question_answer

A dip needle lies initially in the magnetic meridian when it shows an angle of dip q 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 :

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