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:

• Correct answer
  A

  $\frac{4}{3} K$
• B

  $\sqrt{2} K$
• C

  $3 \mathrm{~K}$
• D

  $\frac{2}{3} K$

Solution

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

Hence, the correct answer is option: 'A'.