Question

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

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

Answer 1

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

Let ${\delta }_1$ be the angle of dip when dip circle is in magnetic meridian
$tan{\delta }_1=\frac{V}{H}$
The dip circle when makes an angle $\theta$ , horizontal component of the earth's magnetic field is $Hcos\theta$
The angle of dip when dip circle makes an angle with the magnetic meridian is,
$tan{\delta }_1=\frac{V}{Hcos\theta }$
$tan{\delta }_1=\frac{tan\delta }{cos\theta }$
$tan{\delta }_1=\sqrt{2}$