CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    The efficiency of Carnots heat engine is 0.5 when the temperature of the source is \[{{T}_{1}}\] and that of sink is\[{{T}_{2}}\]. The efficiency of another Carnots heat engine is also 0.5. The temperatures of source and sink of the second engine are respectively

    A)  \[2{{T}_{1}},2{{T}_{2}}\]

    B)  \[2{{T}_{1}},\frac{{{T}_{2}}}{2}\]

    C)  \[{{T}_{1}},5,{{T}_{2}},{{T}_{2}}-5\]

    D)  \[{{T}_{1}}+10,{{T}_{2}}-10\]

    Correct Answer: A

    Solution :

    Efficiency of Carnots heat engine, \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\]Efficiency remains same when both \[{{T}_{1}}\] and \[{{T}_{2}}\]are increased by same factor.


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