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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. 4/3 K
2. 2/3 K
3. √3 K
4. 3 K

Correct answer: 4/3 K

Solution

The equivalent thermal conductivity for two slabs of equal thickness in series is given by the harmonic mean: Keq = 2K1K2 / (K1 + K2). Substituting K1 = K and K2 = 2K, we get Keq = 4K/3.

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