A smooth inclined plane of length L, having an inclination `theta` with horizontal is inside a lift which is moving down with retardation a. The time taken by a block to slide down the inclined plane from rest will be

Options

A. `sqrt((2L)/(sqrt(a sin theta)))`
B. `sqrt((2L)/(g sin theta))`
C. `sqrt((2L)/((g-a) sin theta))`
D. `sqrt((2L)/((g+a) sin theta))`

Explanation

To solve the problem of a block sliding down a smooth inclined plane inside a lift that is moving down with retardation \( a \), we will follow these steps:

### Step 1: Identify the forces acting on the block
- The block experiences gravitational force \( mg \) acting downwards.
- Since the lift is moving down with retardation \( a \), there is a pseudo force \( ma \) acting upwards on the block.

### Step 2: Determine the net acceleration along the inclined plane
- The effective acceleration acting on the block along the incline can be calculated as:
\[ a_{\text{net}} = g \sin \theta + a \cos \theta \]
Here, \( g \sin \theta \) is the component of gravitational force acting down the incline, and \( a \cos \theta \) is the component of the pseudo force acting up the incline.

### Step 3: Use kinematic equations to find the time taken to slide down
- The block starts from rest, so the initial velocity \( u = 0 \).
- The distance \( s \) along the incline is \( L \).
- We can use the kinematic equation:
\[ s = ut + \frac{1}{2} a_{\text{net}} t^2 \]
Substituting the values, we have:
\[ L = 0 + \frac{1}{2} (g \sin \theta + a \cos \theta) t^2 \]
Simplifying gives:
\[ L = \frac{1}{2} (g \sin \theta + a \cos \theta) t^2 \]

### Step 4: Solve for time \( t \)
- Rearranging the equation to solve for \( t^2 \):
\[ t^2 = \frac{2L}{g \sin \theta + a \cos \theta} \]
- Taking the square root gives us the time \( t \):
\[ t = \sqrt{\frac{2L}{g \sin \theta + a \cos \theta}} \]

### Step 5: Conclusion
- The time taken by the block to slide down the inclined plane from rest is:
\[ t = \sqrt{\frac{2L}{g \sin \theta + a \cos \theta}} \]