1. Modeling the Circuit

Let the resistance of the cable from end A to the fault C be $R_x$, and from the fault C to end B be $R_y$. The fault connects point C to earth via a resistance $R_f = 10 \Omega$.

Scenario 1 (Source at A):
Voltage applied at A: $V_{source} = 20$ V.
Voltage measured at B: $V_{measure} = 5$ V.
Since the voltmeter at B is ideal (infinite resistance), no current flows through the segment CB ($R_y$). Therefore, there is no voltage drop across $R_y$, and the potential at C equals the potential measured at B: $V_C = 5$ V.

Current flows from A -> C -> Fault -> Earth. We can write the voltage divider equation:

$$V_C = V_{source} \frac{R_f}{R_x + R_f}$$

$$5 = 20 \frac{10}{R_x + 10} \implies \frac{1}{4} = \frac{10}{R_x + 10}$$

$$R_x + 10 = 40 \implies R_x = 30 \Omega$$

2. Verifying with Scenario 2 (Source at B)

Voltage applied at B: $V_{source} = 30$ V.
Voltage measured at A: $V_{measure} = 5$ V.
Similarly, no current flows through CA ($R_x$), so $V_C = 5$ V.

$$V_C = V_{source} \frac{R_f}{R_y + R_f}$$

$$5 = 30 \frac{10}{R_y + 10} \implies \frac{1}{6} = \frac{10}{R_y + 10}$$

$$R_y + 10 = 60 \implies R_y = 50 \Omega$$

3. Calculating Distance

Total resistance of the cable $R_{total} = R_x + R_y = 30 + 50 = 80 \Omega$.
Total length of the cable $L = 8$ km.

The resistance per unit length is $\lambda = \frac{80 \Omega}{8 \text{ km}} = 10 \Omega/\text{km}$.

The distance of the fault from end A ($x$) corresponds to resistance $R_x = 30 \Omega$.