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BHU PMT BHU PMT (Screening) Solved Paper-2010

  • question_answer
    A magnet is suspended in such a way that it oscillates in the horizontal plane. It makes 20 oscillations per minute at a plane where dip angle is\[30{}^\circ \]and 15 oscillations per minute at a place where dip angle is\[60{}^\circ \]. The ratio of earth's magnetic fields at two places is

    A)  \[3\sqrt{3}:8\]                 

    B)  \[16:9\sqrt{3}\]

    C)  \[4:9\]                                 

    D)  \[2\sqrt{2}:3\]

    Correct Answer: B

    Solution :

                     Time period of suspended magnet \[T=2=\sqrt{\frac{i}{MB\,\cos \delta }}\] \[v=\frac{1}{2\pi }=\sqrt{\frac{MB\,\cos \,\delta }{i}}\] \[v=\sqrt{B\,\cos \delta }\] Or           \[B\propto \frac{{{v}^{2}}}{\cos \delta }\] \[\therefore \]  \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{400}{\cos {{30}^{o}}}\times \frac{\cos {{60}^{o}}}{225}\]                 \[=\frac{16\times 2}{9\times \sqrt{3}}\times \frac{1}{2}=16:9\,\sqrt{3}\]


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