Question

If a magnet is suspended at angle ${30}^o$ to the magnet meridian, the dip of needle makes angle of ${45}^o$ with the horizontal, the real dip is

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

Answer 1

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

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