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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    With a standard rectangular bar magnet of length\[(L)\], breadth \[(b;\,\,b<<l)\] and magnetic moment\[M\], the time period of the magnet in a vibration magnetometer is\[8\,\,s\]. If the magnet is cut normal to its length into 8 equal pieces, then the time period (in second) with one of the pieces is

    A) \[8\,\,s\]                                             

    B) \[2\,\,s\]

    C) \[1\,\,s\]                                             

    D)  \[4\,\,s\]

    Correct Answer: C

    Solution :

    Time period of magnet in vibration magnetometer                 \[T=2\pi \sqrt{\frac{I}{M{{B}_{H}}}}\] where, \[I=\]moment of inertia of magnet \[M=\] magnetic moment \[{{B}_{H}}=\]horizontal component of the earth's magnetic field. \[\therefore \]  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}\times \frac{{{M}_{2}}}{{{M}_{1}}}}\] or     \[\frac{8}{{{T}_{2}}}=\sqrt{\frac{\frac{m{{l}^{2}}}{12}\times M/8}{m/8{{(I/8)}^{2}}/12\times M}}=\sqrt{\frac{8\times {{8}^{2}}}{8}}=8\] or      \[{{T}_{2}}=\frac{8}{8}=1\,\,s\]


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