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Properties of Matter question

2003 · Q165
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Properties of Matter question

2003 · Q165

NEETPhysicsProperties of MatterMCQ+4 / −1
Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities K and 2K, respectively. The equivalent thermal conductivity of the slab is
  1. A
    23K{2 \over 3}K32​K
  2. B
    2K\sqrt 2 K2​K
  3. C
    3 K
  4. D
    43K{4 \over 3}K34​K
View written solutionFree

Correct answer: D

In series, equivalent thermal conductivity

Keq=2K1K2K1+K2{K_{eq}} = {{2{K_1}{K_2}} \over {{K_1} + {K_2}}}Keq​=K1​+K2​2K1​K2​​

⇒Keq=2×K×2KK+2K=43K \Rightarrow {K_{eq}} = {{2 \times K \times 2K} \over {K + 2K}} = {4 \over 3}K⇒Keq​=K+2K2×K×2K​=34​K

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