Q. Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities $K$ and 2$K$ respectively. The equivalent thermal conductivity of the slab is
A) $\sqrt{2K}$
B) 3K
C) $\frac{4}{3}K$
D) $\frac{2}{3}K$
Explanation
(c) Equivalent thermal conductivity of the compound, slab,
$K_{eq} = \ \frac{l_{1} + l_{2}}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}} = \ \frac{l + l}{\frac{l}{K} + \frac{l}{2K}}$
= $\frac{2l}{\frac{3l}{2K}} = \ \frac{4}{3}$ K
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.