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. `g sin theta`
B. `a sin theta`
C. `(g-a) sin theta`
D. `(g+a) sin theta`

Explanation

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 will follow these steps:

### Step 1: Understand the Forces Acting on the Block
When the lift is descending with a retardation \( a \), the effective acceleration acting on the block can be considered in the frame of reference of the lift. The gravitational force acting on the block is \( mg \) downward.

### Step 2: Identify the Effective Acceleration
Since the lift is descending with a retardation \( a \), the effective acceleration \( g' \) acting on the block in the frame of the lift is given by:
\[ g' = g - a \]
This is because the retardation \( a \) reduces the effective gravitational pull felt by the block.

### Step 3: Resolve the Forces Along the Incline
The block is on an inclined plane with an angle \( \theta \). The forces acting on the block can be resolved into two components:
1. The component of the effective weight acting parallel to the incline:
\[ F_{\parallel} = (g - a) \sin \theta \]
2. The component of the effective weight acting perpendicular to the incline:
\[ F_{\perpendicular} = (g - a) \cos \theta \]

### Step 4: Determine the Normal Force
Since the incline is smooth, the normal force \( N \) acting on the block is equal to the perpendicular component of the effective weight:
\[ N = (g - a) \cos \theta \]

### Step 5: Calculate the Acceleration of the Block Relative to the Incline
The net force acting on the block along the incline is the difference between the gravitational component down the incline and the normal force. However, since the normal force does not act along the incline, we only consider the component of the effective weight:
\[ F_{\text{net}} = (g - a) \sin \theta \]
Using Newton's second law, the acceleration \( a_{\text{block}} \) of the block relative to the incline is given by:
\[ a_{\text{block}} = \frac{F_{\text{net}}}{m} = \frac{(g - a) \sin \theta}{m} \]

### Final Expression
Thus, the acceleration of the block relative to the incline is:
\[ a_{\text{block}} = (g - a) \sin \theta \]

### Summary
The acceleration of the block relative to the incline when the lift is descending with a retardation \( a \) is given by:
\[ a_{\text{block}} = (g - a) \sin \theta \]