JEE Physics 2025 Question with Solution

Given:

• Temperature at the free end of rod A is $400 \, \text{K}$.
• Temperature at the free end of rod B is $200 \, \text{K}$.
• $\frac{L_A}{L_B} = \frac{1}{2}$
• $\frac{r_A}{r_B} = 2$
• $\frac{K_A}{K_B} = \frac{1}{2}$

Find: The interface temperature $T$.

For a cylindrical rod, thermal resistance is

$$R_{th} = \frac{L}{KA} = \frac{L}{K\pi r^2}$$

So,

$$R_{th,A} = \frac{L_A}{K_A \pi r_A^2}, \qquad R_{th,B} = \frac{L_B}{K_B \pi r_B^2}$$

Now,

$$\frac{R_{th,A}}{R_{th,B}} = \frac{L_A}{L_B} \cdot \frac{K_B}{K_A} \cdot \left(\frac{r_B}{r_A}\right)^2$$

Substituting the given ratios,

$$\frac{R_{th,A}}{R_{th,B}} = \left(\frac{1}{2}\right) \cdot 2 \cdot \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$

Hence,

$$R_{th,B} = 4R_{th,A}$$

At equilibrium, the heat current through both rods is the same. If the interface temperature is $T$, then

$$\frac{400 - T}{R_{th,A}} = \frac{T - 200}{R_{th,B}}$$

Using $R_{th,B} = 4R_{th,A}$,

$$\frac{400 - T}{R_{th,A}} = \frac{T - 200}{4R_{th,A}}$$

So,

$$4(400 - T) = T - 200$$ $$1600 - 4T = T - 200$$ $$1800 = 5T$$ $$T = 360 \, \text{K}$$

Therefore, the temperature of the rods interface is $360 \, \text{K}$.

Given: The same ratios and end temperatures as above.

Find: The correct interface temperature.

The first approach on the page writes the conduction equation correctly as

$$\frac{K_A \pi r_A^2 (400 - T)}{L_A} = \frac{K_B \pi r_B^2 (T - 200)}{L_B}$$

but simplifies the ratio incorrectly.

Using

$$\frac{K_A}{K_B} = \frac{1}{2}, \qquad \frac{r_A}{r_B} = 2, \qquad \frac{L_A}{L_B} = \frac{1}{2}$$

we get

$$\frac{K_A r_A^2 / L_A}{K_B r_B^2 / L_B} = \frac{K_A}{K_B} \cdot \left(\frac{r_A}{r_B}\right)^2 \cdot \frac{L_B}{L_A}$$ $$= \left(\frac{1}{2}\right) \cdot 4 \cdot 2 = 4$$

Therefore,

$$4(400 - T) = T - 200$$

and not

$$2(400 - T) = T - 200$$

Now solve:

$$1600 - 4T = T - 200$$ $$1800 = 5T$$ $$T = 360 \, \text{K}$$

So the value $333.33 \, \text{K}$ shown in Approach 1 comes from an algebraic simplification error. The correct interface temperature is $360 \, \text{K}$.