10-2.Transmission of Heat
medium

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

A

$\sqrt {2K} $

B

$3K$

C

$\frac{4}{3}K$

D

$\frac{2}{3}K$

(AIPMT-2003)

Solution

(c) $K = \frac{{2{K_1}{K_2}}}{{{K_1} + {K_2}}} = \frac{{2.K.2K}}{{K + 2K}} = \frac{4}{3}K$

Standard 11
Physics

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