A loop of a string of mass per unit length `mu` and radius R is rotated about an axis passing through centre perpendicular to the plane with an angular velocityco. A small disturbance is created in the loop having the same sense of rotation. The linear speed of the disturbance for a stationary observer is :

Options

A. `omegaR`
B. `2omegaR`
C. `3omegaR`
D. Zero

Explanation

To solve the problem step by step, we will analyze the situation involving a rotating loop of string and a small disturbance created in it.

### Step-by-Step Solution:

1. **Identify the Parameters:**
- Mass per unit length of the string: \( \mu \)
- Radius of the loop: \( R \)
- Angular velocity of the loop: \( \omega \)

2. **Consider a Small Element of the Loop:**
- Take a small element of the loop with an angular width \( d\theta \).
- The length of this small element is given by \( dL = R d\theta \).
- The mass of this small element is \( dm = \mu \cdot dL = \mu \cdot R d\theta \).

3. **Centripetal Force Acting on the Element:**
- The centripetal force required to keep this mass moving in a circle is given by:
\[ F_c = dm \cdot R \cdot \omega^2 = \mu \cdot R d\theta \cdot R \cdot \omega^2 = \mu R^2 \omega^2 d\theta \]

4. **Tension in the String:**
- Let \( T \) be the tension in the string. The tension provides the necessary centripetal force.
- The forces acting on the small element must be in equilibrium. The vertical components of the tension must balance the centripetal force:
\[ F_c = 2T \sin\left(\frac{d\theta}{2}\right) \]
- For small angles, \( \sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2} \).

5. **Setting Up the Equation:**
- Substitute the approximation into the force balance equation:
\[ \mu R^2 \omega^2 d\theta = 2T \cdot \frac{d\theta}{2} \]
- Simplifying gives:
\[ \mu R^2 \omega^2 d\theta = T d\theta \]

6. **Solving for Tension:**
- Cancel \( d\theta \) from both sides (assuming \( d\theta \neq 0 \)):
\[ T = \mu R^2 \omega^2 \]

7. **Wave Velocity in the String:**
- The wave velocity \( v \) in the string is given by:
\[ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu R^2 \omega^2}{\mu}} = R \omega \]

8. **Considering the Observer's Frame:**
- The linear speed of the disturbance for a stationary observer is the sum of the speed of the string and the speed of the disturbance:
\[ v_{\text{total}} = v + v = R\omega + R\omega = 2R\omega \]

### Final Answer:
The linear speed of the disturbance for a stationary observer is:
\[ \boxed{2R\omega} \]