Solution: Thermal resistance , Thermal resistance, R= 1k×LAHere, k=thermal conductivity, L= lenght or thickness, A=areaAccording to question, area (A) is constant,so, R∝LkFor, product I:Lk=412=13=0 33For, product II:Lk=66=1For, product III:Lk=24=0 5So, RII>RIII>R1Hence, the product (II) will be the best thermal insulator

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Thermal resistance ,

Thermal resistance, R = \frac{1}{k} \times \frac{L}{A} Here, k = thermal conductivity, L = lenght or thickness, A = area According to question, area (A) is constant, so, R \propto \frac{L}{k} For, product (I) : \frac{L}{k} = \frac{4}{12} = \frac{1}{3} = 0.33 For, product (II) : \frac{L}{k} = \frac{6}{6} = 1 For, product (III) : \frac{L}{k} = \frac{2}{4} = 0.5 So, R_{(II)} > R_{(III)} > R_{(1)} Hence, the product (II) will be the best thermal insulator .

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