Two rods of length `l_(1) and l_(2)` and coefficients of thermal conductivities `k_(1) and K_(2)` are kept touching each other. Both have the same area of cross-section. The equivalent of thermal conductivity is

Options

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

Explanation

To find the equivalent thermal conductivity of two rods that are in series, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have two rods of lengths \( l_1 \) and \( l_2 \) with thermal conductivities \( k_1 \) and \( k_2 \), respectively.
- Both rods have the same cross-sectional area \( A \) and are in contact with each other.

2. **Identify the Thermal Current**:
- The thermal current \( I \) through each rod can be expressed using Fourier's law of heat conduction:
\[ I = \frac{Q}{t} = \frac{k_1 \cdot A \cdot (T_1 - T)}{l_1} \quad \text{(for rod 1)} \]
\[ I = \frac{Q}{t} = \frac{k_2 \cdot A \cdot (T - T_2)}{l_2} \quad \text{(for rod 2)} \]

3. **Set Up the Equations**:
- Since both rods are in series, the thermal current \( I \) through both rods is the same:
\[ \frac{k_1 \cdot A \cdot (T_1 - T)}{l_1} = \frac{k_2 \cdot A \cdot (T - T_2)}{l_2} \]

4. **Express Temperature Differences**:
- Rearranging the equations gives:
\[ T_1 - T = \frac{I \cdot l_1}{k_1 \cdot A} \]
\[ T - T_2 = \frac{I \cdot l_2}{k_2 \cdot A} \]

5. **Combine the Equations**:
- Adding these two equations results in:
\[ T_1 - T_2 = \left( \frac{I \cdot l_1}{k_1 \cdot A} + \frac{I \cdot l_2}{k_2 \cdot A} \right) \]
- Factor out \( I \) and \( A \):
\[ T_1 - T_2 = I \cdot \left( \frac{l_1}{k_1 \cdot A} + \frac{l_2}{k_2 \cdot A} \right) \]

6. **Express Equivalent Thermal Conductivity**:
- The equivalent thermal conductivity \( k_{eq} \) can be defined as:
\[ T_1 - T_2 = \frac{k_{eq} \cdot A \cdot (T_1 - T_2)}{l_1 + l_2} \]
- Equating the two expressions for \( T_1 - T_2 \):
\[ \frac{l_1 + l_2}{k_{eq}} = \frac{l_1}{k_1} + \frac{l_2}{k_2} \]

7. **Solve for \( k_{eq} \)**:
- Rearranging gives:
\[ k_{eq} = \frac{(l_1 + l_2)}{\left( \frac{l_1}{k_1} + \frac{l_2}{k_2} \right)} \]

### Final Answer:
The equivalent thermal conductivity \( k_{eq} \) is given by:
\[ k_{eq} = \frac{(l_1 + l_2)}{\left( \frac{l_1}{k_1} + \frac{l_2}{k_2} \right)} \]