Question:
Choose the correct option
Choose the correct option
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Apparent dip is measured in a vertical plane inclined at angle \(\theta\) to the magnetic meridian, where \(\tan\delta' = \tan\delta/\cos\theta\).
Updated On: Nov 4, 2026
- True dip is always equal to apparent dip.
- True dip is not mathematically related to apparent dip.
- True dip is less than the apparent dip.
- True dip is always greater than the apparent dip.
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The Correct Option is C
Solution and Explanation
Let the true dip be \(\delta\) and let the dip circle be set in a vertical plane making an angle \(\theta\) with the magnetic meridian; the dip measured there is the apparent dip \(\delta'\).
In the true meridian, \(\tan\delta = \dfrac{V}{H}\).
In the inclined plane the vertical component is unchanged (\(V\)), but the horizontal component along the plane is \(H\cos\theta\). So \(\tan\delta' = \dfrac{V}{H\cos\theta} = \dfrac{\tan\delta}{\cos\theta}\).
Since \(\cos\theta \le 1\), \(\tan\delta' \ge \tan\delta\), i.e. \(\delta' \ge \delta\): the true dip is less than the apparent dip (equal only when \(\theta = 0\)).
Hence options (1), (2) and (4) are wrong and option (3) is the intended choice.
Answer: (3) True dip is less than the apparent dip.
In the true meridian, \(\tan\delta = \dfrac{V}{H}\).
In the inclined plane the vertical component is unchanged (\(V\)), but the horizontal component along the plane is \(H\cos\theta\). So \(\tan\delta' = \dfrac{V}{H\cos\theta} = \dfrac{\tan\delta}{\cos\theta}\).
Since \(\cos\theta \le 1\), \(\tan\delta' \ge \tan\delta\), i.e. \(\delta' \ge \delta\): the true dip is less than the apparent dip (equal only when \(\theta = 0\)).
Hence options (1), (2) and (4) are wrong and option (3) is the intended choice.
Answer: (3) True dip is less than the apparent dip.
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