An object is kept on a smooth inclined plane of height 1 unit and length l units. The horizontal acceleration to be imparted to the inclined plane so that the object is stationary relative to the incline is

Options

A. (a)`2 sqrt(r^(2) - 1)`
B. (b)`g (r^(2) - 1)`
C. (c)`(g)/(sqrt(r^(2) - 1)`
D. (d)`(g)/(r^(2) - 1)`

Explanation

To solve the problem of finding the horizontal acceleration that should be imparted to a smooth inclined plane so that an object remains stationary relative to the incline, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have an inclined plane of height \( h = 1 \) unit and length \( l \) units.
- The angle of inclination \( \theta \) can be determined using the relationship between height and length. We have:
\[ \sin \theta = \frac{h}{l} = \frac{1}{l} \]
- We can also find \( \cos \theta \) using the Pythagorean theorem:
\[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{1}{l}\right)^2} \]

2. **Setting Up the Forces**:
- When the inclined plane accelerates horizontally with acceleration \( a \), the object on the incline experiences pseudo forces due to this acceleration.
- The forces acting on the object are:
- Gravitational force \( mg \) acting downwards.
- Normal force \( N \) acting perpendicular to the incline.
- Pseudo force \( ma \) acting horizontally opposite to the direction of acceleration of the incline.

3. **Resolving Forces**:
- The gravitational force can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \theta \)
- Parallel to the incline: \( mg \sin \theta \)
- The pseudo force also has components along the incline:
- Perpendicular to the incline: \( ma \sin \theta \)
- Parallel to the incline: \( ma \cos \theta \)

4. **Equating Forces**:
- For the object to remain stationary relative to the incline, the net force along the incline must be zero:
\[ mg \sin \theta - ma \cos \theta = 0 \]
- Rearranging gives:
\[ mg \sin \theta = ma \cos \theta \]
- Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ g \sin \theta = a \cos \theta \]

5. **Finding Acceleration**:
- Rearranging for \( a \):
\[ a = g \frac{\sin \theta}{\cos \theta} = g \tan \theta \]
- Substituting \( \tan \theta = \frac{1}{\sqrt{l^2 - 1}} \) (from the geometry):
\[ a = g \frac{1}{\sqrt{l^2 - 1}} \]

6. **Final Expression**:
- Thus, the required horizontal acceleration \( a \) to keep the object stationary relative to the incline is:
\[ a = \frac{g}{\sqrt{l^2 - 1}} \]