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JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, \[{{T}_{1}}\]and \[{{T}_{2}}\]. The temperature of the hot reservoir of the first engine is \[{{T}_{1}}\] and the temperature of the cold reservoir of the second engine is\[{{T}_{2}}\]. \[T\] is temperature of the sink of first engine which is also the source for the second engine. How is \[T\] related to \[{{T}_{1}}\] and\[{{T}_{2}}\], if both the engines perform equal amount of work?  [JEE MAIN Held on 07-01-2020 Evening]

    A) \[T=\frac{{{T}_{1}}+{{T}_{2}}}{2}\]

    B) \[T=\sqrt{{{T}_{1}}{{T}_{2}}}\]

    C) \[T=\frac{2{{T}_{1}}{{T}_{2}}}{{{T}_{1}}+{{T}_{2}}}\]    

    D) \[T=0\]

    Correct Answer: A

    Solution :

    [a] Let\[{{Q}_{1}}\]: Heat input to first engine
    \[{{Q}_{C}}\]: Heat rejected by first engine
    \[{{Q}_{2}}\]: Heat rejected by second engine
    \[{{T}_{C}}\]: Lower temperature of first engine
    \[W={{Q}_{1}}{{Q}_{C}}={{Q}_{C}}{{Q}_{2}}\]
    \[\Rightarrow \,\,\,\,2{{Q}_{C}}={{Q}_{1}}+{{Q}_{2}}\]
    \[\Rightarrow 2{{T}_{C}}={{T}_{1}}+{{T}_{2}}\Rightarrow {{T}_{C}}={{T}_{1}}+{{T}_{2}}\]


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