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

  • question_answer
    A long magnet is cut into two equal parts, such that the length of each half is same as that of original magnet. If the period of original magnet is T, the period of new magnet is

    A)  \[T\]              

    B)  \[\frac{T}{2}\]

    C)  \[\frac{T}{4}\]            

    D)  \[2T\]

    Correct Answer: A

    Solution :

    Time period \[T=2\pi \sqrt{\frac{I}{MB}}\] As the magnet is cut into two equal parts along axis, then for each part \[I'=\frac{I}{2}\] \[M'=\frac{M}{2}\] \[\therefore \]   Time period of new magnet, \[T'=\sqrt{\frac{I'}{M'B}}=\sqrt{\frac{I\times 2}{2\times M\times B}}\] \[T'=T\]


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