JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Current Electricity

The full wire has resistance $R$, so each of the four sides of the square carries $\frac{R}{4}$. Since $E$ is the midpoint of $CD$, the piece $EC$ is $\frac{R}{8}$ and $ED$ is also $\frac{R}{8}$. Between $E$ and $C$ there are two routes in parallel: straight along $EC$ with resistance $\frac{R}{8}$, and the long way round $E\to D\to A\to B\to C$ with resistance $\frac{R}{8}+\frac{R}{4}+\frac{R}{4}+\frac{R}{4}=\frac{7R}{8}$. Combining these in parallel, $R_{eff}=\frac{\frac{R}{8}\cdot \frac{7R}{8}}{\frac{R}{8}+\frac{7R}{8}}=\frac{7R^2/64}{R}=\frac{7R}{64}$.