JEE Main & Advanced Physics Transmission of Heat Question Bank Conduction

  • question_answer
    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 slab is                          [CBSE PMT 2003]

    A)            \[\sqrt{2K}\]                        

    B)            \[3K\]

    C)            \[\frac{4}{3}K\]                   

    D)            \[\frac{2}{3}K\]

    Correct Answer: C

    Solution :

               \[K=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}=\frac{2.K.2K}{K+2K}=\frac{4}{3}K\]


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