Question

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

Answer 1

Correct Answer - D
`K_(eq) = (2K_(1)K_(2)) /(K_(1)+K_(2))= (2(K) (2K))/(K+2K) = 4/3 K`