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:

  1. (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., $\cot ^{2} \theta=\cot ^{2} \theta_1+\cot ^{2} \theta_2$



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