A body is kept on a smooth inclined plane having an inclination of 1 in x. Then,

Options

A. slope of inclined plae is `1//x`
B. slope of inclined plane is `(1)/(sqrt(x^(2)-1))`
C. for the body to remain stationary relative to the incline, the incline must be given a horizontal acceleration of `g//sqrt(x^(2)-1)`
D. for the body to remain stationary to the incline, the incline must be offer a normal reaction of `(mgx)/(sqrt(x^(2)-1))`.

Answer

To solve the problem, we need to analyze the forces acting on a body placed on a smooth inclined plane with an inclination of 1 in x. Here’s a step-by-step solution:

### Step 1: Understand the Inclined Plane
The inclined plane has an inclination of 1 in x, which means that the angle of inclination (θ) can be represented as:
\[ \sin \theta = \frac{1}{x} \]
This implies that the opposite side of the triangle formed by the incline is 1 unit and the adjacent side is x units.

### Step 2: Resolve Forces Acting on the Body
When a body of mass \( m \) is placed on the inclined plane, the gravitational force acting on it is \( mg \) (where \( g \) is the acceleration due to gravity). This force can be resolved into two components:
- **Parallel to the incline**: \( F_{\parallel} = mg \sin \theta \)
- **Perpendicular to the incline**: \( F_{\perpendicular} = mg \cos \theta \)

### Step 3: Calculate the Acceleration of the Body
Since the inclined plane is smooth, there is no friction. The net force acting on the body along the incline will be equal to the component of gravitational force acting down the incline:
\[ F_{\parallel} = m a \]
Substituting the expression for \( F_{\parallel} \):
\[ mg \sin \theta = m a \]
Dividing both sides by \( m \):
\[ g \sin \theta = a \]
Thus, the acceleration \( a \) of the body is given by:
\[ a = g \sin \theta \]

### Step 4: Find \( \cos \theta \)
To find \( \cos \theta \), we can use the Pythagorean identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Substituting \( \sin \theta = \frac{1}{x} \):
\[ \left(\frac{1}{x}\right)^2 + \cos^2 \theta = 1 \]
This simplifies to:
\[ \frac{1}{x^2} + \cos^2 \theta = 1 \]
Rearranging gives:
\[ \cos^2 \theta = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} \]
Taking the square root:
\[ \cos \theta = \frac{\sqrt{x^2 - 1}}{x} \]

### Step 5: Calculate \( \tan \theta \)
Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{x}}{\frac{\sqrt{x^2 - 1}}{x}} = \frac{1}{\sqrt{x^2 - 1}} \]

### Step 6: Substitute \( \tan \theta \) into the Acceleration Formula
Now we substitute \( \tan \theta \) back into the acceleration formula:
\[ a = g \tan \theta = g \cdot \frac{1}{\sqrt{x^2 - 1}} = \frac{g}{\sqrt{x^2 - 1}} \]

### Final Results
1. **Slope of the inclined plane**:
\[ \tan \theta = \frac{1}{\sqrt{x^2 - 1}} \]
2. **Acceleration of the body**:
\[ a = \frac{g}{\sqrt{x^2 - 1}} \]