JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        If a magnet is suspended at angle\[30{}^\circ \]to the magnetic meridian, the dip needle makes an angle of\[45{}^\circ \]with the horizontal. The real dip is

    A)  \[{{\tan }^{-1}}(\sqrt{3/2})\]     

    B)  \[{{\tan }^{-1}}(\sqrt{3})\]

    C)  \[{{\tan }^{-1}}(\sqrt{2/3})\]     

    D)  \[{{\tan }^{-1}}(2/\sqrt{3})\]

    Correct Answer: D

    Solution :

                    \[\tan \delta =\frac{\tan \delta }{\cos \theta }\] \[=\frac{\tan 45{}^\circ }{\cos 30{}^\circ }\] \[\tan \delta =\frac{1}{\sqrt{3}/2}=\frac{2}{\sqrt{3}}\] \[\therefore \]  \[\delta ={{\tan }^{-1}}\left( \frac{2}{\sqrt{3}} \right)\]


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