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 will analyze the forces acting on the bead when the wire is accelerated. The bead will find a new equilibrium position due to the acceleration of the wire.

### Step-by-Step Solution:

1. **Understand the System**:
The wire is bent in the shape of a parabola described by the equation \( y = kx^2 \). The bead is initially at the lowest point of the parabola when the wire is at rest.

2. **Identify Forces Acting on the Bead**:
When the wire is accelerated with a constant acceleration \( a \) parallel to the x-axis, the bead experiences two main forces:
- The gravitational force acting downwards, \( \vec{F_g} = mg \).
- A pseudo force acting on the bead due to the acceleration of the wire, \( \vec{F_p} = ma \), acting in the opposite direction of the acceleration.

3. **Set Up the Coordinate System**:
Let \( \theta \) be the angle that the tangent to the parabola makes with the horizontal at the position of the bead. The forces can be resolved into components along the tangent and normal directions.

4. **Resolve Forces**:
- The component of the weight acting along the tangent to the parabola is \( mg \sin(\theta) \).
- The component of the pseudo force acting along the tangent is \( ma \cos(\theta) \).

5. **Apply Equilibrium Condition**:
For the bead to be in equilibrium, the net force along the tangent must be zero:
\[ mg \sin(\theta) = ma \cos(\theta) \]
Rearranging gives:
\[ \tan(\theta) = \frac{a}{g} \]

6. **Relate \( \tan(\theta) \) to the Parabola**:
The slope of the parabola at any point is given by the derivative:
\[ \tan(\theta) = \frac{dy}{dx} = \frac{d(kx^2)}{dx} = 2kx \]
Therefore, we have:
\[ 2kx = \frac{a}{g} \]

7. **Solve for \( x \)**:
Rearranging the equation gives:
\[ x = \frac{a}{2gk} \]

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

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