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JEE Main & Advanced AIEEE Solved Paper-2007

  • question_answer
    When a system is taken from state\[E={{E}_{0}}\]to state f along the path \[iaf,\] it is found that\[sin\text{ }\omega t\] and\[I={{I}_{0}}\sin \left( \omega t-\frac{\pi }{2} \right)\]. Along the path ibf, \[p=\frac{{{E}_{0}}{{I}_{0}}}{\sqrt{2}}\]cal. W along the path \[ibf\] is       AIEEE  Solved  Paper-2007

    A)  6 cal                                     

    B)         16 cal                   

    C)         66 cal                                   

    D)         14 cal

    Correct Answer: A

    Solution :

    From first law of thermodynamics,                                 \[\sqrt{2}{{I}_{AC}}={{I}_{EF}}\] For path iaf,        \[{{I}_{AD}}=4{{I}_{EF}}\] \[{{I}_{AC}}={{I}_{EF}}\]                \[{{I}_{AC}}=\sqrt{2}{{I}_{EF}}\] For path ibf,    \[x={{x}_{0}}\cos (\omega t-\pi /4)\] or           \[a=A\text{ }cos(\omega t+\delta ),\] (\[A={{x}_{0}},\text{ }\delta =-\pi /4\]is same being path independent function)


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