A slab consists of two layers of different materials of the same thickness and having thermal conductivities `K_(1)` and `K_(2)`. The equivalent thermal conductivity of the slab is

Options

A. `K_(1) + K_(2)`
B. `(K_(1) K_(2))/(K_(1) + K_(2))`
C. `sqrt(K_(1) K_(2))`
D. `(2K_(1) K_(2))/(K_(1) + K_(2))`

Answer

D

Explanation

(4) `R_(eq) = R_(1) + R_(2)`
`(2L)/(K_(eq) A) = (L)/(K_(1) A) + (L)/(K_(2) A)`
`K_(eq) = (2 K_(1) K_(2))/(K_(1) + K_(2))`