Consider a compound slab consisting of two different material having equal thickness and thermal conductivities `K` and `2K` respectively. The equivalent thermal conductivity of the slab is

Options

A. `(2//3)K`
B. `sqrt(2)K`
C. `3 K`
D. `(4//3)K`

Answer

D

Explanation

The equivalent thermal condutivity of two slabs connected in series is given by
`K=(2K_(1)K_(2))/(K_(1)+K_(2)) = (2Kxx2K)/(K+2K) = (4)/(3) K`