JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    Efficiency of Carnot engine is \[20%\] If temperature of sink \[{{27}^{o}}C\], then the temperature of source will be:

    A)  \[{{102}^{o}}C\]                              

    B)  \[102K\] 

    C)  \[{{129}^{o}}C\]                              

    D)  \[129K\]

    Correct Answer: A

    Solution :

    \[\eta =1-\frac{{{T}_{1}}}{{{T}_{2}}}=1-\frac{27+273}{T+273}=20%=0.2\]


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