Consider a compound slab consisting of two different materials having equal thickness and thermal…

Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities $K$ and $2 K$, respectively. The equivalent thermal conductivity of the slab is:
  1. $\frac{4}{3} K$
  2. $\sqrt{2} K$
  3. $3 \mathrm{~K}$
  4. $\frac{2}{3} K$

Solution

$\therefore \frac{2 l}{K_{e g}(A)}=\frac{l}{2 K A}+\frac{l}{K A}$ for series connection $\mathrm{R}=\mathrm{R}_1+\mathrm{P}_2$ $\Rightarrow K_{e g}=\frac{4}{3} K$ .

Asked in: NEET 2003

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