Magnetism and Matter - Result Question 11
15. If $\theta_1$ and $\theta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of $dip \theta$ is given by :-
(a) $\tan ^{2} \theta=\tan ^{2} \theta_1+\tan ^{2} \theta_2$
[2017]
(b) $\cot ^{2} \theta=\cot ^{2} \theta_1-\cot ^{2} \theta_2$
(c) $\tan ^{2} \theta=\tan ^{2} \theta_1-\tan ^{2} \theta_2$
(d) $\cot ^{2} \theta=\cot ^{2} \theta_1+\cot ^{2} \theta_2^{2}$
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Answer:
Correct Answer: 15. (d)
Solution:
- (d) If $\theta_1$ and $\theta_2$ are opparent angles of dip Let $\alpha$ be the angle which one of the plane make with the magnetic meridian.
$\tan \theta_1=\frac{v}{H \cos \alpha}$
i.e., $\cos \alpha=\frac{v}{H \tan \theta_1}$
$\tan \theta_2=\frac{v}{H \sin \alpha}$,
i.e., $\sin \alpha=\frac{v}{H \tan \theta_2}$
Squaring and adding (i) and (ii), we get
$\cos ^{2} \alpha+\sin ^{2} \alpha=(\frac{V}{H})^{2}(\frac{1}{\tan ^{2} \theta_1}+\frac{1}{\tan ^{2} \theta_2})$ i.e., $1=\frac{V^{2}}{H^{2}}[\cot ^{2} \theta_1+\cot ^{2} \theta_2]$
or $\frac{H^{2}}{V^{2}}=\cot ^{2} \theta_1+\cot ^{2} \theta_2$
i.e., $\cot ^{2} \theta=\cot ^{2} \theta_1+\cot ^{2} \theta_2$
