Two slabs A and B of different materials but of the same thickness are joined as shown in figure. The thermal conductivities of A and B are k1 and k2 respectively. The thermal conductivity of the composite slab will be
The thermal resistance of a slab of length l,area A and thermal conductivity kis given by
R=lkAFor slab A: R1=12k1AFor slab B:R2=12k2A
Since the slabs are joined in series, the thermal resistance of the composite slab is
Or 1kcA=12k1A+12k2AOr 1kc=12(1k1+1k2)Which gives kc=2k1k2(k1+k2)
Hence the correct choice is (d).