A given length of a wire is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?
Am. Length becomes one-fourth of the original length and area of cross-section becomes four times that of original.
i.e., $$l_{2} = \dfrac{1}{4}l_{1} $$ and $$A_{2} = 4A_{1} $$
$$ \therefore$$ $$ \dfrac{R_2}{R_1} = \dfrac{l_2}{l_1} \times \dfrac{A_1}{A_2} = \dfrac{1}{4} \times \dfrac{1}{4} = \dfrac{1}{16} $$
$$\Rightarrow$$ $$R_{2} = \dfrac{1}{16}R_{1} $$
So, new resistance is $$(1/16)th $$ of original resistance.