A wire of resistance R is bent to form a square A B C D as shown in the figure.

(2) Here $\mathrm{R}_{\mathrm{DA}}=\mathrm{R}_{\mathrm{AB}}=\mathrm{R}_{\mathrm{BC}}=\mathrm{R} / 4$

and $\mathrm{R}_{\mathrm{DE}}=\mathrm{R}_{\mathrm{EC}}=\mathrm{R} / 8$

Now $R_{E D}, R_{D A}, R_{A B}, R_{B C}$ are in series.

$\therefore R_{s}=\frac{R}{8}+\frac{R}{4}+\frac{R}{4}+\frac{R}{4}=\frac{R+2 R+2 R+2 R}{8}=\frac{7 R}{8}$

$\therefore \mathrm{R}_{\mathrm{eq}}=\frac{\left(\frac{7 R}{8}\right)\left(\frac{R}{8}\right)}{R}=\frac{7 R}{64}$