A large cylindrical rod of length L is made by joining two identical rods of copper and steel of length `((L)/(2))` each. The rods are completely insulated from the surroundings. If the free end of copper rod is maintained at `100^(@)C` and that of steel at `0^(@)C` then the temperature of junction is (Thermal conductivity of copper is 9 times that of steel)

Options

A. `90^(@)C`
B. `50^(@)C`
C. `10^(@)C`
D. `67^(@)C`

Answer

To solve the problem, we need to find the temperature of the junction (T) between the copper and steel rods. We will use the concept of thermal conductivity and the principle of heat transfer through the rods.

### Step-by-Step Solution:

1. **Identify the given values:**
- Length of each rod = \( \frac{L}{2} \)
- Temperature at the free end of the copper rod = \( 100^\circ C \)
- Temperature at the free end of the steel rod = \( 0^\circ C \)
- Let the thermal conductivity of steel be \( K \). Therefore, the thermal conductivity of copper is \( 9K \).

2. **Calculate the thermal resistance of each rod:**
- The thermal resistance \( R \) of a rod can be calculated using the formula:
\[ R = \frac{L}{kA} \]
- For the steel rod:
\[ R_{steel} = \frac{\frac{L}{2}}{KA} = \frac{L}{2KA} \]
- For the copper rod:
\[ R_{copper} = \frac{\frac{L}{2}}{9KA} = \frac{L}{18KA} \]

3. **Set up the equation for heat transfer:**
- Since the rods are in series, the heat flow through both rods is the same. Thus, we can write:
\[ \frac{T - 0}{R_{steel}} = \frac{100 - T}{R_{copper}} \]
- Substituting the thermal resistances:
\[ \frac{T}{\frac{L}{2KA}} = \frac{100 - T}{\frac{L}{18KA}} \]

4. **Simplify the equation:**
- Cross-multiplying gives:
\[ T \cdot \frac{L}{18KA} = (100 - T) \cdot \frac{L}{2KA} \]
- Cancel \( \frac{L}{KA} \) from both sides:
\[ \frac{T}{18} = \frac{100 - T}{2} \]

5. **Solve for T:**
- Cross-multiplying again:
\[ 2T = 18(100 - T) \]
\[ 2T = 1800 - 18T \]
\[ 20T = 1800 \]
\[ T = \frac{1800}{20} = 90^\circ C \]

6. **Conclusion:**
- The temperature of the junction is \( T = 90^\circ C \).