Introduction:
In this scenario, we have a smooth inclined plane inside a lift that is moving down with retardation. The inclined plane has a length L and an inclination angle β with the horizontal. We need to determine the time it takes for a block to slide down from rest on this inclined plane.

Given:
- Length of the inclined plane: L
- Inclination angle with the horizontal: β
- Lift moving down with retardation: a

Analysis:
To solve this problem, we can divide it into two parts:
1. Finding the acceleration of the block on the inclined plane.
2. Using the acceleration to calculate the time taken to slide down.

Part 1: Finding the acceleration:
The force acting on the block on the inclined plane can be resolved into two components:
- The component parallel to the inclined plane: mg sin(β)
- The component perpendicular to the inclined plane: mg cos(β)

The force parallel to the inclined plane is responsible for the acceleration of the block. According to Newton's second law, this force can be related to the acceleration as:
mg sin(β) = ma
where m is the mass of the block.

Part 2: Calculating the time taken to slide down:
The distance traveled by the block on the inclined plane is L. Using the first equation of motion, we can relate the distance, acceleration, and time as:
L = ut + (1/2)at^2
where u is the initial velocity (0 in this case), a is the acceleration, and t is the time taken.

Rearranging the equation, we get:
t^2 = 2L/a

Taking the square root of both sides, we find:
t = √(2L/a)

Conclusion:
The time taken for the block to slide down from rest on the smooth inclined plane inside the lift can be calculated using the equation t = √(2L/a), where L is the length of the inclined plane and a is the acceleration of the block.

Time= √[{2LsinB}/{g+a}]