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A dip needle arranged to move freely in the magnetic meridian dips by an angle $θ$ . If the vertical plane in which the needle moves is rotated through an angle α to the magnetic meridian than the needle will dip by an angle.

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$θ$

More than $θ$

Less than $θ$

answer is C.

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Detailed Solution

$Where dip circle is set at on angle α to the magnetic meridian$

$δ^{'} is the apparent dip$

$δ is actual dip$

$\begin{array}{l} \tan δ = \frac{B_{v}}{B_{H}} \\ \tan δ^{'} = \frac{B_{v}}{B_{H} \cos α} \end{array}$

In this case;

$\cos α < 1$

So, the apparent dip is less than the actual dip.

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