1) If a magnet is suspened at angle $30^{0}$ to the magnet meridian , the dip of needle makes angle of $45^{0}$ with the horizontal , the real dip is A) $\tan^{-1}\left(\frac{\sqrt{3}}{2}\right)$ B) $\tan^{-1}(\sqrt{3})$ C) $\tan^{-1}\left(\sqrt{\frac{3}{2}}\right)$ D) $\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$ Answer: Option DExplanation:Hence, $tan \delta'=\frac{\tan\delta}{\cos \theta}=\frac{\tan 45^{0}}{\cos 30^{0}}$ $\tan \delta=\frac{1}{\sqrt{3}/2}=\frac{2}{\sqrt{3}}$ or $\delta=\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$
