A wheel is rolling uniformly along a level road without slipping . Velocity of the highest point on its rim about the road is V . Then magnitude of velocity of a point on its rim which is at the same level as that of the centre is

Options

A. `sqrt(2)V`
B. `V//2`
C. 2V
D. `V//sqrt(2)`

Explanation

To solve the problem step-by-step, we need to analyze the motion of the wheel and the velocities of the points on its rim.

### Step 1: Understand the motion of the wheel
The wheel is rolling uniformly along a level road without slipping. This means that every point on the rim of the wheel has both translational and rotational motion.

### Step 2: Identify the velocities
1. The velocity of the highest point on the rim (let's call it point A) is given as \( V \).
2. We need to find the magnitude of the velocity of a point on the rim that is at the same level as the center of the wheel (let's call this point B).

### Step 3: Relate translational and rotational motion
Let \( V_0 \) be the translational velocity of the center of the wheel. In pure rolling motion, the relationship between the translational velocity \( V_0 \) and the angular velocity \( \omega \) is given by:
\[ V_0 = \omega R \]
where \( R \) is the radius of the wheel.

### Step 4: Analyze the velocities of point B
Point B is at the same level as the center of the wheel. It has two components of velocity:
1. **Translational velocity**: This is equal to \( V_0 \).
2. **Rotational velocity**: The tangential velocity due to rotation about the center of the wheel. Since point B is at a distance \( R \) from the center, the tangential velocity is also \( V_0 \).

### Step 5: Calculate the resultant velocity of point B
The total velocity of point B is the vector sum of its translational and rotational velocities. Since both velocities are in the same direction, we can add them directly:
\[ V_B = V_0 + V_0 = 2V_0 \]

### Step 6: Substitute \( V_0 \) in terms of \( V \)
From the information given, we know that the velocity of the highest point (point A) is \( V \). Since point A is also moving with the same angular velocity, we can express \( V_0 \) in terms of \( V \):
\[ V = V_0 + V_0 = 2V_0 \implies V_0 = \frac{V}{2} \]

### Step 7: Substitute \( V_0 \) back into the equation for point B
Now substituting \( V_0 \) back into the equation for the velocity of point B:
\[ V_B = 2V_0 = 2 \left(\frac{V}{2}\right) = V \]

### Final Answer
The magnitude of the velocity of the point on the rim which is at the same level as that of the center is:
\[ \boxed{V} \]