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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2010

  • question_answer
    A condenser having a capacity 2.0 \[\mu F\] is charged to 200 Volts and then the plates of the capacitor are connected to a resistance wire. The heat produced in joule will be

    A)  \[4\times {{10}^{4}}J\]          

    B)  \[4\times {{10}^{10}}J\]

    C)  \[4\times {{10}^{-2}}J\]         

    D)  \[2\times {{10}^{-2}}J\]

    Correct Answer: C

    Solution :

     Heat produced\[U=\frac{1}{2}C{{V}^{2}}\] \[U=\frac{1}{2}\times 2\times {{10}^{-6}}{{(200)}^{2}}=4\times {{10}^{-2}}J\]


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