Question

The correct value of dip angle at a place is ${45}^{\circ }$. If the dip circle is rotated by ${45}^{\circ }$ out of the meridian, then the tangent of the angle of apparent dip at the place is

• A. $\sqrt{2}$
• B. $\sqrt{\frac{1}{2}}$
• C. $1$
• D. $\sqrt{3}$

Answer 1

The correct option is A $\sqrt{2}$

Given,

The angle of dip when dip circle is in magnetic meridian is

$\tan \delta =\frac{B_V}{B_H}$



When dip is rotated through an angle $\theta$, then horizontal component of the earth's magnetic field becomes

$B_H^{\prime }=B_H\cos \alpha$

Since, the vertical component does not change, so apparent dip will be,

$\tan {\delta }^{{}^{\prime }}=\frac{B_V^{\prime }}{B_H^{\prime }}=\frac{B_V}{B_H\cos \alpha }=\frac{\tan \delta }{\cos \alpha }$

Here, $\delta ={45}^{\circ }$ and $\alpha ={45}^{\circ }$

$\therefore \tan {\delta }^{{}^{\prime }}=\frac{\tan {45}^{\circ }}{\cos {45}^{\circ }}=\sqrt{2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.