A body of mass `m` is kept on a rough fixed inclined plane of angle of inclination `theta=30^(@)`. It remains stationary. Then magnitude of force acting on the body by the inclined plane is equal to `:`

Options

A. `mg`
B. `mg sin theta`
C. `mg cos theta`
D. none of these

Explanation

To solve the problem of finding the magnitude of the force acting on a body of mass \( m \) on a rough fixed inclined plane at an angle of inclination \( \theta = 30^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- The weight of the body \( W = mg \) acts vertically downward.
- This weight can be resolved into two components:
- Perpendicular to the inclined plane: \( W_{\perpendicular} = mg \cos \theta \)
- Parallel to the inclined plane: \( W_{\parallel} = mg \sin \theta \)

2. **Determine the Normal Force**:
- The normal force \( N \) exerted by the inclined plane on the body balances the perpendicular component of the weight:
\[ N = mg \cos \theta \]

3. **Determine the Frictional Force**:
- Since the body is stationary, the frictional force \( f \) must balance the parallel component of the weight:
\[ f = mg \sin \theta \]

4. **Calculate the Magnitudes**:
- Substitute \( \theta = 30^\circ \):
- \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
- \( \sin 30^\circ = \frac{1}{2} \)
- Thus, we can calculate:
\[ N = mg \cos 30^\circ = mg \cdot \frac{\sqrt{3}}{2} \]
\[ f = mg \sin 30^\circ = mg \cdot \frac{1}{2} \]

5. **Determine the Resultant Force**:
- The resultant force \( R \) acting on the body by the inclined plane is the vector sum of the normal force and the frictional force. Since these two forces are perpendicular to each other, we can use the Pythagorean theorem:
\[ R = \sqrt{N^2 + f^2} \]
- Substituting the values of \( N \) and \( f \):
\[ R = \sqrt{\left(mg \cdot \frac{\sqrt{3}}{2}\right)^2 + \left(mg \cdot \frac{1}{2}\right)^2} \]
\[ R = \sqrt{m^2g^2 \left(\frac{3}{4} + \frac{1}{4}\right)} = \sqrt{m^2g^2} = mg \]

### Final Answer:
The magnitude of the force acting on the body by the inclined plane is \( mg \).