Let the common length of each conductor be $L$.

Take area of cross-section of conductor 1 and 2 as $A$. Then for conductor 3, area is $2A$.

In the shown arrangement, conductor 1 and 2 are connected between the same two temperature points, so they are in parallel, and conductor 3 is connected in series with this parallel combination.

We are to find the junction temperature $\theta$ in steady state.

1. Thermal conductance of each rod

Thermal conductance is $G=\frac{kA}{L}$

So,

• For conductor 1: $G_1=\frac{k_1A}{L}=\frac{60A}{L}$

• For conductor 2: $G_2=\frac{k_2A}{L}=\frac{120A}{L}$

• For conductor 3: $G_3=\frac{k_3(2A)}{L}=\frac{135\cdot 2A}{L}=\frac{270A}{L}$

2. Equivalent conductance of conductors 1 and 2 in parallel

For parallel combination, $G_{12}=G_1+G_2=\frac{60A}{L}+\frac{120A}{L}=\frac{180A}{L}$

3. Apply steady-state heat current condition

From the figure, the parallel combination is between $100^\circ \mathrm{C}$ and $\theta$, and conductor 3 is between $\theta$ and $0^\circ \mathrm{C}$.

Hence,

• Heat current through the parallel part: $\dot Q_{12}=G_{12}(100-\theta)=\frac{180A}{L}(100-\theta)$

• Heat current through conductor 3: $\dot Q_3=G_3(\theta-0)=\frac{270A}{L}\theta$

In steady state, these must be equal: $\frac{180A}{L}(100-\theta)=\frac{270A}{L}\theta$

Cancel $\frac{A}{L}$: $180(100-\theta)=270\theta$

$18000-180\theta=270\theta$

$18000=450\theta$

$\theta=40^\circ \mathrm{C}$

4. Final answer

$\boxed{40}$

The derived answer matches the stored correct answer.