JEE Physics 2026 Question with Solution

Given: Temperatures at the ends are $100^\circ\text{C}$ at $A$ and $40^\circ\text{C}$ at $F$. All rods have equal length and area. Thermal conductivity $k_x = 3k_y$.

Find: The temperatures at junctions $B$ and $E$.

Use the thermal resistance relation

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

Since $k_x = 3k_y$, we get

$$R_x = \frac{L}{k_x A} = \frac{1}{3}\frac{L}{k_y A} = \frac{1}{3}R_y$$

Between $B$ and $E$, heat can flow through two parallel branches:

$$B \rightarrow C \rightarrow E$$

and

$$B \rightarrow D \rightarrow E$$

The upper branch has resistance $2R_y$ and the lower branch has resistance $2R_x$. Therefore, the equivalent resistance between $B$ and $E$ is

$$R_{BE} = \left(\frac{1}{2R_y} + \frac{1}{2R_x}\right)^{-1}$$

Substituting $R_x = \frac{1}{3}R_y$,

$$R_{BE} = \left(\frac{1}{2R_y} + \frac{3}{2R_y}\right)^{-1} = \left(\frac{4}{2R_y}\right)^{-1} = \frac{R_y}{2}$$

Now the full path from $A$ to $F$ is a series combination:

$$R_{\text{total}} = R_x + R_{BE} + R_y$$

So,

$$R_{\text{total}} = \frac{R_y}{3} + \frac{R_y}{2} + R_y = \frac{11R_y}{6}$$

The total temperature difference is

$$\Delta T = 100 - 40 = 60^\circ\text{C}$$

The temperature drop across $AB$ is proportional to $R_x$:

$$\Delta T_{AB} = 60 \times \frac{R_x}{R_{\text{total}}} = 60 \times \frac{\frac{R_y}{3}}{\frac{11R_y}{6}} = 60 \times \frac{2}{11} \approx 20^\circ\text{C}$$

Hence,

$$T_B \approx 100 - 20 = 80^\circ\text{C}$$

The temperature drop across $EF$ is proportional to $R_y$:

$$\Delta T_{EF} = 60 \times \frac{R_y}{R_{\text{total}}} = 60 \times \frac{R_y}{\frac{11R_y}{6}} = 60 \times \frac{6}{11} \approx 30^\circ\text{C}$$

Therefore,

$$T_E \approx 40 + 30 = 70^\circ\text{C}$$

So the temperatures are $80^\circ\text{C}$ and $70^\circ\text{C}$ respectively. The correct option is C.