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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    A heat source at \[T={{10}^{3}}\] K is connected to another heat reservoir at \[T={{10}^{2}}\] K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 \[W{{K}^{-}}^{1}{{m}^{-}}^{1}\], the energy flux through it in the steady state is - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) 200 \[W{{m}^{-}}^{2}\]                                

    B) 65 \[W{{m}^{-}}^{2}\]

    C) 120 \[W{{m}^{-}}^{2}\]                    

    D)   90 \[W{{m}^{-}}^{2}\]

    Correct Answer: D

    Solution :

    Heat current \[H\,\,=\,\,\frac{kA\Delta T}{L}\] Heat flux \[=\,\frac{H}{A}\,\,=\,\frac{k\,\Delta \,T}{L}\,=\,\frac{0.1(900)}{1}\] \[=\text{ }90\text{ }W{{m}^{-}}^{2}\]


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