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JCECE Medical JCECE Medical Solved Paper-2015

  • question_answer
    With a standard rectangular bar magnet of length (L), breadth \[(b;b<<I)\] 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 \[\text{M =}\]magnetic moment \[{{\text{B}}_{\text{H}}}\text{=}\]horizontal component of the earths magnetic field. \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}\times \frac{{{M}_{2}}}{{{M}_{1}}}}\] or \[\frac{8}{{{T}_{2}}}=\sqrt{\frac{\frac{m{{l}^{2}}}{12}\times M/8}{m/8{{(l/8)}^{2}}/12\times M}}=\sqrt{\frac{8\times {{8}^{2}}}{8}}=8\] or \[{{T}_{2}}=\frac{8}{8}=1s\]


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