RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • question_answer
    An electric heater of resistance\[6\,\Omega \] is run for 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,\[R=6\Omega ,\,\,t=10\min =600s\] \[V=120\,\,volt\] Energy liberated\[=\frac{{{V}^{2}}}{R}\cdot t\]                                \[=\frac{120\times 120\times 600}{6}\]                                 \[=144\times {{10}^{4}}J\]                                 \[=14.4\times {{10}^{5}}J\]


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