A) 40°
B) 30°
C) More than 40°
D) Less than 40°
Correct Answer: C
Solution :
\[\tan \theta =\frac{B_{V}^{{}}}{{{B}_{H}}}\] ... (i) If apparent dip is \[\theta '\] then \[\tan \theta '=\frac{B{{'}_{V}}}{{B}'_{H}^{{}}}=\frac{{{B}_{V}}}{{{B}_{H}}\cos 30{}^\circ }=\frac{B_{V}^{{}}}{B_{H}^{{}}\times \frac{\sqrt{3}}{2}}\] Þ \[\tan \theta '=\left( \frac{2}{\sqrt{3}} \right)\tan \theta \Rightarrow \,\tan \theta '>\tan \theta \] Þ \[\theta '>\theta \]
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