JEE Physics 2025 Question with Solution

Given: The three conductors have the same length. Their thermal conductivities are $k_1 = 60 \, \text{Js}^{-1}\text{m}^{-1}\text{K}^{-1}$, $k_2 = 120 \, \text{Js}^{-1}\text{m}^{-1}\text{K}^{-1}$, and $k_3 = 135 \, \text{Js}^{-1}\text{m}^{-1}\text{K}^{-1}$. Also, $A_1 = A_2 = A$ and $A_3 = 2A$.

Find: The steady-state junction temperature $\theta$.

In steady state, the total heat current entering the junction equals the heat current leaving the junction.

For a conductor,

$$\dot Q = \frac{kA\Delta T}{L}$$

So for the three branches,

$$\dot Q_1 = \frac{k_1A(100-\theta)}{L}$$ $$\dot Q_2 = \frac{k_2A(100-\theta)}{L}$$ $$\dot Q_3 = \frac{k_3(2A)(\theta-0)}{L}$$

Applying heat balance at the junction,

$$\dot Q_1 + \dot Q_2 = \dot Q_3$$

Therefore,

$$\frac{k_1A(100-\theta)}{L} + \frac{k_2A(100-\theta)}{L} = \frac{2k_3A\theta}{L}$$

Cancelling the common factor $\frac{A}{L}$,

$$(k_1+k_2)(100-\theta)=2k_3\theta$$

Substitute the given values:

$$(60+120)(100-\theta)=2\times 135\theta$$ $$180(100-\theta)=270\theta$$ $$18000-180\theta=270\theta$$ $$18000=450\theta$$ $$\theta=40$$

Therefore, the value of $\theta$ is $40^\circ\text{C}$.

Given: Thermal conductance of a rod is $G=\frac{kA}{L}$. Here $A_1=A_2=A$ and $A_3=2A$.

Find: The junction temperature $\theta$.

Write the conductances:

$$G_1=\frac{k_1A}{L}, \qquad G_2=\frac{k_2A}{L}, \qquad G_3=\frac{2k_3A}{L}$$

The two left rods are connected to $100^\circ\text{C}$ and the right rod is connected to $0^\circ\text{C}$. Hence,

$$\dot Q_1=G_1(100-\theta), \qquad \dot Q_2=G_2(100-\theta), \qquad \dot Q_3=G_3\theta$$

At steady state,

$$\dot Q_1+\dot Q_2=\dot Q_3$$

So,

$$\frac{k_1A}{L}(100-\theta)+\frac{k_2A}{L}(100-\theta)=\frac{2k_3A}{L}\theta$$

After cancelling $\frac{A}{L}$,

$$k_1(100-\theta)+k_2(100-\theta)=2k_3\theta$$

Substituting $k_1=60$, $k_2=120$, and $k_3=135$,

$$(60+120)(100-\theta)=270\theta$$ $$180(100-\theta)=270\theta$$ $$18000-180\theta=270\theta$$ $$\theta=40$$

Thus, the required temperature is $40^\circ\text{C}$.