A block of mass m is placed on a rough inclined plane of inclination `theta` kept on the floor of the lift. The coefficient of friction between the block and the inclined plane is `mu`. With what acceleration will the block slide down the inclined plane when the lift falls freely ?

Options

A. Zero
B. `g sin theta-mug cos theta`
C. `g sin theta-mug cos theta`
D. None of these

Answer

To solve the problem, we need to analyze the forces acting on the block when the lift is in free fall.

### Step-by-Step Solution:

1. **Understanding the Situation**:
When the lift falls freely, it accelerates downwards with an acceleration equal to \( g \) (acceleration due to gravity). In this scenario, the block inside the lift will also experience this acceleration.

2. **Identifying Forces**:
The forces acting on the block are:
- The gravitational force acting downwards, \( F_g = mg \).
- The normal force \( N \) exerted by the inclined plane on the block.
- The frictional force \( F_f \) which opposes the motion of the block down the incline.

3. **Analyzing the Normal Force**:
In the frame of the lift, the block is in free fall, so it experiences a condition of weightlessness. This means that the normal force \( N \) acting on the block becomes zero:
\[ N = 0 \]

4. **Frictional Force Calculation**:
Since the normal force is zero, the frictional force, which is given by \( F_f = \mu N \), also becomes:
\[ F_f = \mu \cdot 0 = 0 \]

5. **Net Force Acting on the Block**:
The only force acting on the block is its weight component along the incline. The component of the gravitational force acting down the incline can be calculated as:
\[ F_{\text{down the incline}} = mg \sin(\theta) \]

6. **Applying Newton's Second Law**:
Since there are no opposing forces (friction is zero), the net force acting on the block is equal to the gravitational force component down the incline. Therefore, we can write:
\[ F_{\text{net}} = ma = mg \sin(\theta) \]
Here, \( a \) is the acceleration of the block down the incline.

7. **Solving for Acceleration**:
We can simplify the equation:
\[ ma = mg \sin(\theta) \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ a = g \sin(\theta) \]

### Final Answer:
The acceleration of the block sliding down the inclined plane when the lift falls freely is:
\[ a = g \sin(\theta) \]