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?

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{(l / 4)}{4 A}=\frac{1}{4 \times 4}\left(\rho \frac{l}{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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