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AMU Medical AMU Solved Paper-2014

  • question_answer
    An iron rod is subjected to the cycles of magnetisation at the rate of 50 Hz. Given the density of the rod is 8 x 103 kg/m3 and specific heat is \[0.11\times {{10}^{-3}}\] \[cal/k{{g}^{o}}C\]. The rise in temperature per minute, if the area enclosed by the B-H loop corresponds to energy of \[{{10}^{-2}}J\] is

    A)  \[{{78}^{o}}C\]

    B)  \[{{88}^{o}}C\]

    C)  \[8.1\,C\]

    D)  None of the above

    Correct Answer: B

    Solution :

                    Energy loss per unit volume per cycle = area of hysteresis loop Energy loss per second per unit volume                 \[={{10}^{-2}}\times 50\,J\] Hence, heat produced in one minute                 \[Q={{10}^{-2}}\times 50\times 80\,J\]                 \[=\frac{30}{4.2}J\] Let \[\theta \] is rise in temperature, then \[8\times {{10}^{3}}\times 0.11\times {{10}^{-3}}\times \theta =\frac{30}{4.2}\]                 \[\theta =\frac{30}{4.2\times 8\times 0.11}\]                 \[\theta =8.1{{\,}^{o}}C\]


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