VMMC VMMC Medical Solved Paper-2008

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

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

    B) \[{{\tan }^{-1}}\left( \sqrt{3} \right)\]                    

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

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

    Correct Answer: D

    Solution :

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


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