Question:

Throw and heave of a bed offset by a normal fault are 100 m and 200 m, respectively. The dip of the fault plane is __________degree (rounded off to one decimal place).

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The dip of a fault plane can be calculated using the ratio of throw to heave. A higher dip angle indicates a steeper fault.
Updated On: Dec 11, 2025
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Correct Answer: 26

Solution and Explanation

Step 1: Understanding Throw and Heave.
Throw refers to the vertical displacement on the fault, while heave refers to the horizontal displacement. These values are related to the dip angle (\(\delta\)) of the fault plane. The relationship between the throw (T), heave (H), and dip (\(\delta\)) is given by the formula: \[ \tan(\delta) = \frac{\text{Throw}}{\text{Heave}}. \] Step 2: Calculation of Dip Angle.
Substituting the values for throw (100 m) and heave (200 m): \[ \tan(\delta) = \frac{100}{200} = 0.5. \] Now, calculate the dip angle: \[ \delta = \tan^{-1}(0.5) = 63.4^\circ. \] Step 3: Conclusion.
The dip of the fault plane is 63.4°.
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