Wires A and B have have identical lengths and have circular cross-sections. The radius of A is twice the radius of B i.e. `R_A = 2R_B`. For a given temperature difference between the two ends, both wires conduct heat at the same rate. The relation between the thermal conductivities is given by-

Options

A. `K_(A) = 4K_(B)`
B. `K_(A) = 2K_(B)`
C. `K_(A)= K_(B)/2`
D. `K_(A) = K_(B)/4`

Explanation

To solve the problem, we need to analyze the heat conduction through the two wires A and B, given their dimensions and the relationship between their thermal conductivities.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Wires A and B have identical lengths (L).
- The radius of wire A is twice that of wire B: \( R_A = 2R_B \).
- Both wires conduct heat at the same rate for a given temperature difference.

2. **Understand the Heat Conduction Formula:**
The rate of heat transfer (Q) through a wire is given by the formula:
\[ Q = \frac{K \cdot A \cdot (T_1 - T_2) \cdot t}{L} \]
where:
- \( K \) is the thermal conductivity,
- \( A \) is the cross-sectional area,
- \( T_1 - T_2 \) is the temperature difference,
- \( t \) is the time,
- \( L \) is the length of the wire.

3. **Calculate the Cross-Sectional Areas:**
The cross-sectional area \( A \) of a circular wire is given by:
\[ A = \pi R^2 \]
For wire A:
\[ A_A = \pi R_A^2 = \pi (2R_B)^2 = 4\pi R_B^2 \]
For wire B:
\[ A_B = \pi R_B^2 \]

4. **Set Up the Heat Transfer Equations:**
Since both wires conduct heat at the same rate, we can set their heat transfer equations equal to each other:
\[ \frac{K_A \cdot A_A \cdot (T_1 - T_2) \cdot t}{L} = \frac{K_B \cdot A_B \cdot (T_1 - T_2) \cdot t}{L} \]
The terms \( (T_1 - T_2) \), \( t \), and \( L \) cancel out:
\[ K_A \cdot A_A = K_B \cdot A_B \]

5. **Substitute the Areas:**
Substitute the expressions for \( A_A \) and \( A_B \):
\[ K_A \cdot (4\pi R_B^2) = K_B \cdot (\pi R_B^2) \]
Cancel \( \pi R_B^2 \) from both sides (assuming \( R_B \neq 0 \)):
\[ 4K_A = K_B \]

6. **Rearrange to Find the Relation Between Conductivities:**
Rearranging gives:
\[ K_A = \frac{K_B}{4} \]

### Conclusion:
The relation between the thermal conductivities of wires A and B is:
\[ K_A = \frac{K_B}{4} \]