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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    An electric heater of resistance 70, heats 0.1 kg of water of\[20{}^\circ C\]for 3 min. If it carries a current of 4 A, then what will be the final temperature (specific heat of water\[=4.2\times {{10}^{3}}J/kg\text{ }K\])

    A)  \[28{}^\circ C\]              

    B)  \[48{}^\circ C\]

    C)  \[52{}^\circ C\]             

    D)  \[68{}^\circ C\]

    Correct Answer: D

    Solution :

     \[{{I}^{2}}Rt=ms\Delta t\] \[{{(4)}^{2}}\times (7)\times 3\times 60=0.1\times 4.2\times {{10}^{4}}\times (T-20)\] \[T=68{}^\circ C\]


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