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. (a)No heat is transmitted across the copper–aluminium or aluminium–brass junctions.
B. (b)More heat is transmitted across the copper–aluminium junction than across the aluminium–brass junction
C. (c)More heat is transmitted across the aluminium–brass junction than the copper–aluminium junction
D. (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 thermal conductivity of the materials involved and how heat transfer occurs through the rods when they reach a steady state.

### Step-by-Step Solution:

1. **Identify Thermal Conductivities**:
Let the thermal conductivity of aluminum be \( k \). According to the problem:
- Thermal conductivity of copper, \( k_{Cu} = 2k \)
- Thermal conductivity of brass, \( k_{Br} = \frac{k}{4} \)

2. **Set Up the Heat Transfer Equation**:
In steady state, the rate of heat transfer \( Q \) through each rod must be equal. The heat transfer through a rod can be expressed using Fourier's law:
\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]
where:
- \( k \) = thermal conductivity of the material
- \( A \) = cross-sectional area of the rod
- \( \Delta T \) = temperature difference across the rod
- \( L \) = length of the rod

3. **Calculate Cross-Sectional Area**:
The diameter of the rods is 2 cm, so the radius \( r = 1 \) cm = 0.01 m. The cross-sectional area \( A \) is given by:
\[ A = \pi r^2 = \pi (0.01)^2 = \pi \times 10^{-4} \, \text{m}^2 \]

4. **Set Up Heat Transfer Equations for Each Rod**:
- For copper rod (from 100°C to the aluminum junction):
\[ Q = \frac{(2k) \cdot A \cdot (100 - T_{Al})}{L} \]
- For aluminum rod (from \( T_{Al} \) to \( T_{Br} \)):
\[ Q = \frac{k \cdot A \cdot (T_{Al} - T_{Br})}{L} \]
- For brass rod (from \( T_{Br} \) to 0°C):
\[ Q = \frac{\left(\frac{k}{4}\right) \cdot A \cdot (T_{Br} - 0)}{L} \]

5. **Equate Heat Transfer Rates**:
Since the heat transfer rate \( Q \) is the same through all rods, we can set the equations equal to each other:
\[ \frac{(2k) \cdot A \cdot (100 - T_{Al})}{L} = \frac{k \cdot A \cdot (T_{Al} - T_{Br})}{L} = \frac{\left(\frac{k}{4}\right) \cdot A \cdot (T_{Br} - 0)}{L} \]

6. **Simplify and Solve**:
We can cancel \( A \) and \( L \) from all equations:
\[ 2(100 - T_{Al}) = T_{Al} - T_{Br} = \frac{1}{4} T_{Br} \]

From the first part:
\[ 200 - 2T_{Al} = T_{Al} - T_{Br} \implies 200 = 3T_{Al} - T_{Br} \]

From the second part:
\[ T_{Al} - T_{Br} = \frac{1}{4} T_{Br} \implies T_{Al} = \frac{5}{4} T_{Br} \]

Substitute \( T_{Al} \) in the first equation:
\[ 200 = 3\left(\frac{5}{4} T_{Br}\right) - T_{Br} \implies 200 = \frac{15}{4} T_{Br} - T_{Br} \]
\[ 200 = \frac{15}{4} T_{Br} - \frac{4}{4} T_{Br} = \frac{11}{4} T_{Br} \]
\[ T_{Br} = \frac{200 \times 4}{11} \approx 72.73°C \]

Now substitute \( T_{Br} \) back to find \( T_{Al} \):
\[ T_{Al} = \frac{5}{4} \times 72.73 \approx 90.91°C \]

7. **Conclusion**:
The steady state temperatures are \( T_{Cu} = 100°C \), \( T_{Al} \approx 90.91°C \), and \( T_{Br} \approx 72.73°C \). The correct statement is that equal amounts of heat are transmitted through the copper, aluminum, and brass junctions.