A

Step 1: Analyze the forces acting on the bob of the plumb line. The gravitational force acts downwards with magnitude mg and the tension in the thread acts along the thread.
Step 2: The bob makes an angle \( \alpha \) with the vertical. Therefore, the components of the forces can be described as:
- The vertical component of tension: \( T \cos(\alpha) = mg \)
- The horizontal component of tension providing centripetal force: \( T \sin(\alpha) = \frac{mv^2}{r} \)
Here, \( v \) is the tangential speed and \( r \) is the distance from the axis of rotation to the point of suspension.
Step 3: Substitute \( T \) from the first equation into the second equation:
From \( T = \frac{mg}{\cos(\alpha)} \), substitute into the horizontal force equation:
\( \frac{mg \sin(\alpha)}{\cos(\alpha)} = \frac{mv^2}{r} \)
Step 4: Simplify the equation:
\( g \tan(\alpha) = \frac{v^2}{r} \)
This leads to: \( v^2 = g r \tan(\alpha) \)
Step 5: The relation between linear velocity \( v \) and angular velocity \( \omega \) is \( v = r \omega \). Therefore, substituting this into the equation gives:
\( (r \omega)^2 = gr \tan(\alpha) \)
Step 6: Rearranging gives: \( \omega^2 = \frac{g \tan(\alpha)}{r} \)
Step 7: Taking the square root to find angular velocity: \( \omega = \sqrt{\frac{g \tan(\alpha)}{r}} \)
Hence, the angular velocity of rotation is \( \sqrt{\frac{g \tan(\alpha)}{r}} \).
Therefore, A.