In the following figure all surfaces are smooth. The system is released from rest , then

A. acceleration of wedge greater then `g tanθ`
B. acceleration of `m` is `g√(1+2cos⁡^2θ)`
C. acceleration of `m` is `g`
D. acceleration of wedge is `g sinθ`

Options

A. acceleration of wedge greater then `g tanθ`
B. acceleration of `m` is
`g√(1+2cos⁡^2θ)`
C. acceleration of `m` is `g`
D. acceleration of wedge is `g sinθ`

Explanation

To solve the problem, we need to analyze the forces acting on the block and the wedge when the system is released from rest. Let's break it down step by step:

### Step 1: Identify the Forces Acting on the Block
- The block (mass \( m \)) is on an inclined plane with angle \( \theta \).
- The gravitational force acting on the block is \( mg \) (downward).
- The component of this force acting parallel to the incline is \( mg \sin \theta \).
- The component of the gravitational force acting perpendicular to the incline is \( mg \cos \theta \).

### Step 2: Analyze the Motion of the Block
- Since the surfaces are smooth, there is no friction.
- The net force acting on the block along the incline is \( mg \sin \theta \).
- According to Newton's second law, the acceleration \( a \) of the block can be expressed as:
\[ ma = mg \sin \theta \]
Therefore, the acceleration of the block \( a \) is:
\[ a = g \sin \theta \]

### Step 3: Identify the Forces Acting on the Wedge
- The wedge (mass \( M \)) experiences a normal force \( N \) from the block.
- This normal force can be resolved into two components: one acting perpendicular to the incline and the other acting parallel to the incline.
- The component of the normal force acting parallel to the incline is \( N \sin \theta \), which causes the wedge to accelerate horizontally.

### Step 4: Apply Newton's Second Law to the Wedge
- The net force acting on the wedge in the horizontal direction is equal to the horizontal component of the normal force:
\[ F = N \sin \theta \]
- The acceleration \( A \) of the wedge can be expressed as:
\[ MA = N \sin \theta \]

### Step 5: Relate the Normal Force to the Block's Acceleration
- From the block's perspective, the normal force \( N \) can be related to the acceleration of the block and the wedge.
- Since the block is accelerating down the incline with \( g \sin \theta \), we can express \( N \) in terms of the accelerations:
\[ N = m(g \sin \theta - A) \]

### Step 6: Solve for the Accelerations
- Substitute \( N \) back into the equation for the wedge:
\[ MA = m(g \sin \theta - A) \sin \theta \]
- Rearranging gives:
\[ MA + mA \sin \theta = mg \sin^2 \theta \]
- Factor out \( A \):
\[ A(M + m \sin \theta) = mg \sin^2 \theta \]
- Thus, the acceleration of the wedge \( A \) is:
\[ A = \frac{mg \sin^2 \theta}{M + m \sin \theta} \]

### Conclusion
After analyzing the forces and applying Newton's laws, we conclude that:
- The acceleration of the block \( m \) is \( g \sin \theta \).
- The acceleration of the wedge \( M \) is dependent on the mass ratio and angle.

### Final Answers
- The correct options are:
- C: acceleration of \( m \) is \( g \sin \theta \)
- D: acceleration of the wedge is \( g \sin \theta \)