A smooth wire of length `2pir` is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed about the vertical diameter AB, as shown in the figure, the bead is at rest with respect to the circular ring at potion P as shown. Then the value of `omega^(2)` is equal to:

Options

A. `(sqrt(3)g)/(2r)`
B. `2g//r`
C. `2g//(rsqrt(3))`
D. `(gsqrt(3))//r`

Explanation

To solve the problem, we need to analyze the forces acting on the bead and apply the principles of circular motion and equilibrium. Here’s a step-by-step solution:

### Step 1: Understand the setup
The wire is bent into a circle of radius \( r \) and rotates about a vertical diameter. The bead is at rest with respect to the circular ring at position \( P \).

### Step 2: Identify forces acting on the bead
1. **Weight of the bead (mg)**: This force acts downward.
2. **Centripetal force**: This is required to keep the bead moving in a circular path. It acts towards the center of the circle.
3. **Normal force (N)**: This acts perpendicular to the surface of the wire at the position of the bead.

### Step 3: Analyze the geometry
In the triangle formed by the center of the circle (O), the point where the bead is located (P), and the point directly below the bead on the vertical diameter (C):
- The length \( OC \) is \( r/2 \) (the radius of the circle).
- The length \( OP \) is \( r \) (the radius of the original circle).
- The angle \( \theta \) is formed at point O.

Using the sine function:
\[ \sin \theta = \frac{OC}{OP} = \frac{r/2}{r} = \frac{1}{2} \]
This gives us \( \theta = 30^\circ \).

### Step 4: Set up equations for equilibrium
1. **Vertical direction**: The vertical component of the normal force balances the weight of the bead:
\[ N \cos \theta = mg \tag{1} \]

2. **Horizontal direction**: The horizontal component of the normal force provides the necessary centripetal force:
\[ N \sin \theta = m \frac{v^2}{r/2} \tag{2} \]
where \( v \) is the tangential speed of the bead.

### Step 5: Relate tangential speed to angular speed
The relationship between tangential speed \( v \) and angular speed \( \omega \) is given by:
\[ v = \omega \cdot \frac{r}{2} \]
Substituting this into equation (2):
\[ N \sin \theta = m \frac{(\omega \cdot \frac{r}{2})^2}{r/2} = m \omega^2 \frac{r}{4} \]

### Step 6: Substitute and solve
Now we have two equations:
1. \( N \cos \theta = mg \)
2. \( N \sin \theta = m \omega^2 \frac{r}{4} \)

Dividing equation (2) by equation (1):
\[ \frac{N \sin \theta}{N \cos \theta} = \frac{m \omega^2 \frac{r}{4}}{mg} \]
This simplifies to:
\[ \tan \theta = \frac{\omega^2 r}{4g} \]

Substituting \( \theta = 30^\circ \) (where \( \tan 30^\circ = \frac{1}{\sqrt{3}} \)):
\[ \frac{1}{\sqrt{3}} = \frac{\omega^2 r}{4g} \]

### Step 7: Solve for \( \omega^2 \)
Rearranging gives:
\[ \omega^2 = \frac{4g}{r \sqrt{3}} \]

### Final Answer
Thus, the value of \( \omega^2 \) is:
\[ \omega^2 = \frac{4g}{r \sqrt{3}} \]