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,\[\frac{\tan \theta '}{\tan \theta }\]will be

A) \[\cos x\]                            

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

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

D)  \[\frac{1}{\tan x}\]

Correct Answer: B

Solution :

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