Question

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 $100^\circ$C and $40^\circ$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): 

• A. $60^\circ$C and $45^\circ$C respectively
• B. $89^\circ$C and $73^\circ$C respectively
• C. $80^\circ$C and $70^\circ$C respectively
• D. $80^\circ$C and $60^\circ$C respectively

Hint:

In steady-state heat conduction networks, treat rods like resistors and apply series–parallel combinations.

Answer 1

The Correct Option is C

Step 1: Thermal resistance concept.
Thermal resistance: \[ R = \frac{L}{kA} \] Given: \[ k_x = 3k_y \Rightarrow R_x = \frac{1}{3}R_y \] Step 2: Analyze heat flow.
Between $B$ and $E$, heat flows through two parallel paths: \[ B \rightarrow C \rightarrow E \quad (y\text{-rods}) \] \[ B \rightarrow D \rightarrow E \quad (x\text{-rods}) \] Effective resistance between $B$ and $E$: \[ R_{BE} = \left(\frac{1}{2R_y} + \frac{1}{2R_x}\right)^{-1} \] \[ = \left(\frac{1}{2R_y} + \frac{3}{2R_y}\right)^{-1} = \frac{R_y}{2} \] Step 3: Series combination from $A$ to $F$.
Total resistance: \[ R_{\text{total}} = R_x + R_{BE} + R_y = \frac{R_y}{3} + \frac{R_y}{2} + R_y = \frac{11R_y}{6} \] Total temperature difference: \[ \Delta T = 100 - 40 = 60^\circ\text{C} \] Step 4: Temperature drops.
Drop across $AB$: \[ \Delta T_{AB} = 60 \times \frac{R_x}{R_{\text{total}}} = 60 \times \frac{2}{11} \approx 20 \] \[ T_B \approx 100 - 20 = 80^\circ\text{C} \] Drop across $EF$: \[ \Delta T_{EF} = 60 \times \frac{6}{11} \approx 30 \] \[ T_E \approx 40 + 30 = 70^\circ\text{C} \] Final Answer: $\boxed{80^\circ\text{C and }70^\circ\text{C}}$