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) $\tan\delta = \dfrac{\sqrt{3}}{2}$
Concept. In a vertical plane at angle $\theta$ to the magnetic meridian, the dip needle feels only the component $B_H\cos\theta$ of the horizontal field, while the vertical component $B_V$ is unchanged.
Step 1. Apparent dip.
$$\tan\delta' = \frac{B_V}{B_H\cos\theta} = \frac{\tan\delta}{\cos\theta}$$
Step 2. Solve for the true dip.
$$\tan\delta = \tan\delta'\cos\theta = \tan 45° \times \cos 30° = \frac{\sqrt{3}}{2}$$
Answer. $\tan\delta = \dfrac{\sqrt{3}}{2}$ (Option D).
Solution by Sreeraj P, M.Sc Physics