Physics - Current Electricity Question with Solution | TestHub

For the wire bent into an equilateral triangle, each side has a resistance $\frac{R}{3}$.$R_{\text{eq }}=\frac{\left(\frac{2R}{3}\right)\left(\frac{R}{3}\right)}{\frac{2R}{3}+\frac{R}{3}}=\frac{2R}{9}=R_1$ (lets say) For the wire bent into a square, each side has a resistance $\frac{R}{4}$.$\begin{array}{rl}& R_{\text{eq }}=\frac{\left(\frac{3R}{4}\right)\left(\frac{R}{4}\right)}{\frac{3R}{4}+\frac{R}{4}}=\frac{3R}{16}=R_3\text{ (lets say) }\\ & \Rightarrow \frac{R_1}{R_3}=\frac{\frac{2R}{9}}{\frac{3R}{16}}=\frac{32}{27}\end{array}$