MCQ
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
    $\Big(\frac23\Big)\text{K}$
  • B
    $\sqrt{2}\text{K}$
  • C
    $3\text{K}$
  • D
    $\Big(\frac{4}{3}\Big)\text{K}$

Answer

  1. $\Big(\frac{4}{3}\Big)\text{K}$

Explanation:

When slabs are connected in series, the equivalent thermal conductivity will be given by,

$\text{K}'=\frac{\sum\text{x}_\text{i}}{\sum\frac{\text{x}_\text{i}}{\text{K}_\text{i}}}$

$=\frac{\text{x}+\text{x}}{\frac{\text{x}}{\text{K}}+\frac{\text{x}}{2\text{K}}}=\frac{\text{2x}}{\frac{\text{3x}}{\text{2K}}}=\frac43\text{K}$

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