A body kept on a smooth inclined plane having inclination 1 in x will remain stationary relative to the inclined plane if the plane is given a horizontal accelertion equal to

Options

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

Answer

To solve the problem, we need to find the horizontal acceleration of a smooth inclined plane such that a body resting on it remains stationary relative to the inclined plane. The inclination of the plane is given as 1 in x.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- We have a body of mass \( m \) on a smooth inclined plane.
- The inclination of the plane is given as \( \frac{1}{x} \), which means the angle \( \theta \) can be derived from the sine function.

2. **Setting Up the Forces:**
- When the inclined plane accelerates horizontally with acceleration \( a \), the body experiences a pseudo force acting down the incline due to this acceleration.
- The components of forces acting on the body are:
- Down the incline: \( m g \sin \theta \)
- Up the incline (due to pseudo force): \( m a \cos \theta \)

3. **Applying Newton's Second Law:**
- For the body to remain stationary relative to the inclined plane, the net force acting on it must be zero.
- Therefore, we can set up the equation:
\[ m a \cos \theta = m g \sin \theta \]

4. **Cancelling Mass:**
- Since \( m \) appears on both sides of the equation, we can cancel it out:
\[ a \cos \theta = g \sin \theta \]

5. **Rearranging the Equation:**
- Rearranging gives us:
\[ a = g \frac{\sin \theta}{\cos \theta} = g \tan \theta \]

6. **Finding \( \tan \theta \):**
- Given that \( \sin \theta = \frac{1}{x} \), we can find \( \cos \theta \) using the Pythagorean identity:
\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{1}{x}\right)^2 = \frac{x^2 - 1}{x^2} \]
- Thus, \( \cos \theta = \frac{\sqrt{x^2 - 1}}{x} \).

7. **Calculating \( \tan \theta \):**
- Now, we can find \( \tan \theta \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{x}}{\frac{\sqrt{x^2 - 1}}{x}} = \frac{1}{\sqrt{x^2 - 1}} \]

8. **Substituting Back:**
- Substitute \( \tan \theta \) back into the equation for \( a \):
\[ a = g \tan \theta = g \cdot \frac{1}{\sqrt{x^2 - 1}} = \frac{g}{\sqrt{x^2 - 1}} \]

9. **Final Answer:**
- The horizontal acceleration \( a \) required for the body to remain stationary relative to the inclined plane is:
\[ a = \frac{g}{\sqrt{x^2 - 1}} \]

### Conclusion:
The correct option for the horizontal acceleration is \( \frac{g}{\sqrt{x^2 - 1}} \).