A given length of a wire is doubled on itself and this process is repeated once again. By the factor of [[1]] the resistance changes.

Solution:

A given length of a wire is doubled on itself and this process is repeated once again. By the factor of

\frac{1}{16}

the resistance changes.
The resistance is directly proportional to the length of the wire and inversely proportional to the area of the cross-section. From the data given, the length of the new wire becomes one-fourth and the area of the cross-section becomes four times.

R \propto \frac{l}{A}

\Rightarrow \frac{R_{2}}{R_{1}} = \frac{l_{2}}{l_{1}} \times \frac{A_{1}}{A_{2}}

\Rightarrow \frac{R_{2}}{R_{1}} = \frac{(\frac{l_{1}}{4})}{l_{1}} \times \frac{A_{1}}{{4 A}_{1}}

\Rightarrow \frac{R_{2}}{R_{1}} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

\Rightarrow \frac{R_{2}}{R_{1}} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

\Rightarrow R_{2} = \frac{R_{1}}{16}

Hence, the new resistance is

\frac{1}{16}

of the old resistance.

A given length of a wire is doubled on itself and this process is repeated once again. By the factor of [[1]] the resistance changes.

Solution:

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