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 conductivity when the slabs are connected in series is
`K_(eq) = (sumx_(n))/((sumx_(n))/(sumK_(n)))= (x+x)/(x/K+x/(2K)) = (2x(2K^(2)))/((2Kx+Kx))=4/3 K`