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KVPY Sample Paper KVPY Stream-SX Model Paper-4

  • question_answer
    A bar magnet of dipole moment \[1 A-{{m}^{2}}\]and moment of inertia \[{{10}^{-5}}kg-{{m}^{2}}\]when suspended freely has a time period of \[\sqrt{10}s\] at a certain place. Then the earth's magnetic field at that place if angle of dip is \[60{}^\circ \] will be (Take \[{{\pi }^{2}}= 10\])

    A) \[0.2\times {{10}^{-4}}\operatorname{tesla}\]   

    B) \[0.4\times {{10}^{-4}}\operatorname{tesla}\]

    C) \[0.6\times {{10}^{-4}}\operatorname{tesla}\]   

    D) \[0.8\times {{10}^{-4}}\operatorname{tesla}\]

    Correct Answer: D

    Solution :

    [d]
    Using, \[T=2\pi \sqrt{\frac{I}{M\,\,{{B}_{H}}}}\]
    \[\sqrt{10}=2\pi \sqrt{\frac{{{10}^{-5}}}{1\times {{B}_{H}}}}\]
    or \[10=4{{\pi }^{2}}\times \frac{{{10}^{-5}}}{B{{ & }_{H}}}\]
    or \[{{B}_{H}}=4\times {{10}^{-5}}T\]
    \[\therefore B=\frac{{{B}_{H}}}{\cos \theta }=\frac{4\times {{10}^{-5}}}{\cos 60{}^\circ }=0.8\times {{10}^{-4}}T\]


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