Three conductors of same length having thermal conductivity $$k_1, k_2 \text{ and } K_3$$ are connected as shown in figure.

Area of cross sections of $$ 1^{st} \text{ and }2^{nd}$$ 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 0 is _______ $$^oC$$.
$$(\text{Given }:k_1=60Js^{-1}m^{-1}K^{-1},k_2= 120Js^{-1}m^{-1}K^{-1}, k_3= 135Js^{-1}m^{-1}K^{-1})$$

Treat it as two conductors $$k_1,k_2$$​ in parallel, connected in series with conductor $$k_3$$

Let area of conductors 1 and 2 be A.

Given for conductor 3:

$$A_3=2A$$

All lengths are same l.

Thermal resistance:

$$R=\frac{l}{kA}$$

For conductor 1:

$$R_1=\frac{l}{60A}$$

For conductor 2:

$$R_2=\frac{l}{120A}$$

These are in parallel:

$$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{60A}{l}+\frac{120A}{l}=l60A​+l120A​$$

$$=\frac{180A}{l}$$

So

$$R_p=\frac{l}{180A}$$

Equivalent conductivity of left section:

$$k_{eq}=180$$

Now conductor 3:

$$R_3=\frac{l}{135(2A)}=\frac{l}{270A}$$

Suppose junction temperature is θ\thetaθ.

Heat flow through left section:

$$Q_1=\frac{(100-\theta)}{R_p}$$

$$=(100-\theta)\frac{180A}{l}$$

Heat flow through right section:

$$Q_2=\frac{\theta}{R_3}=\theta\frac{270A}{l}$$

Steady state:

$$Q_1=Q_2$$

$$180(100-\theta)=270\theta$$

$$18000=450\theta$$

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