Magnetism Question 257
Question: A dip circle is so set that its needle moves freely in the magnetic meridian. In this position, the angle of dip is $ 40{}^\circ $ . Now the dip circle is rotated so that the plane in which the needle moves makes an angle of $ 30{}^\circ $ with the magnetic meridian. In this position, the needle will dip by an angle
Options:
A) $ 40^{o} $
B) $ 30{}^\circ $
C) more than $ 40{}^\circ $
D) less than $ 40^{o} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ {\delta _{1}}=40^{o} $ , $ {\delta _{2}}=30^{o} $
$ \delta =? $
$ \cot ,\delta =\sqrt{{{\cot }^{2}}{\delta _{1}}+{{\cot }^{2}}{\delta _{2}}} $
$ =\sqrt{{{\cot }^{2}}40^{o}+{{\cot }^{2}}30^{o}} $
$ \cot ,\delta =\sqrt{{{1.19}^{2}}+3}=2.1 $
$ \
Therefore ,,,,\delta =25^{o} $ i.e. $ \delta <40^{o} $ .
BETA
