Three rods `A,B` and `C` have the same dimensions Their conductivities are `K_(A)` K and `K_(C)` respectively `A` and `B` are placed end to end with their free ends kept at certain temperature difference `C` is placed separately with its ends kept at same temperature difference The two arrangements conduct heat at the same rate `K_(c)` must be equal to .

Options

A. `K_(A)+K_(B)`
B. `(K_(A)+K_(B))/(K_(A)K_(B))`
C. `(1)/(2)(K_(A)+K_(B))`
D. `(K_(A)+K_(B))/(K_(A)K_(B))`

Explanation

To solve the problem, we need to find the thermal conductivity \( K_C \) of rod C such that the heat conduction rates through the two arrangements (rods A and B in series, and rod C separately) are the same.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Rods A and B are placed end to end, and both have the same dimensions. Their thermal conductivities are \( K_A \) and \( K_B \).
- Rod C is placed separately with the same temperature difference across its ends.

2. **Thermal Resistance of Rods A and B in Series**:
- The total thermal resistance \( R_{AB} \) of rods A and B in series can be expressed as:
\[ R_{AB} = R_A + R_B \]
- The thermal resistance \( R \) of a rod is given by:
\[ R = \frac{L}{K \cdot A} \]
where \( L \) is the length, \( K \) is the thermal conductivity, and \( A \) is the cross-sectional area.
- Therefore, for rods A and B:
\[ R_A = \frac{L}{K_A \cdot A} \quad \text{and} \quad R_B = \frac{L}{K_B \cdot A} \]
- Thus, the total resistance becomes:
\[ R_{AB} = \frac{L}{K_A \cdot A} + \frac{L}{K_B \cdot A} = \frac{L}{A} \left( \frac{1}{K_A} + \frac{1}{K_B} \right) \]

3. **Thermal Resistance of Rod C**:
- The thermal resistance \( R_C \) of rod C is given by:
\[ R_C = \frac{L}{K_C \cdot A} \]

4. **Setting the Resistances Equal**:
- For the heat conduction rates to be the same, the thermal resistances must be equal:
\[ R_{AB} = R_C \]
- Substituting the expressions for \( R_{AB} \) and \( R_C \):
\[ \frac{L}{A} \left( \frac{1}{K_A} + \frac{1}{K_B} \right) = \frac{L}{K_C \cdot A} \]

5. **Simplifying the Equation**:
- Canceling \( \frac{L}{A} \) from both sides (assuming \( L \) and \( A \) are not zero):
\[ \frac{1}{K_A} + \frac{1}{K_B} = \frac{1}{K_C} \]

6. **Finding \( K_C \)**:
- Rearranging the equation gives:
\[ \frac{1}{K_C} = \frac{1}{K_A} + \frac{1}{K_B} \]
- Taking the reciprocal:
\[ K_C = \frac{K_A \cdot K_B}{K_A + K_B} \]

### Final Answer:
\[ K_C = \frac{K_A \cdot K_B}{K_A + K_B} \]