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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    The temperature of source and sink of a Carnot engine are\[327{}^\circ C\]and\[27{}^\circ C\] respectively. The efficiencies of the engine is

    A)  \[1-\left( \frac{27}{327} \right)\]      

    B)  \[\frac{27}{327}\]

    C)  \[0.5\]

    D)  \[0.7\]

    Correct Answer: C

    Solution :

     \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] Given, \[{{T}_{1}}=327+273=600\text{ }K\] \[{{T}_{2}}=27+273=300K\] \[\therefore \] \[\eta =1-\frac{300}{600}\] \[\eta =1-\frac{1}{2}=0.5\]


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