A block can slide on a smooth inclined plane of inclination `theta` kept on the floor of a lift. When the lift is descending with a retardation a , the acceleration of the block relative to the incline is

Options

A. The acceleration of block relative to inclined is `( g - a) sin theta `
B. The normal force on block by the incline is `m(g + a) costheta`
C. The acceleration of block relative to ground is g
D. The acceleration of block relative to incline is (` g + a )sin theta `

Answer

To solve the problem of finding the acceleration of the block relative to the incline when the lift is descending with a retardation \( a \), we can follow these steps:

### Step 1: Understand the System
We have a block on a smooth inclined plane with an angle \( \theta \). The inclined plane is inside a lift that is descending with a retardation \( a \). The forces acting on the block include gravitational force and a pseudo force due to the retardation of the lift.

### Step 2: Draw the Free Body Diagram
1. The gravitational force acting on the block is \( mg \) (downward).
2. The pseudo force acting on the block due to the retardation of the lift is \( ma \) (upward).
3. Resolve these forces into components parallel and perpendicular to the incline.

### Step 3: Resolve Forces
- The component of gravitational force acting down the incline:
\[ F_{\text{gravity, parallel}} = mg \sin \theta \]
- The component of the pseudo force acting up the incline:
\[ F_{\text{pseudo, parallel}} = ma \sin \theta \]

### Step 4: Write the Equation of Motion
Since the block is sliding down the incline, we can write the equation of motion along the incline:
\[ mg \sin \theta - ma \sin \theta = m b \]
Where \( b \) is the acceleration of the block relative to the incline.

### Step 5: Simplify the Equation
Factoring out \( m \sin \theta \):
\[ (mg - ma) \sin \theta = mb \]
Dividing through by \( m \):
\[ (g - a) \sin \theta = b \]

### Step 6: Final Expression
Thus, the acceleration of the block relative to the incline is:
\[ b = (g - a) \sin \theta \]

### Conclusion
The acceleration of the block relative to the incline is given by:
\[ b = (g - a) \sin \theta \]