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

Answer

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

### Step-by-Step Solution:

1. **Understanding the Parabola**:
The equation of the parabola is given as \( y = kx^2 \). This means that the parabola opens upwards with its vertex at the origin (0,0).

2. **Forces Acting on the Bead**:
When the wire is accelerated, the bead experiences a normal force \( N \) from the wire and its weight \( mg \) acting downwards. The normal force can be resolved into two components:
- Horizontal component: \( N \cos \theta \)
- Vertical component: \( N \sin \theta \)

3. **Equations of Motion**:
In the horizontal direction, the bead must satisfy:
\[ N \cos \theta = ma \]
In the vertical direction, the bead must satisfy:
\[ N \sin \theta = mg \]

4. **Dividing the Equations**:
Dividing the horizontal equation by the vertical equation gives:
\[ \frac{N \cos \theta}{N \sin \theta} = \frac{ma}{mg} \]
This simplifies to:
\[ \cot \theta = \frac{a}{g} \]

5. **Finding the Slope of the Parabola**:
The slope of the parabola at any point is given by the derivative \( \frac{dy}{dx} \). For the parabola \( y = kx^2 \):
\[ \frac{dy}{dx} = 2kx \]
This slope at the bead's position can also be expressed in terms of the angle \( \alpha \) as:
\[ \tan \alpha = 2kx \]

6. **Relating Angles**:
From geometry, we know that:
\[ \alpha + \theta = \frac{\pi}{2} \]
Therefore, we can express \( \tan \alpha \) in terms of \( \cot \theta \):
\[ \tan \alpha = \cot \theta = \frac{g}{a} \]

7. **Setting Up the Equation**:
From the previous steps, we have:
\[ 2kx = \frac{g}{a} \]

8. **Solving for \( x \)**:
Rearranging gives:
\[ x = \frac{g}{2ka} \]

### Final Answer:
The distance of the new equilibrium position of the bead from the y-axis is:
\[ x = \frac{a}{2kg} \]