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BHU PMT BHU PMT Solved Paper-2004

  • question_answer
    For a Carnot engine, the source is at 500 K and the sink at 300 K. What is efficiency of this engine?

    A)  0.2                                        

    B)  0.4

    C)  0.6                                        

    D)  0.3

    Correct Answer: B

    Solution :

                     Key Idea: \[\eta =\frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}}\]where, \[\eta =\]efficiency of engine \[{{T}_{1}}=\]temperature of source = 500 K \[{{T}_{2}}=\]temperature of sink = 300 K \[\eta =\frac{500-300}{500}=\frac{200}{500}=0.4\]


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