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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the lengths of the two rods be \( d_1 \) and \( d_2 \).
- Let the thermal conductivities of the two rods be \( K_1 \) and \( K_2 \).
- Both rods have the same area of cross-section, denoted as \( A \).

2. **Calculate Thermal Resistance**:
- The thermal resistance \( R \) for a rod is given by the formula:
\[ R = \frac{d}{K \cdot A} \]
- For rod 1:
\[ R_1 = \frac{d_1}{K_1 \cdot A} \]
- For rod 2:
\[ R_2 = \frac{d_2}{K_2 \cdot A} \]

3. **Total Thermal Resistance**:
- Since the rods are in series, the total thermal resistance \( R_{total} \) is the sum of the individual resistances:
\[ R_{total} = R_1 + R_2 = \frac{d_1}{K_1 \cdot A} + \frac{d_2}{K_2 \cdot A} \]

4. **Combine the Resistances**:
- Factor out \( A \) from the equation:
\[ R_{total} = \frac{1}{A} \left( \frac{d_1}{K_1} + \frac{d_2}{K_2} \right) \]

5. **Relate Total Resistance to Equivalent Conductivity**:
- The equivalent thermal resistance can also be expressed in terms of the equivalent thermal conductivity \( K_{eq} \):
\[ R_{total} = \frac{d_{total}}{K_{eq} \cdot A} \]
- Here, \( d_{total} = d_1 + d_2 \).

6. **Set the Two Expressions for Resistance Equal**:
- Equating the two expressions for \( R_{total} \):
\[ \frac{d_1 + d_2}{K_{eq} \cdot A} = \frac{1}{A} \left( \frac{d_1}{K_1} + \frac{d_2}{K_2} \right) \]

7. **Cancel out Area \( A \)**:
- Since \( A \) is the same for both rods, we can cancel it out:
\[ \frac{d_1 + d_2}{K_{eq}} = \frac{d_1}{K_1} + \frac{d_2}{K_2} \]

8. **Solve for \( K_{eq} \)**:
- Rearranging gives:
\[ K_{eq} = \frac{(d_1 + d_2) \cdot (K_1 \cdot K_2)}{d_1 \cdot K_2 + d_2 \cdot K_1} \]

### Final Result:
The equivalent thermal conductivity \( K_{eq} \) is given by:
\[ K_{eq} = \frac{(d_1 + d_2) \cdot (K_1 \cdot K_2)}{d_1 \cdot K_2 + d_2 \cdot K_1} \]