A) \[{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]
B) \[{{\tan }^{-1}}(\sqrt{3})\]
C) \[{{45}^{0}}\]
D) \[{{30}^{0}}\]
Correct Answer: A
Solution :
Let \[{{B}_{e}}\]be the magnetic field at some point. H and V be the horizontal and vertical components of \[{{B}_{e}}\]and \[\theta \] is the actual angle of dip at the same place. \[H=B{{}_{e}}\cos \theta \] and \[V={{B}_{e}}\sin \theta \] \[\therefore \] \[\frac{V}{H}=\tan \theta \] or \[\tan \theta =\frac{V}{H}\] ?(i) In a vertical plane at \[{{30}^{o}}\] from the magnetic meridian, the horizontal component is \[H=H\cos {{30}^{o}}=\frac{H\sqrt{3}}{2}\] While vertical component is still V. Therefore, apparent dip will be given by \[\tan \theta =\frac{V}{H}=\frac{V}{H\sqrt{3}/2}=\frac{2V}{H\sqrt{3}}\] ?(ii) Dividing Eq.(i) by Eq.(ii),we have \[\frac{\tan \theta }{\tan \theta }=\frac{\frac{V}{H}}{\frac{2V}{H\sqrt{3}}}=\frac{\sqrt{3}}{2}\] or \[\tan \theta =\frac{\sqrt{3}}{2}\tan \theta \] \[=\frac{\sqrt{3}}{2}\tan {{45}^{o}}\] \[(\because \,\theta ={{45}^{o}})\] \[=\frac{\sqrt{3}}{2}\] \[\therefore \] \[\theta ={{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]
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