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Correct option is C)

$H=H_{1}+H_{2}$

Given that two slab are in parallel and have same dimensions.

$LKΓAΓ(T_{2}βT_{1})β=LK_{1}Γ2AβΓ(T_{2}βT_{1})β+LK_{2}Γ2AβΓ(T_{2}βT_{1})β$,

whereΒ $K$ is the effective thermal conductivity, $T_{2}βT_{1}$ is the temperature difference between the ends of the slab, $A$ is area of the slab and $L$ is the thickness of the each layer of the slab

So, $LKΓAβ=LK_{1}Γ2Aββ+LK_{2}Γ2Aββ$

or, $K=2K_{1}+K_{2}β$

Solve any question of Thermal Properties Of Matter with:-

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