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Question

If a dip circle is placed in a vertical plane at an angle of $$30^{\underline{o}}$$ to the magnetic meridian, the dip needle makes an angle of $$45^{\underline{o}}$$ with the horizontal. The real dip at that place is?


A
tan1(3/2)
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B
tan1(3)
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C
tan1(3/2)
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D
tan1(2/3)
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Solution

The correct option is A $$\tan^{-1}(\sqrt{3}/2)$$
Let vertical component and horizontal component of earth's magnetic field at magnetic meridian be $$V$$ and $$H$$ respectively.
Angle of dip:$$\tan\theta=\dfrac V{H}\rightarrow(1)$$
For $$30°$$ to meridian, $$40°$$ to horizontal,
$$\tan 45°=\dfrac V{H\cos 30°}\implies \dfrac V{H}=\cos 30°\rightarrow (2)$$
Compare $$(1)$$ and $$(2)$$
$$\implies \tan\theta=\cos 30°$$
$$\implies \theta=\tan^{-1}\left(\dfrac{\sqrt{3}}{2}\right )$$

Physics

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