Three identical metal rods `A`, `B` and `C` are placed end to end and a temperature difference is maintained between the free ends of `A` and `C`. If the thermal conductivity of `B(K_(B))` is thrice that of `C(K_(C ))` and half that of `A(K_(A))`, `(K_(A)=49w//mK)` calculate the effective thermal conductivity of the system ?

Options

A. `1/3 K_(A)`
B. `3K_(A)`
C. `2K_(A)`
D. `2/3 K_(A)`

Explanation

To solve the problem, we need to calculate the effective thermal conductivity of three identical metal rods A, B, and C placed end to end, given their thermal conductivities and the relationships between them.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Thermal conductivity of rod A, \( K_A = 49 \, \text{W/mK} \)
- Thermal conductivity of rod B, \( K_B = \frac{1}{2} K_A = \frac{1}{2} \times 49 = 24.5 \, \text{W/mK} \)
- Thermal conductivity of rod C, \( K_C = \frac{1}{3} K_B = \frac{1}{3} \times 24.5 = 8.17 \, \text{W/mK} \)

2. **Calculate the Thermal Resistances:**
- The thermal resistance \( R \) for each rod is given by the formula:
\[ R = \frac{L}{K \cdot A} \]
- Since the rods are identical in length \( L \) and cross-sectional area \( A \), we can express the resistances as:
\[ R_A = \frac{L}{K_A \cdot A}, \quad R_B = \frac{L}{K_B \cdot A}, \quad R_C = \frac{L}{K_C \cdot A} \]

3. **Substituting the Values:**
- Substitute the values of \( K_A \), \( K_B \), and \( K_C \):
\[ R_A = \frac{L}{49 \cdot A}, \quad R_B = \frac{L}{24.5 \cdot A}, \quad R_C = \frac{L}{8.17 \cdot A} \]

4. **Calculate the Total Thermal Resistance:**
- The total thermal resistance \( R_{\text{total}} \) for the series combination is:
\[ R_{\text{total}} = R_A + R_B + R_C = \frac{L}{49 \cdot A} + \frac{L}{24.5 \cdot A} + \frac{L}{8.17 \cdot A} \]
- Factor out \( \frac{L}{A} \):
\[ R_{\text{total}} = \frac{L}{A} \left( \frac{1}{49} + \frac{1}{24.5} + \frac{1}{8.17} \right) \]

5. **Calculate Each Resistance Value:**
- Calculate the individual resistance values:
\[ \frac{1}{49} \approx 0.0204, \quad \frac{1}{24.5} \approx 0.0408, \quad \frac{1}{8.17} \approx 0.122 \]
- Sum these values:
\[ \frac{1}{49} + \frac{1}{24.5} + \frac{1}{8.17} \approx 0.0204 + 0.0408 + 0.122 \approx 0.1832 \]

6. **Calculate the Effective Thermal Conductivity:**
- The effective thermal conductivity \( K_{\text{eff}} \) can be found using the formula:
\[ K_{\text{eff}} = \frac{L}{R_{\text{total}} \cdot A \cdot 3L} = \frac{1}{R_{\text{total}} \cdot 3} \]
- Thus:
\[ K_{\text{eff}} = \frac{1}{\left( \frac{L}{A} \cdot 0.1832 \right) \cdot 3} = \frac{K_A}{3} = \frac{49}{3} \approx 16.33 \, \text{W/mK} \]

### Final Answer:
The effective thermal conductivity of the system is approximately \( K_{\text{eff}} \approx 16.33 \, \text{W/mK} \).