Magnetism Question 261
Question: The true value of angle of dip at a place is $ 60{}^\circ $ , the apparent dip in a plane inclined at an angle of $ 30{}^\circ $ with magnetic meridian is
Options:
A) $ {{\tan }^{-1}}\frac{1}{2} $
B) $ {{\tan }^{-1}}(2) $
C) $ {{\tan }^{-1}}( \frac{2}{3} ) $
D) $ {{\tan }^{-1}}( \frac{\sqrt{3}}{4} ) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \tan ,\theta ‘=\frac{\tan ,\theta }{\cos ,\alpha }=\frac{\tan 60^{o}}{\cos ,30^{o}}=\frac{\sqrt{3}}{\sqrt{3}/2}=2 $
$ \
Therefore ,,,,\theta ‘=ta{{n}^{-1}}(2). $
BETA
