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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is :

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

    B)  \[3K\]

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

    D)  \[\frac{2}{3}K\]

    Correct Answer: C

    Solution :

                     Equivalent thermal conductivity of the compound slab, \[{{K}_{eq}}=\frac{{{l}_{1}}+{{l}_{2}}}{\frac{{{l}_{1}}}{{{K}_{1}}}+\frac{{{l}_{2}}}{{{K}_{2}}}}=\frac{l+l}{\frac{l}{K}+\frac{l}{2K}}\] \[=\frac{2l}{3l}=\frac{4}{3}K\]


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