A wire of given length is doubled on itself and this process is r… - Vidyadip

Question

A wire of given length is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?

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Answer

As the length of the wire become $\frac{1}{4}$ of the original length its area of cross-section become $4$ times.
So, new resistance
$R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{(I / 4)}{4 A}=\frac{1}{4 \times 4}\left(\rho \frac{1}{A}\right)=\frac{R}{16}\left(\because R =\rho \frac{l}{A}\right)$
Thus, the new resistance is $\frac{1}{16}$ of the original resistance.

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