The temperature θ at the junction of two insulating sheets, having thermal resistances R1 and R2 as well as top and bottom temperatures θ 1 and θ 2 (as shown in figure) is given by: (a) (θ₁R₂+θ₂R₁)/(R₁+R₂) (b) (θ₁R₁+θ₂R₂)/(R₁+R₂) (c) (θ₁R₂-θ₂R₁)/(R₂-R₁) (d) (θ₂R₂-θ₁R₁)/(R₂-R₁)

Answer: (a). Same heat current through both sheets: (θ₂ - θ)/(R₂) = (θ - θ₁)/(R₁) θ₂R₁ - θ R₁ = θ R₂ - θ₁R₂ θ = (θ₁R₂ + θ₂R₁)/(R₁+R₂), so the answer is (a).

The temperature θ at the junction of two insulating sheets, having… | JEE Main 2021 (26 Feb, Shift 1)

Q39JEE Main 2021 · 26 FEB · SHIFT 1Medium

Heat Conduction

The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances R1 and R2 as well as top and bottom temperatures $\theta$ 1 and $\theta$ 2 (as shown in figure) is given by:

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Answer: (a)

Same heat current through both sheets: $\frac{\theta_2 - \theta}{R_2} = \frac{\theta - \theta_1}{R_1}$
$\theta_2R_1 - \theta R_1 = \theta R_2 - \theta_1R_2$
$\theta = \frac{\theta_1R_2 + \theta_2R_1}{R_1+R_2}$ , so the answer is (a).

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Heat Conduction question from JEE Main 2021 (26 Feb, Shift 1)

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