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Consider a compound slab consisting of two different material having equal thickness and thermal conductivities `K` and `2K` respectively. The equivalent thermal conductivity of the slab isA. `sqrt(2K)`B. `3K`C. `4/3K`D. `2/3 K` |
Answer» Correct Answer - C `K=(2K_(1)K_(2))/(K_(1)+K_(2)) = (2.K.2K)/(K+2K) = 4/3 K`. |
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