You have two copper cables of equal length for carrying current. One of them has a single wire of area of across section `A`, the other has ten wires each of cross section area `A//10`. Judge their suitability for transporting `ac` and `dc`.

Options

A. Only multiple strands for A.C. either for D.C.
B. Only mutiple stands for A.C, only single strand for D.C.
C. Only single strand for D.C. either for A.C.
D. Only single strand for A.C, either for D.C.

Answer

To determine the suitability of two copper cables for transporting AC and DC, we will analyze their resistances and surface areas.

### Step-by-Step Solution:

1. **Identify the Cables**:
- Cable 1: A single wire with a cross-sectional area \( A \).
- Cable 2: Ten wires, each with a cross-sectional area \( \frac{A}{10} \).

2. **Calculate Resistance for DC**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material and \( L \) is the length of the wire.
- For Cable 1:
\[ R_1 = \frac{\rho L}{A} \]
- For Cable 2, since there are 10 parallel wires, we first calculate the resistance of one wire:
\[ R_{\text{single}} = \frac{\rho L}{\frac{A}{10}} = \frac{10 \rho L}{A} \]
Since they are in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{single}}} + \frac{1}{R_{\text{single}}} + \ldots + \frac{1}{R_{\text{single}}} \quad (10 \text{ times}) \]
\[ \frac{1}{R_{\text{eq}}} = 10 \cdot \frac{A}{10 \rho L} = \frac{A}{\rho L} \]
Thus,
\[ R_{\text{eq}} = \frac{\rho L}{A} \]

3. **Comparison of Resistances**:
- Both cables have the same resistance \( R_1 = R_{\text{eq}} = \frac{\rho L}{A} \).
- Therefore, both cables are equally suitable for transporting DC current.

4. **Calculate Surface Area for AC**:
- For AC current, the effective surface area of the wire is important since AC tends to travel on the surface of conductors.
- The surface area \( S \) of a cylindrical wire is given by:
\[ S = 2 \pi r h \]
where \( r \) is the radius and \( h \) is the length.
- For Cable 1:
- The radius \( r_1 \) can be calculated from the area \( A \):
\[ r_1 = \sqrt{\frac{A}{\pi}} \]
Therefore, the surface area \( S_1 \) is:
\[ S_1 = 2 \pi \left(\sqrt{\frac{A}{\pi}}\right) L = 2L \sqrt{A \pi} \]

- For Cable 2:
- Each wire has a radius \( r_2 = \sqrt{\frac{A/10}{\pi}} \).
- The surface area of one wire is:
\[ S_{\text{single}} = 2 \pi \left(\sqrt{\frac{A/10}{\pi}}\right) L = 2L \sqrt{\frac{A}{10} \pi} \]
- Since there are 10 wires:
\[ S_2 = 10 \cdot S_{\text{single}} = 10 \cdot 2L \sqrt{\frac{A}{10} \pi} = 2L \sqrt{10A \pi} \]

5. **Comparison of Surface Areas**:
- Comparing \( S_1 \) and \( S_2 \):
\[ S_2 = 2L \sqrt{10A \pi} \quad \text{and} \quad S_1 = 2L \sqrt{A \pi} \]
- Clearly, \( S_2 \) is greater than \( S_1 \) since \( \sqrt{10} > 1 \).

6. **Conclusion**:
- For DC, both cables are equally suitable as they have the same resistance.
- For AC, the cable with multiple strands (Cable 2) is more suitable due to its larger surface area.

### Final Answer:
- **Both cables are suitable for DC.**
- **The cable with multiple strands (Cable 2) is more suitable for AC.**