Three conductions of same length having thermal conductivity k₁, k₂ and k₃ are connected as shown in figure

Area of cross sections of $1^{st}$ and $2^{nd}$ conductor are same and for $3^{rd}$conductor it is double of the $1^{st}$ conductor. The temperatures are given in the figure. In steady state condition, the value of θ is _________${ }^{\circ }\mathrm{C}$
$\text{(Given :}{\mathrm{k}}_1=60{\mathrm{Js}}^{-1}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1},{\mathrm{k}}_2=120{\mathrm{Js}}^{-1}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1},{\mathrm{k}}_{\mathrm{a}}=135{\mathrm{Js}}^{-1}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}\text{)}$

Given:

1. Thermal conductivities: $$k_1 = 60 \, \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}, \quad k_2 = 120 \, \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}, \quad k_3 = 135 \, \mathrm{J\,s^{-1}\,m^{-1}\,K^{-1}}$$
2. Length of all conductors is the same.
3. Cross-sectional areas:
   • For $k_1$ and $k_2$, let say the area is $A$.
   • For $k_3$, the area is $2A$.
4. Temperatures:
   • At one end: $100^\circ \mathrm{C}$
   • At the other end: $0^\circ \mathrm{C}$
   • Intermediate temperature: $\theta$.

The heat flow in the system is in steady state, so the heat current remains constant.

Step 1: Identify Heat Flow Paths

1. The conductors $k_1$ and $k_2$ are connected in parallel between $100^\circ \mathrm{C}$ and $\theta$.
2. The conductor $k_3$ is connected in series with the combination of $k_1$ and $k_2$.

Step 2: Heat Current for Parallel Combination

For conductors $k_1$ and $k_2$ in parallel, the effective thermal conductivity $k_\text{eff}$  with Area $2A$ is given by:

$k_{\text{eff}}=\frac{k_1+k_2}{2}$

Substitute the given values:

$k_{\text{eff}}=\frac{60+120}{2}=90\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$

Step 3: Heat Current for the Series Combination

The total heat current $H$ remains constant throughout the system. Using Fourier’s law of heat conduction:

$H=\frac{k_{\text{eff}}\left(2A\right)(100-\theta )}{L}=\frac{k_3\left(2A\right)(\theta -0)}{L}$

Simplify:

$\frac{90(100-\theta )}{L}=\frac{135\theta }{L}$

  $$\theta = \frac{18000}{450} = 40^\circ \mathrm{C}$$