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JEE Main & Advanced Sample Paper JEE Main Sample Paper-41

  • question_answer
    For a cyclic process \[A\to B\to C\to A\], the following information is given
    Path \[\Delta Q\](Heat supplied) \[\Delta U\](Increase in internal energy) Work done by system on surrounding
    \[A\to B\] 600 J 200 J 400 J
    \[B\to C\] -100 J 100 J -200 J
    \[C\to A\] -100 J -300 J 200 J
    Calculate the heat of the cycle and efficiency of cycle.

    A)  400 J, 100%              

    B)  600 J, 66.67%

    C)  400 J, 66.67%           

    D)  600 J, 100%

    Correct Answer: C

    Solution :

    Heat of cycle.= Net heat= 600-200 = 400 J Efficiency\[\eta =\frac{\text{Work}\,\text{done}}{\text{Heat}\,\text{supplied}}=\frac{400}{600}\]\[=66.67%\]


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