A piece of wire is bent in the shape of a parabola `y=kx^(2)` (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is:

Options

A. a/gk
B. a/2gk
C. 2a/gk
D. a/4gk

Explanation

To solve the problem, we need to find the new equilibrium position of the bead on the parabolic wire when the wire is accelerated parallel to the x-axis with a constant acceleration \( a \).

### Step-by-Step Solution:

1. **Understand the Setup**:
The wire is shaped like a parabola given by the equation \( y = kx^2 \). Initially, when the wire is at rest, the bead is at the lowest point of the parabola (at the origin, \( (0, 0) \)).

2. **Acceleration of the Wire**:
When the wire is accelerated with a constant acceleration \( a \) in the positive x-direction, the bead will move to a new position along the wire.

3. **Free Body Diagram**:
At the new position, we can analyze the forces acting on the bead:
- The weight of the bead \( mg \) acts downward.
- The normal force \( N \) acts perpendicular to the surface of the wire.
- A pseudo force \( F_{\text{pseudo}} = ma \) acts on the bead in the opposite direction of the acceleration of the wire (to the left).

4. **Determine Forces**:
Let \( \theta \) be the angle that the tangent to the parabola makes with the horizontal at the new position of the bead. The components of the normal force can be resolved as:
- \( N \sin \theta = ma \) (horizontal direction)
- \( N \cos \theta = mg \) (vertical direction)

5. **Relate Forces**:
From the equations above, we can derive:
\[ \tan \theta = \frac{a}{g} \]

6. **Slope of the Parabola**:
The slope of the parabola at any point is given by the derivative:
\[ \frac{dy}{dx} = 2kx \]
Therefore, at the new position, we can also express \( \tan \theta \) as:
\[ \tan \theta = 2kx \]

7. **Equate the Two Expressions for \( \tan \theta \)**:
Setting the two expressions for \( \tan \theta \) equal gives:
\[ 2kx = \frac{a}{g} \]

8. **Solve for \( x \)**:
Rearranging the equation to solve for \( x \):
\[ x = \frac{a}{2kg} \]

9. **Conclusion**:
The distance of the new equilibrium position of the bead from the y-axis is:
\[ x = \frac{a}{2kg} \]

### Final Answer:
The distance of the new equilibrium position of the bead from the y-axis is \( \frac{a}{2kg} \).