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`

Answer

To solve the problem, we need to establish the relationship between the thermal conductivities of wires A and B based on their dimensions and the condition that they conduct heat at the same rate.

### Step-by-Step Solution:

1. **Identify the given information:**
- Length of both wires (L) is the same.
- Radius of wire A (R_A) is twice the radius of wire B (R_B), i.e., R_A = 2R_B.
- Both wires conduct heat at the same rate for a given temperature difference (ΔT).

2. **Use the formula for heat conduction:**
The rate of heat conduction (Q/t) through a wire is given by the formula:
\[ \frac{Q}{t} = \frac{k \cdot A \cdot \Delta T}{L} \]
where:
- \( k \) is the thermal conductivity,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference,
- \( L \) is the length of the wire.

3. **Express the cross-sectional area:**
The cross-sectional area \( A \) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]
Therefore, for wires A and B:
- \( A_A = \pi R_A^2 = \pi (2R_B)^2 = 4\pi R_B^2 \)
- \( A_B = \pi R_B^2 \)

4. **Set up the equation for heat conduction:**
Since both wires conduct heat at the same rate:
\[ \frac{k_A \cdot A_A \cdot \Delta T}{L} = \frac{k_B \cdot A_B \cdot \Delta T}{L} \]
The lengths and temperature differences cancel out:
\[ k_A \cdot A_A = k_B \cdot A_B \]

5. **Substitute the areas into the equation:**
Substitute \( A_A \) 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. **Express the relationship between thermal conductivities:**
Rearranging gives:
\[ k_B = 4k_A \]

### Final Answer:
The relation between the thermal conductivities is:
\[ k_B = 4k_A \]