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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    The efficiency of a Carnot engine operating with reservoir temperature of \[100{}^\circ C\]and \[-23{}^\circ C\] will be:

    A)  \[\frac{100-23}{373}\]  

    B)         \[\frac{100+23}{373}\]

    C)  \[\frac{100+23}{100}\] 

    D)         \[\frac{100-23}{100}\]

    Correct Answer: B

    Solution :

    Using the relation for efficiency of Carnots engine \[\eta =\frac{{{T}_{1}}={{T}_{2}}}{{{T}_{1}}}=\frac{373-250}{373}\]     \[\frac{123}{373}=\frac{100+23}{373}\]


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