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JEE Main & Advanced Sample Paper JEE Main Sample Paper-46

  • question_answer
    A rod of length \[l\] (laterally thermally insulated) of uniform cross-sectional area A consists of a material whose thermal conductivity varies with temperature as \[K=\frac{{{K}_{0}}}{a+bT},\]where \[{{K}_{0}}\], a and b are constants. \[{{T}_{1}}\]and\[{{T}_{2}}\,(<{{T}_{1}})\]are the temperature of two ends of rod. Then, rate of flow of heat across the rod is

    A)  \[\frac{A{{K}_{0}}}{bl}\,\left( \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right)\] 

    B)  \[\frac{A{{K}_{0}}}{bl}\left( \frac{a+b{{T}_{2}}}{a+b{{T}_{1}}} \right)\]

    C) \[\frac{A{{K}_{0}}}{bl}\,In\,\left[ \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right]\]         

    D)  \[\frac{A{{K}_{0}}}{al}\,In\,\left[ \frac{a+b{{T}_{2}}}{a+b{{T}_{1}}} \right]\]

    Correct Answer: C

    Solution :

     As, we know that, \[\Rightarrow \]            \[\frac{dQ}{dt}=-KA\frac{dT}{dx}\] \[\Rightarrow \]            \[\frac{dQ}{dt}=-\frac{{{K}_{0}}A}{a+bT}\frac{dT}{dx}\] On entegrating both sides within the proper limits. \[\frac{dQ}{dt}\,\int_{0}^{l}{dx=-{{K}_{0}}A}\,\int_{{{T}_{1}}}^{{{T}_{2}}}{\frac{dT}{a+bT}}\] This gives,        \[\frac{dQ}{dt}=\frac{A{{K}_{0}}}{b\,l}\,In\,\,\left[ \frac{a+b{{T}_{1}}}{a+b{{T}_{2}}} \right]\]


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