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JEE Main & Advanced Physics Transmission of Heat Question Bank Critical Thinking

  • question_answer
    The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1 (T2  > T1). The rate of heat transfer through the slab, in a steady state is \[\left( \frac{A({{T}_{2}}-{{T}_{1}})K}{x} \right)f\], with ¦ which equal to      [AIEEE 2004]

    A)            1

    B)            \[\frac{1}{2}\]

    C)            \[\frac{2}{3}\]

    D)            \[\frac{1}{3}\]

    Correct Answer: D

    Solution :

                       Equation of thermal conductivity of the given combination \[{{K}_{eq}}=\frac{{{l}_{1}}+{{l}_{2}}}{\frac{{{l}_{1}}}{{{K}_{1}}}+\frac{{{l}_{2}}}{{{K}_{2}}}}=\frac{x+4x}{\frac{x}{K}+\frac{4x}{2K}}=\frac{5}{3}K\]. Hence rate of flow of heat through the given combination is \[\frac{Q}{t}=\frac{{{K}_{eq}}.A({{T}_{2}}-{{T}_{1}})}{(x+4x)}=\frac{\frac{5}{3}K\,A\,({{T}_{2}}-{{T}_{1}})}{5x}\]=\[\frac{\frac{1}{3}K\,A\,({{T}_{2}}-{{T}_{1}})}{x}\] On comparing it with given equation we get \[f=\frac{1}{3}\]


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