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Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d respectively are joined face to face to form a slab, the heat flowing straight through both. The outer surface of the 3K material is at temperature θ₂ and the outer surface of the K material is at temperature θ₁, with θ₂>θ₁. The temperature at the interface is

Asked in JEE Main 9th April 1st Shift 2019 · Junction temperature

• (1) (θ₁)/(10)+(9θ₂)/(10)
• (2) (θ₁)/6+(5θ₂)/6
• (3) (θ₂+θ₁)/2
• (4) (θ₁)/3+(2θ₂)/3

Step-by-step solution

Why the other options are wrong

• (2) comes from taking the resistance ratio as 1:5, which is what you get by dividing the thicknesses d and 3d by the conductivities 3K and K but then adding rather than comparing — a 1/3:3/1 pair misread as 1/3:5/3. The true ratio is d/(3K):(3d)/K=1:9.
• (3) comes from is the average of the two face temperatures, the classic trap. Averaging would be right only if the two layers had equal thermal resistance; here one has nine times the resistance of the other.
• (4) comes from using the thicknesses alone, d:3d=1:3, and forgetting that the good conductor also has three times the conductivity. Both effects push the same way, and together they give 1:9, not 1:3.

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