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

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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