Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively. The equivalent thermal conductivity of the slab is
(4) 43K
The two layers are in series (same heat current through both). Let each have thickness d and area A; total thickness = 2d.
Thermal resistance of a layer = dKA
⇒ 2dKeqA = dKA + d2KA
⇒ 2Keq = 1K + 12K = 32K
∴ Keq = 43K
Answer: (4)
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