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JCECE Engineering JCECE Engineering 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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