JEE Main 2025 Jan 22 Shift 1, Physics Q23: Heat Transfer

JEE Main2025Jan 22, Shift 1Physics

Three conductors of same length having thermal conductivity $k_{1}$ , $k_{2}$ and $k_{3}$ are connected as shown in figure.

Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of $θ$ is __________ $^{\circ}$ C.

(Given : $k_{1} = 60$ Js $^{- 1}$ m $^{- 1}$ K $^{- 1}$ , $k_{2} = 120$ Js $^{- 1}$ m $^{- 1}$ K $^{- 1}$ , $k_{3} = 135$ Js $^{- 1}$ m $^{- 1}$ K $^{- 1}$ )

Your answer

Solution

Thermal resistances: $R_{i} = \frac{L}{k _{i} A _{i}}$ . With unit length $L$ and cross-section $A$ (or $2 A$ for rod 3):

$R_{1} = \frac{L}{60 A}$ , $R_{2} = \frac{L}{120 A}$ , $R_{3} = \frac{L}{135 \cdot 2 A} = \frac{L}{270 A}$ .

Rods 1 and 2 in parallel: $\frac{1}{R _{12}} = \frac{60 A}{L} + \frac{120 A}{L} = \frac{180 A}{L}$ , so $R_{12} = \frac{L}{180 A}$ .

At the junction (temperature $θ$ ), heat flowing in from $10 0^{\circ}$ C equals heat flowing out to $0^{\circ}$ C:

$\frac{100 - θ}{R _{12}} = \frac{θ - 0}{R _{3}}$ .

$(100 - θ) \cdot 180 = θ \cdot 270 \Rightarrow 18000 - 180 θ = 270 θ \Rightarrow 450 θ = 18000 \Rightarrow θ = 4 0^{\circ}$ C.

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JEE Main 2025 Jan 22 Shift 1, Physics Q23: Heat Transfer

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