Question

If a magnet is suspended at angle ${30}^o$ to the magnetic meridian, the dip needle makes an angle of ${45}^o$ 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. ${\tan }^{-1}\left(2/\sqrt{3}\right)$

Answer 1

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

Angle of dip, $\delta ={45}^o$

$\therefore \tan {\delta }^{\prime }=\frac{\tan \delta }{\cos \theta }$
$=\frac{\tan 45}{\cos {30}^o}=\frac{1}{\sqrt{3}/2}=\frac{2}{\sqrt{3}}$

$\therefore$ Real dip ${\delta }^{\prime }={\tan }^{-1}\left(2/\sqrt{3}\right)$