Two slabs A and B of equal surface area are placed one over the other such that their surfaces are completely in contact. The thickness and the coefficient of thermal conductivity of slab A are twice those of B. The first surface of slab A is maintained at $100°$C, while the second surface of slab B is maintained at $25°$C. The temperature at the contact of their surfaces is
Answer: (B) $62.5°$C
Concept. Slabs in series: the temperature drop across each is proportional to its thermal resistance $\dfrac{l}{KA}$.
Step 1. Compare resistances. A has twice the thickness and twice $K$ of B, so the two resistances are equal.
Step 2. Interface temperature. The drop is shared equally:
$$\theta = \frac{100 + 25}{2} = 62.5°\text{C}$$
Answer. $62.5°$C (Option B).
Solution by Sreeraj P, M.Sc Physics