Question

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

• A. ${{\tan }^{-1}}(\sqrt{3}/2)$
• B. ${{\tan }^{-1}}(\sqrt{3})$
• C. ${{\tan }^{-1}}(3/\sqrt{2})$
• D. ${{\tan }^{-1}}(2/\sqrt{3})$

Answer 1

Answer: (D)

${{\tan }^{-1}}(2/\sqrt{3})$

Answer 2

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