Question

A dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position, the angle of dip is
​${40}^{\circ }$. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of ​${30}^{\circ }$ with the magnetic meridian. In this position the needle will dip by an angle

• A. ${40}^{\circ }$
• B. ${30}^{\circ }$
• C. $Morethan{40}^{\circ }$
• D. $Lessthan{40}^{\circ }$

Answer 1

The correct option is C $Morethan{40}^{\circ }$

$tan\theta =\frac{B_V}{B_H}$ ...(i)
If apparent dip is ${\theta }^{\prime }$ then
$tan{\theta }^{\prime }=\frac{B_V^{\prime }}{B_H^{\prime }}=\frac{B_V}{B_{H}cos{30}^{\circ }}=\frac{B_V}{B_H\times \frac{\sqrt{3}}{2}}\Rightarrow tan{\theta }^{\prime }=\left(\frac{2}{\sqrt{3}}\right)tan\theta \Rightarrow tan{\theta }^{\prime }>tan\theta \Rightarrow {\theta }^{\prime }\theta$