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BHU PMT BHU PMT (Mains) Solved Paper-2008

  • question_answer
    A bar magnet suspended freely in a uniform magnetic field is vibrating with a time period of 3 s. If the field strength is increased to 4 times of the earlier field strength, the time period will be (in seconds)

    A)  12                                         

    B)  6

    C)  1.5                                        

    D)  0.75

    Correct Answer: C

    Solution :

                     \[T=2\pi \sqrt{\frac{I}{MB}}\] Or \[T\propto \frac{1}{\sqrt{B}}\]             or            \[\frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{B}_{1}}}{{{B}_{2}}}}\] Or           \[\frac{{{T}_{2}}}{3}=\sqrt{\frac{1}{4}}\] Or           \[{{T}_{2}}=1.5\,s\]


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