Assertion: A body of weight 10N (W) is at rest on an inclined plane `(mu=sqrt(3)/(2))` making an angle of `30^(@)` with the horizontal. The force of friction acting on it is 5N
Reason: In above situation, the limiting force of friction is given by `f_("limitting")=mu W cos theta=7.5N`.

Options

A. If both assertion `&` Reason are True `&` the Reason is a corrrect explanation of the Asserion.
B. If both Assertion `&` Reason are True but Reason is not correct explanation of the Assertion.
C. If Assertion is Trie but the Reason is False.
D. If both Assertion `&` Reason are false

Explanation

To solve the problem, we need to analyze the forces acting on the body resting on the inclined plane and verify the assertion and reason provided.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Weight of the body, \( W = 10 \, \text{N} \)
- Coefficient of friction, \( \mu = \frac{\sqrt{3}}{2} \)
- Angle of inclination, \( \theta = 30^\circ \)
- Force of friction acting on it, \( f = 5 \, \text{N} \)

2. **Calculate the Normal Force:**
The normal force \( N \) acting on the body can be calculated using the formula:
\[ N = W \cos \theta \]
Substituting the values:
\[ N = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{N} \]

3. **Calculate the Limiting Force of Friction:**
The limiting force of friction \( f_{\text{limiting}} \) is given by:
\[ f_{\text{limiting}} = \mu N \]
Substituting the values:
\[ f_{\text{limiting}} = \frac{\sqrt{3}}{2} \times 5\sqrt{3} = \frac{3 \times 5}{2} = \frac{15}{2} = 7.5 \, \text{N} \]

4. **Calculate the Component of Weight Parallel to the Incline:**
The component of the weight acting down the incline is:
\[ W \sin \theta \]
Substituting the values:
\[ W \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{N} \]

5. **Analyze the Forces:**
Since the body is at rest, the force of friction must balance the component of the weight acting down the incline. Here:
\[ f = W \sin \theta = 5 \, \text{N} \]
The force of friction acting (5 N) is less than the limiting force of friction (7.5 N), which means the friction force can equal the applied force (5 N).

6. **Conclusion:**
- The assertion states that the force of friction acting on it is 5 N, which is true.
- The reason states that the limiting force of friction is 7.5 N, which is also true.
- However, the reason does not explain why the friction force is 5 N; it is because the applied force (5 N) is less than the limiting friction (7.5 N).

### Final Answer:
- The assertion is true, and the reason is true, but the reason does not adequately explain the assertion.