Q.

A slab consists of two layers of two different materials of same thickness having thermal conductivities K1 and K2, the equivalent conductivity of the combination in series is

Moderate

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K1+K2

K1+K22

2K1K2K1+K2

K1+K22K1K2

answer is 3.

(Detailed Solution Below)

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Detailed Solution

In series slabs are,for 2 slabs of same thickness l having coefficient of thermal conductivities K1,K2 is since, in series effective resistance is Rseries=R1+R2 Resistance is R=lKA, here l = thickness, K=coefficient of thermal conductivity, A=area of cross section Rseries=R1+R2 l+lKseriesA=lK1A+lK2A ⇒Kseries=2K1K2K1+K2

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