question_answer

A dip needle lies initially in the magnetic meridian when it shows an angle of dip \[\theta \] at a place. The dip circle is rotated through an angle \[x\] in the horizontal plane and then it shows an angle of dip\[\theta '\]. Then \[\frac{\tan \theta '}{\tan \theta }\]is

A) \[\frac{1}{\cos x}\]        

B) \[\frac{1}{\sin x}\]

C) \[\frac{1}{\tan x}\]        

D) \[\cos x\]

Correct Answer: A

Solution :

[a] In first case \[\tan \theta =\frac{Bv}{{{B}_{H}}}\] Second case \[\tan \theta '=\frac{Bv}{{{B}_{H}}\cos x}\] From Eqs. (i) and (ii), \[\frac{\tan \theta '}{\tan \theta }=\frac{1}{\cos x}\]