Q 12-05-057JEE MainJEE Main 2021 (20 Jul, Shift 2)Easy
At an angle of $30^\circ$ to the magnetic meridian, the apparent dip is $45^\circ$. Find the true dip:
Answer: (D) $\tan^{-1}\frac{\sqrt3}{2}$
In a vertical plane at angle $\theta$ to the meridian, the horizontal component seen is $B_H\cos\theta$, so $\tan\delta' = \dfrac{\tan\delta}{\cos\theta}$.
$$\tan\delta = \tan45^\circ\cos30^\circ = \frac{\sqrt3}{2} \Rightarrow \delta = \tan^{-1}\frac{\sqrt3}{2}$$
Solution by Sreeraj P, M.Sc Physics