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Consider a compound slab consisting of two different materials having equal lengths, thickness and thermal conductivities $K$ and $2 K$ respectively. The equivalent thermal conductivity of the slab is
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Verified Answer
The correct answer is:
$\frac{4}{3} K$
Equivalent thermal conductivity of the compound slab,
$$
K_{\mathrm{eq}}=\frac{l_1+l_2}{\frac{l_1}{K_1}+\frac{l_2}{K_2}}=\frac{l+l}{\frac{l}{K}+\frac{l}{2 K}}=\frac{2 l}{\frac{3 l}{2 K}}=\frac{4 K}{3}
$$
$$
K_{\mathrm{eq}}=\frac{l_1+l_2}{\frac{l_1}{K_1}+\frac{l_2}{K_2}}=\frac{l+l}{\frac{l}{K}+\frac{l}{2 K}}=\frac{2 l}{\frac{3 l}{2 K}}=\frac{4 K}{3}
$$
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