Q.

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 slabs taken in series is

Moderate

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a

25K

b

3K

c

2K

d

43K

answer is 4.

(Detailed Solution Below)

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Detailed Solution

resistance in series is Rseries=R1+R2 where R = lKA, here l is thickness or length, A is area of cross section,  K is coefficient of thermal conductivity when 2 slabs of different coefficient of thermal conductivity K,2K ,  having same thickness and area of cross section are taken in series effective thermal conductivity is Kseries, given by, l+lKseries=lK+l2K ⇒2Kseries=32K ⇒Kseries=43K
ctaimg
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