A circular loop of rope of length L rotates with uniform angular velocity `omega` about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with resppect to rope ) produce due to slight radiul displacement is given by `(omegaL)/2(pi)` ,if both are in opposite direction. Then the velocity of the pulse w.r.t. to groun will be:

Options

A. `omegaL`
B. `(omegaL)/(2pi)`
C. `(omegaL)/(pi)`
D. `0`

Answer

To solve the problem, we need to find the velocity of the pulse with respect to the ground when a circular loop of rope of length \( L \) rotates with uniform angular velocity \( \omega \) and the pulse moves in the opposite direction to the rotation of the rope.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circular loop of rope of length \( L \) rotating about its center with angular velocity \( \omega \).
- The velocity of the pulse with respect to the rope is given as \( \frac{\omega L}{2\pi} \).

2. **Identifying the Directions**:
- Let's assume the pulse moves in one direction (say clockwise) and the rope rotates in the opposite direction (counterclockwise).

3. **Velocity of the Rope**:
- The linear velocity \( v \) of a point on the rope at radius \( r \) is given by:
\[ v = r \cdot \omega \]
- The radius \( r \) of the circular loop can be calculated from its circumference:
\[ 2\pi r = L \implies r = \frac{L}{2\pi} \]
- Therefore, the velocity of the rope at this radius is:
\[ v_{\text{rope}} = r \cdot \omega = \left(\frac{L}{2\pi}\right) \cdot \omega = \frac{\omega L}{2\pi} \]

4. **Velocity of the Pulse with Respect to the Ground**:
- The velocity of the pulse with respect to the ground \( v_{\text{pulse, ground}} \) can be expressed as:
\[ v_{\text{pulse, ground}} = v_{\text{pulse, rope}} + v_{\text{rope}} \]
- Here, \( v_{\text{pulse, rope}} = \frac{\omega L}{2\pi} \) (in the clockwise direction) and \( v_{\text{rope}} = \frac{\omega L}{2\pi} \) (in the counterclockwise direction). Since they are in opposite directions, we subtract the rope's velocity from the pulse's velocity:
\[ v_{\text{pulse, ground}} = \frac{\omega L}{2\pi} - \frac{\omega L}{2\pi} = 0 \]

5. **Conclusion**:
- The velocity of the pulse with respect to the ground is \( 0 \).

### Final Answer:
The velocity of the pulse with respect to the ground is **0**.

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