J & K CET Engineering J and K - CET Engineering Solved Paper-2007

  • question_answer
    A bar magnet of magnetic moment At and moment of inertia I is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle \[\theta ,\] angular acceleration is (magnetic induction of earth's horizontal field \[={{B}_{H}}\])

    A)  \[\frac{M{{B}_{H}}\theta }{I}\]

    B)  \[\frac{I{{B}_{H}}\theta }{M}\]

    C)  \[\frac{M\theta }{I{{B}_{H}}}\]

    D)  \[\frac{I\theta }{M{{B}_{H}}}\]

    Correct Answer: A

    Solution :

    If a is the angular acceleration produced, then \[I\alpha \,=\,M{{B}_{H}}\,\,\sin \,\theta \] If \[\theta \] is small, then \[\theta \approx \theta \] and hence the angular acceleration is given by \[\alpha =\frac{M{{B}_{H}}\theta }{I}\]


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