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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    An electric heater of resistance \[6\,\Omega \] is run are 10 min on a 120 V line. The energy liberated in this period of time is

    A)  \[7.2\times {{10}^{3}}J\]                             

    B)  \[14.4\times {{10}^{5}}J\]

    C)  \[43.2\times {{10}^{4}}J\]                           

    D)  \[28.8\times {{10}^{4}}J\]

    Correct Answer: B

    Solution :

    Given,    \[R=6\Omega ,\]  \[t=10\min =600s\] \[V=120volt\] Energy liberated \[=\frac{{{V}^{2}}}{R}.t\]                                 \[=\frac{120\times 120\times 600}{6}\]                                 \[=144\times {{10}^{4}}J\]


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