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  4. 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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Q. A wire of resistance $R_1$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:

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Solution:

$ R _1=\rho \frac{ L _1}{ A _1} $
$R _2=\rho\left(\frac{3 L _1}{ A _1 / 3}\right)=9 \rho \frac{ L _1}{ A _1} $
$\therefore \frac{ R _2}{ R _1}=9$