A body is placed on an inclined plane and has to be pushed down. The angle made by the normal reaction with the vertical will be:-

Options

A. Equal to angle of repose
B. Equal to the angle of friciton
C. Less than the angle of repose
D. More than the angle of friction

Explanation

To solve the problem, we need to analyze the forces acting on the body placed on the inclined plane and determine the relationship between the angle of inclination (θ) and the angle of repose (α).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- The weight of the body (mg) acts vertically downward.
- The normal reaction (N) acts perpendicular to the surface of the inclined plane.
- The frictional force (f) acts parallel to the inclined plane, opposing the motion.

2. **Resolve the Weight into Components**:
- The weight can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)

3. **Apply Newton's Second Law**:
- For the body to be pushed down the incline, the net force acting down the incline must be greater than the frictional force.
- The frictional force can be expressed as: \( f = \mu N \), where \( \mu \) is the coefficient of friction.

4. **Set Up the Equations**:
- The normal force (N) is equal to the perpendicular component of the weight:
\[ N = mg \cos \theta \]
- The frictional force is:
\[ f = \mu mg \cos \theta \]
- For the body to move down, we have:
\[ mg \sin \theta > \mu mg \cos \theta \]

5. **Simplify the Inequality**:
- Dividing through by \( mg \) (assuming \( mg \neq 0 \)):
\[ \sin \theta > \mu \cos \theta \]
- This can be rewritten as:
\[ \tan \theta > \mu \]

6. **Relate to the Angle of Repose**:
- The angle of repose (α) is defined as the angle at which the frictional force is maximum and is given by:
\[ \mu = \tan \alpha \]
- Therefore, we can substitute:
\[ \tan \theta > \tan \alpha \]
- This implies:
\[ \theta > \alpha \]

7. **Conclusion**:
- The angle made by the normal reaction with the vertical is related to the angle of repose. Since we found that \( \theta > \alpha \), the angle made by the normal reaction with the vertical will be greater than the angle of repose.

### Final Answer:
The angle made by the normal reaction with the vertical will be **more than the angle of repose**.

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