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Q.

A composite slab consists of two plates of thicknesses L1 and L2 and thermal conductivities K1 and K2. The cross-sectional areas are equal. The equivalent thermal conductivity is:

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a

K1K2(L1+L2)K2L1+K1L2

b

K2L1+K1L2K1K2(L1+L2)

c

L1L2(K1+K2)(K2L1+K1L2)

d

K2L1+K1L2L1L2(K1+K2)

answer is A.

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Detailed Solution

In series connectionKeq=∑li∑liKi=L1+L2L1K1+L2K2=K1K2(L1+L2)K2L1+K1L2
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