Q 12-05-017NEETJEE MainMedium
A dip circle is placed in a vertical plane at $30°$ to the magnetic meridian. The apparent dip is $45°$. The true dip $\delta$ satisfies
Answer: (D) $\tan\delta = \dfrac{\sqrt{3}}{2}$
In a plane at angle $\theta$ to the meridian, the horizontal component seen is $B_H\cos\theta$ while the vertical is unchanged: $\tan\delta' = \dfrac{\tan\delta}{\cos\theta}$.
$$\tan\delta = \tan 45° \times \cos 30° = \frac{\sqrt{3}}{2}$$
Solution by Sreeraj P, M.Sc Physics