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

  • question_answer
    The time period of a freely suspended bar magnet in a field is 2 s. It is cut into two equal parts along its axis, (hen the time period is

    A)  \[4\,\,s\]          

    B)  \[0.5\,\,s\]

    C)  \[2\,\,s\]          

    D)  \[0.25\,\,s\]

    Correct Answer: C

    Solution :

    \[T=2\pi \,\sqrt{\frac{I}{MB}}\] \[T=2s,\,\,I'=\frac{1}{2},\,\,M'=\frac{M}{2}\] \[\therefore \] \[T'=T\] \[\Rightarrow \] \[T'=2s\]


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