1. Interpret the geometry

A wire of total resistance $R$ is bent into a square $ABCD$.

So each side of the square has resistance $\frac{R}{4}.$

Point $E$ is the midpoint of side $CD$, so side $CD$ is split into two equal parts: $CE = ED = \frac{1}{2}\cdot \frac{R}{4} = \frac{R}{8}.$

We need the effective resistance between $E$ and $C$.

2. Identify the two paths between $E$ and $C$

From $E$ to $C$, there are two possible branches:

• Direct branch: along segment $EC$ $R_1 = \frac{R}{8}.$

• Other branch: go from $E \to D \to A \to B \to C$

Its resistance is $R_2 = ED + DA + AB + BC = \frac{R}{8} + \frac{R}{4} + \frac{R}{4} + \frac{R}{4}.$

So, $R_2 = \frac{R}{8} + \frac{3R}{4} = \frac{R}{8} + \frac{6R}{8} = \frac{7R}{8}.$

Thus, between $E$ and $C$, we have two resistances in parallel: $\frac{R}{8} \quad \text{and} \quad \frac{7R}{8}.$

3. Compute the equivalent resistance

For two resistors in parallel, $R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2}.$

Substitute: $R_{\text{eq}} = \frac{\left(\frac{R}{8}\right)\left(\frac{7R}{8}\right)}{\frac{R}{8}+\frac{7R}{8}}.$

Since $\frac{R}{8}+\frac{7R}{8}=R,$ we get $R_{\text{eq}} = \frac{\frac{7R^2}{64}}{R} = \frac{7R}{64}.$

4. Match with the options

$R_{\text{eq}} = \frac{7}{64}R$ which corresponds to Option A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So the derived answer agrees with the stored correct answer.