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JEE Main & Advanced Sample Paper JEE Main Sample Paper-18

  • question_answer
    A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B. Under thermal equilibrium, the temperature difference across the layer A, if the temperature across the free surfaces is \[36{}^\circ C\], is

    A)  \[6{}^\circ C\]                                  

    B)  \[12{}^\circ C\]

    C)  \[18{}^\circ C\]                

    D)  \[24{}^\circ C\]

    Correct Answer: B

    Solution :

     \[\frac{2kA\Delta T}{x}=\frac{kA(36-\Delta T)}{x}\] \[\Rightarrow \] \[2\Delta T=36-\Delta T\] \[\Rightarrow \] \[\Delta T=12{}^\circ C\]


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