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JCECE Medical JCECE Medical Solved Paper-2011

  • question_answer
    A doubled layered wall has layer A, 10 cm thick and B, 20 cm thick. The thermal conductivity of A is thrice that of B. In the steady state, the  temperature difference across the wall is \[35{{\,}^{o}}C.\] The temperature difference across the layer A  is

    A) \[28{{\,}^{o}}C\]             

    B) \[14{{\,}^{o}}C\]

    C)  \[7{{\,}^{o}}C\]              

    D) \[5{{\,}^{o}}C\]

    Correct Answer: D

    Solution :

     In the steady state, rate of heat across each layer of the wall is the same \[\frac{{{K}_{A}}A(\Delta T)}{10}=\frac{{{K}_{B}}A(35-\Delta T)}{20}\] using         \[{{K}_{A}}=3{{K}_{B}}\] we get        \[\Delta T=5{{\,}^{o}}C\]


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