A block of mass m is placed on a rough plane inclined at an angle `theta` with the horizontal. The coefficient of friction between the block and inclined plane is `mu` . If`theta lt tan^(-1) (mu)` , then net contact force exerted by the plane on the block is :

Options

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

Explanation

To solve the problem, we need to analyze the forces acting on the block placed on the inclined plane. We are given that the angle of inclination is \( \theta \) and the coefficient of friction is \( \mu \). The condition \( \theta < \tan^{-1}(\mu) \) indicates that the block will not slip down the incline.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The weight of the block \( W = mg \) acts vertically downward.
- The normal force \( N \) acts perpendicular to the inclined plane.
- The frictional force \( f_r \) acts parallel to the inclined plane and opposes the motion (up the incline).

2. **Resolve the Weight into Components:**
- The component of the weight acting parallel to the incline is \( W_{\parallel} = mg \sin \theta \).
- The component of the weight acting perpendicular to the incline is \( W_{\perpendicular} = mg \cos \theta \).

3. **Set Up the Equations:**
- In the direction perpendicular to the incline, the normal force balances the perpendicular component of the weight:
\[ N = mg \cos \theta \]
- In the direction parallel to the incline, since the block is not slipping, the frictional force equals the parallel component of the weight:
\[ f_r = mg \sin \theta \]
- The frictional force can also be expressed in terms of the normal force:
\[ f_r = \mu N = \mu (mg \cos \theta) \]

4. **Equate the Two Expressions for Friction:**
- Set the two expressions for friction equal to each other:
\[ mg \sin \theta = \mu (mg \cos \theta) \]
- Simplifying this gives:
\[ \sin \theta = \mu \cos \theta \]
- This confirms the condition \( \theta < \tan^{-1}(\mu) \).

5. **Calculate the Resultant Contact Force:**
- The net contact force \( R \) exerted by the plane on the block is the vector sum of the normal force and the frictional force:
\[ R = \sqrt{N^2 + f_r^2} \]
- Substitute \( N \) and \( f_r \):
\[ R = \sqrt{(mg \cos \theta)^2 + (mg \sin \theta)^2} \]
- Factor out \( mg \):
\[ R = mg \sqrt{\cos^2 \theta + \sin^2 \theta} \]
- Using the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ R = mg \cdot 1 = mg \]

### Final Answer:
The net contact force exerted by the plane on the block is \( R = mg \).