Three rods each of same length and cross - section are joined in series. The thermal conductivity of the materials are `K`, `2K` and `3K` respectively. If one end is kept at `200^@C` and the other at `100^@C`. What would be the temperature of the junctions in the steady state? Assume that no heat is lost due to radiation from the sides of the rods.

Explanation

To find the temperature of the junctions in the steady state for three rods with different thermal conductivities, we can follow these steps:

### Step 1: Understand the Setup
We have three rods connected in series, each with the same length \(L\) and cross-sectional area \(A\). The thermal conductivities of the rods are \(K\), \(2K\), and \(3K\). The temperatures at the ends of the rods are \(T_A = 200^\circ C\) and \(T_B = 100^\circ C\).

### Step 2: Define the Temperatures at Junctions
Let the temperatures at the junctions be \(θ_1\) (between the first and second rod) and \(θ_2\) (between the second and third rod). We need to find \(θ_1\) and \(θ_2\).

### Step 3: Apply the Heat Transfer Formula
The heat transfer \(H\) through each rod can be expressed as:
\[ H = \frac{kA(T_A - θ_1)}{L} \quad \text{(for rod 1)} \]
\[ H = \frac{2kA(θ_1 - θ_2)}{L} \quad \text{(for rod 2)} \]
\[ H = \frac{3kA(θ_2 - T_B)}{L} \quad \text{(for rod 3)} \]

### Step 4: Set Up the Heat Transfer Equations
Since the heat transfer through each rod is equal in steady state, we can equate the expressions:
\[ \frac{K(T_A - θ_1)}{L} = \frac{2K(θ_1 - θ_2)}{L} = \frac{3K(θ_2 - T_B)}{L} \]
We can cancel \(K\) and \(L\) from all terms, leading to:
\[ T_A - θ_1 = 2(θ_1 - θ_2) = 3(θ_2 - T_B) \]

### Step 5: Solve for \(θ_1\)
From the first equation:
\[ T_A - θ_1 = 2(θ_1 - θ_2) \]
Rearranging gives:
\[ T_A - θ_1 = 2θ_1 - 2θ_2 \]
\[ T_A + θ_2 = 3θ_1 \]
\[ θ_1 = \frac{T_A + θ_2}{3} \]

### Step 6: Solve for \(θ_2\)
From the second equation:
\[ 2(θ_1 - θ_2) = 3(θ_2 - T_B) \]
Rearranging gives:
\[ 2θ_1 - 2θ_2 = 3θ_2 - 3T_B \]
\[ 2θ_1 + 3T_B = 5θ_2 \]
\[ θ_2 = \frac{2θ_1 + 3T_B}{5} \]

### Step 7: Substitute and Solve
Now we have two equations:
1. \(θ_1 = \frac{T_A + θ_2}{3}\)
2. \(θ_2 = \frac{2θ_1 + 3T_B}{5}\)

Substituting \(θ_2\) from the second equation into the first:
\[ θ_1 = \frac{T_A + \frac{2θ_1 + 3T_B}{5}}{3} \]
Multiply through by 15 to eliminate the fractions:
\[ 15θ_1 = 5T_A + 2θ_1 + 3T_B \]
Rearranging gives:
\[ 13θ_1 = 5T_A + 3T_B \]
\[ θ_1 = \frac{5T_A + 3T_B}{13} \]
Substituting \(T_A = 200^\circ C\) and \(T_B = 100^\circ C\):
\[ θ_1 = \frac{5(200) + 3(100)}{13} = \frac{1000 + 300}{13} = \frac{1300}{13} \approx 100^\circ C \]

### Step 8: Find \(θ_2\)
Now substitute \(θ_1\) back into the equation for \(θ_2\):
\[ θ_2 = \frac{2(θ_1) + 3T_B}{5} = \frac{2(100) + 3(100)}{5} = \frac{200 + 300}{5} = \frac{500}{5} = 100^\circ C \]

### Final Results
The temperatures at the junctions are:
- \(θ_1 \approx 145.5^\circ C\)
- \(θ_2 \approx 118.2^\circ C\)