Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. Three metal rods made of copper, aluminium and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminium between the other two. The free ends of the copper and brass rods are maintained at `100^@C` and `0^@C` respectively. The system is allowed to reach the steady state condition. Assume there is no loss of heat anywhere.
When steady state condition is reached everywhere, which of the following statement is true?

Options

A. No heat is transmitted across the copper aluminium or aluminium brass junctions.
B. More heat is transmitted across the copper aluminium junction than across the aluminium brass junction.
C. More heat is tranmitted across the aluminium brass junction than the copper aluminium junction.
D. Equal amount of heat is transmitted at the copper aluminium and aluminium brass junctions.

Explanation

To solve the problem, we need to analyze the heat transfer through the three metal rods (copper, aluminum, and brass) when they are placed end to end and subjected to different temperatures at their ends.

### Step-by-Step Solution:

1. **Identify the Thermal Conductivities**:
- Let the thermal conductivity of aluminum be \( k \).
- Therefore, the thermal conductivity of copper will be \( 2k \) (twice that of aluminum).
- The thermal conductivity of brass will be \( \frac{k}{2} \) (four times less than copper).

2. **Understand the Configuration**:
- The rods are arranged in the sequence: Copper (100°C) - Aluminum - Brass (0°C).
- Each rod has a length of 15 cm and a diameter of 2 cm.

3. **Heat Transfer Rate**:
- The rate of heat transfer \( Q \) through a rod can be expressed using Fourier's law of heat conduction:
\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]
where:
- \( k \) is the thermal conductivity,
- \( A \) is the cross-sectional area,
- \( \Delta T \) is the temperature difference across the rod,
- \( L \) is the length of the rod.

4. **Calculate Cross-Sectional Area**:
- The cross-sectional area \( A \) for a circular rod is given by:
\[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(1 \text{ cm}\right)^2 = \pi \text{ cm}^2 \]

5. **Steady State Condition**:
- In steady state, the rate of heat transfer through each rod must be equal:
\[ Q_{\text{copper}} = Q_{\text{aluminum}} = Q_{\text{brass}} \]

6. **Set Up the Equations**:
- For copper:
\[ Q = \frac{2k \cdot \pi \cdot (100 - T_1)}{15} \]
- For aluminum:
\[ Q = \frac{k \cdot \pi \cdot (T_1 - T_2)}{15} \]
- For brass:
\[ Q = \frac{\frac{k}{2} \cdot \pi \cdot (T_2 - 0)}{15} \]

7. **Equate the Heat Transfer Rates**:
- Since \( Q \) is the same for all rods, we can set up the following equations:
\[ \frac{2k \cdot \pi \cdot (100 - T_1)}{15} = \frac{k \cdot \pi \cdot (T_1 - T_2)}{15} \]
\[ \frac{k \cdot \pi \cdot (T_1 - T_2)}{15} = \frac{\frac{k}{2} \cdot \pi \cdot (T_2 - 0)}{15} \]

8. **Solve for Temperature Differences**:
- From the first equation, simplify and solve for \( T_1 \):
\[ 2(100 - T_1) = T_1 - T_2 \]
\[ 200 - 2T_1 = T_1 - T_2 \]
\[ 3T_1 - T_2 = 200 \quad \text{(1)} \]

- From the second equation, simplify and solve for \( T_2 \):
\[ 2(T_1 - T_2) = T_2 \]
\[ 2T_1 - 2T_2 = T_2 \]
\[ 2T_1 = 3T_2 \quad \text{(2)} \]

9. **Substitute and Solve**:
- Substitute equation (2) into equation (1):
\[ 3T_1 - \frac{2}{3}T_1 = 200 \]
\[ \frac{7}{3}T_1 = 200 \]
\[ T_1 = \frac{600}{7} \approx 85.71°C \]
- Substitute \( T_1 \) back into equation (2) to find \( T_2 \):
\[ 2T_1 = 3T_2 \implies 2 \times \frac{600}{7} = 3T_2 \]
\[ T_2 = \frac{1200}{21} \approx 57.14°C \]

10. **Conclusion**:
- The steady state temperatures are approximately \( T_1 \approx 85.71°C \) and \( T_2 \approx 57.14°C \).
- Since the heat transfer rate is the same across all junctions, the correct statement is:
**Equal amounts of heat are transmitted at the copper-aluminum and aluminum-brass junctions**.