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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    1\[1.56\times {{10}^{5}}\,J\] of heat is conducted through a \[\text{2}\,{{\text{m}}^{\text{2}}}\]wall of 12 cm thick in 1h. Temperature difference between the two sides of the wall is \[\text{20}{{\,}^{\text{o}}}\text{C}\text{.}\] The thermal conductivity of the material of the wall is (in \[\text{W}\,{{\text{m}}^{-1}}{{K}^{-1}}\])

    A)  0.11                                      

    B)  0.13

    C)  0.15                                      

    D)  1.2

    Correct Answer: B

    Solution :

                     Given that \[1.56\times {{10}^{5}}\,J\]of heat is conducted through a \[2{{m}^{2}}\] wall having a thickness of 12 cm in a time interval of 1 h. Hence, we have \[\frac{dQ}{dt}=\frac{KA\Delta T}{x}\] Thus, we get      \[\frac{1.56\times {{10}^{5}}}{1\times 60\times 60}=\frac{K\times 2\times 20}{12\times {{10}^{-2}}}\] So, \[K=\frac{1.56\times {{10}^{5}}\times 12\times {{10}^{-12}}}{1\times 60\times 60\times 2\times 20}=\frac{1.56}{12}=0.13\]


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