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JEE Main 2025 (22 Jan, Shift 1) · Heat Conduction

Heat Conduction question from JEE Main 2025 (22 Jan, Shift 1)

Q17JEE Main 2025 · 22 JAN · SHIFT 1Difficult

Heat Conduction

Three conductors of same length having thermal conductivity k1,k2k_1, k_2 and k3k_3 are connected as shown in figure.

Figure

Area of cross sections of 11st and 22nd conductor are same and for 33rd conductor it is double of the 11st conductor. The temperatures are given in the figure. In steady state condition, the value of θ\theta is _________ ∘C^\circ\text{C}. (Given: k1=60 J s−1 m−1 K−1, k2=120 J s−1 m−1 K−1, k3=135 J s−1 m−1 K−1k_1 = 60\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1},\ k_2 = 120\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1},\ k_3 = 135\text{ J s}^{-1}\text{ m}^{-1}\text{ K}^{-1})

Numerical answer type. Work it out, then open the answer.

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Answer: 40

  1. Rods 1 and 2 are in parallel between 100∘100^\circC and θ\theta; rod 3 (area 2A2A) carries the same total heat from θ\theta to 0∘0^\circC.
  2. (k1+k2)A(100−θ)l=k3(2A)θl\dfrac{(k_1+k_2)A(100-\theta)}{l} = \dfrac{k_3(2A)\theta}{l}
  3. 180(100−θ)=270θ⇒450θ=18000180(100-\theta) = 270\theta \Rightarrow 450\theta = 18000
  4. θ=40∘\theta = 40^\circC
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