Magnetism Question 49
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°. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30° with the magnetic meridian. In this position the needle will dip by an angle
[DCE 2005]
Options:
A) 40°
B) 30°
C) More than 40°
D) Less than 40°
Show Answer
Answer:
Correct Answer: C
Solution:
$ \tan \theta =\frac{B _{V}^{{}}}{B _{H}} $ … (i)
If apparent dip is $ \theta ’ $ then
$ \tan \theta ‘=\frac{B’_V}{B’_H}=\frac{B_V}{B_H\cos 30{}^\circ }=\frac{B_V}{B_H\times \frac{\sqrt{3}}{2}} $
Therefore $ \tan \theta ‘=( \frac{2}{\sqrt{3}} )\tan \theta \Rightarrow ,\tan \theta ‘>\tan \theta $
Therefore $ \theta ‘>\theta $
BETA
