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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 15

  • question_answer
    A magnet is suspended in such a way that it oscillates in the horizontal plane if it makes 20 oscillations per minute at a place-1 where dip angle is \[30{}^\circ \]and 15 oscillations per minute at a place-2 where dip angle is \[60{}^\circ \]. The ratio of earth's total magnetic field \[\left( \frac{{{B}_{1}}}{{{B}_{2}}} \right)\]at the two places is

    A) \[16:9\sqrt{3}\]

    B) \[16\sqrt{3}:9\]             

    C) \[16:1\]       

    D) \[1:16\]

    Correct Answer: A

    Solution :

    [a] \[T=2\pi \sqrt{\frac{I}{MB\,\,\cos \theta }}\] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{B}_{2}}\cos {{\theta }_{2}}}{{{B}_{1}}\cos {{\theta }_{1}}}}\Rightarrow \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{16}{9\sqrt{3}}\]


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