Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is (A) √2K (B) 3K (C) (4/3)K (D) (2/3)K

The thermal resistances can be added as they are in series. Here the width of the combined slab is twice the width of individual slab

Hence, we have:

2L/KeqA = (L/KA) + {L/(2K)A}

⇒ (2K + k)/ 2K2 = 2/Keq

⇒ Keq = (4/3)K

Therefore,

The correct option is (C)

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