}}{A_{1}} a n d R_{2} = ρ \frac{l_{2}}{A_{2}}$
Step 3 Finding the new resistance:
Since, the volume of metal wire remains same.
Therefore,
$\begin{array}{rcl} A_{1} l_{1} & = & A_{2} l_{2} \\ \frac{l_{2}}{l_{1}} & = & \frac{A_{1}}{A_{2}} . . . . . . . (1) \\ D i v i d e R_{1} a n d R_{2} \\ \frac{R_{2}}{R_{1}} & = & \frac{l_{2}}{A_{2}} \times \frac{A_{1}}{l_{1}} \\ = & \frac{l_{2}}{l_{1}} \times \frac{A_{1}}{A_{2}} \\ = & \frac{l_{2}}{l_{1}} \times \frac{l_{2}}{l_{1}} (F r o m 1) \\ R_{2} & = & R_{1} {(\frac{l_{2}}{l_{1}})}^{2} \\ = & 6 {(\frac{2 l}{l})}^{2} \\ = & 24 Ω \end{array}$
Therefore, the new resistance is $24 Ω$ .

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