NEET NEET SOLVED PAPER 2017

  • question_answer
    A Carnot engine having an efficiency of\[\frac{1}{10}\] as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

    A)  100 J                    

    B)  1 J

    C)   90 J                                      

    D)   99 J

    Correct Answer: C

    Solution :

                     \[\beta =\frac{1-\eta }{\eta }\]                 \[=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}\]                                 \[\beta =9\]                                 \[\beta =\frac{{{Q}_{2}}}{W}\]                                 \[{{Q}_{2}}=9\times 10=90\,J\]


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