Rods $$x$$ and $$y$$ of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $$A$$ and $$F$$ are maintained at $$100 ^{o}C$$ and $$40 ^{o}C$$ respectively. Given the thermal conductivity of rod $$x$$ is three times of that of rod $$y$$, the temperature at junction points $$B$$ and $$E$$ are (close to):

We can model the thermal conduction network as an electrical circuit where thermal resistance is $$R = \frac{L}{KA}$$. Since dimensions are equal and $$K_x = 3K_y$$, the resistance of rod $$y$$ is three times that of rod $$x$$. Let $$R_x = R$$ and $$R_y = 3R$$.

1. Equivalent Resistance of the Network 

The middle section consists of two parallel paths between junctions $$B$$ and $$E$$:

• Path BCE: $$R_y + R_y = 3R + 3R = 6R$$
• Path BDE: $$R_x + R_x = R + R = 2R$$

Equivalent resistance $$R_{BE} = \frac{6R \times 2R}{6R + 2R} = 1.5R$$.

The total resistance of the series system $$A \to B \to E \to F$$ is: 

$$R_t​​=R+1.5R+3R=5.5R$$

2. Temperature at Junction B ($$T_B$$) 

The temperature drop across rod $$AB$$ is proportional to its resistance relative to the total:

$$100 - T_B = \frac{R_{AB}}{R_{total}}(100 - 40)$$

$$100 - T_B = \frac{R}{5.5R}(60) \approx 10.9^\circ\text{C}$$

$$T_B \approx 89.1^\circ\text{C}$$

3. Temperature at Junction E ($$T_E$$) 

Similarly, the temperature at $$E$$ can be found by calculating the drop from $$A$$ to $$E$$:

$$T_E = 100 - \frac{R_{AB} + R_{BE}}{R_{total}}(60)$$

$$T_E = 100 - \frac{2.5R}{5.5R}(60) \approx 100 - 27.3^\circ\text{C}$$

$$T_E \approx 72.7^\circ\text{C}$$

Rounding to the nearest whole numbers, we get $$89^\circ\text{C}$$ and $$73^\circ\text{C}$$.