JEE Main 2025 Apr 7 Shift 2, Physics Q23: Heat Transfer

JEE Main2025Apr 7, Shift 2Physics

Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are :

$\frac{L _{A}}{L _{B}} = \frac{1}{2}$ , $\frac{r _{A}}{r _{B}} = 2$ and $\frac{K _{A}}{K _{B}} = \frac{1}{2}$ . The free ends of rods A and B are maintained at 400 K, 200 K, respectively. The temperature of rods interface is ____________ K, when equilibrium is established.

Your answer

Solution

In steady state, the same rate of heat flow $\frac{d Q}{d t} = \frac{Δ T}{R _{t h}}$ passes through both rods, where the thermal resistance of a cylinder is $R_{t h} = \frac{L}{K A}$ and $A = π r^{2}$ .

The ratio of thermal resistances is $\frac{R _{A}}{R _{B}} = \frac{L _{A}}{L _{B}} \cdot \frac{r _{B}^{2}}{r _{A}^{2}} \cdot \frac{K _{B}}{K _{A}} = \frac{1}{2} \cdot \frac{1}{4} \cdot 2 = \frac{1}{4}$ .

Equating heat flows across the two rods: $\frac{400 - T}{R _{A}} = \frac{T - 200}{R _{B}} \Rightarrow \frac{400 - T}{T - 200} = \frac{R _{A}}{R _{B}} = \frac{1}{4}$ .

Cross-multiplying: $4 (400 - T) = T - 200 \Rightarrow 1600 - 4 T = T - 200 \Rightarrow 5 T = 1800 \Rightarrow T = 360$ K.

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JEE Main 2025 Apr 7 Shift 2, Physics Q23: Heat Transfer

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