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A magnetic needle is free to rotate in a vertical plane which makes an angle $60 °$ with the magnetic meridian. If the needle stays in a direction making an angle of $\tan^{- 1} (\frac{2}{\sqrt{3}})$ with the horizontal, what would be the true dip at that place ?

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$60^{0}$

$37^{0}$

$30^{0}$

$45^{0}$

answer is C.

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Detailed Solution

Given, apparent dip

$θ' = \tan^{- 1} (\frac{2}{\sqrt{3}})$

and $α = 60 °$

We have $\tan θ' = \frac{\tan θ}{\cos α}$

or $\tan θ = \tan θ' \cos α$

$= \frac{2}{\sqrt{3}} \times \frac{1}{2} = \frac{1}{\sqrt{3}}$

or $θ = 30 °$

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