Question

If a dip circle is placed in a vertical plane at an angle of ${30}^{\underset{–}{o}}$ to the magnetic meridian, the dip needle makes an angle of ${45}^{\underset{–}{o}}$ with the horizontal. The real dip at that place 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 A ${\tan }^{-1}\left(\sqrt{3}/2\right)$

Let vertical component and horizontal component of earth's magnetic field at magnetic meridian be $V$ and $H$ respectively.

Angle of dip:$\tan \theta =\frac{V}{H}\to \left(1\right)$

For $30°$ to meridian, $40°$ to horizontal,

$\tan 45°=\frac{V}{H\cos 30°}⟹\frac{V}{H}=\cos 30°\to \left(2\right)$

Compare $\left(1\right)$ and $\left(2\right)$

$⟹\tan \theta =\cos 30°$

$⟹\theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$