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

Answer 1

The correct option is D ${\tan }^{-1}\left(2/\sqrt{3}\right)$

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