A wheel of radius `r` rolls without slipping with a speed `v` on a horizontal road. When it is at a point `A` on the road, a small blob of mud separates from'the wheel at- its highest point and lands at point `B` on the road:

Options

A. (a)`AB=vsqrt(r/g)`
B. (b)`AB=2vsqrt(r/g)`
C. (c)`AB=4vsqrt(r/g)`
D. (d)`vgtsqrt(4rg)`,the blob of mud will land on the wheel and not on the road

Explanation

To solve the problem step by step, we will analyze the motion of the blob of mud that separates from the wheel and lands on the road.

### Step 1: Determine the initial conditions of the blob of mud
When the blob of mud separates from the wheel at its highest point, it has two key characteristics:
- **Initial velocity (u)**: Since the wheel is rolling without slipping, the velocity of the blob at the highest point is equal to the linear velocity of the wheel, which is `v`. However, since it is at the highest point, it also has a vertical component due to the rolling motion. The total velocity of the blob when it separates is `2v` (the horizontal component is `v` and the vertical component is also `v`).
- **Initial height (h)**: The height from which the blob falls is equal to the diameter of the wheel, which is `2r`.

### Step 2: Calculate the time of flight (t)
The motion of the blob can be analyzed using the equations of motion under gravity. The vertical motion can be described by the equation:
\[ h = ut + \frac{1}{2}gt^2 \]
where:
- \( h = 2r \) (the height),
- \( u = 0 \) (the initial vertical velocity),
- \( g \) is the acceleration due to gravity.

Substituting the values, we have:
\[ 2r = 0 + \frac{1}{2}gt^2 \]
This simplifies to:
\[ 2r = \frac{1}{2}gt^2 \]
Multiplying both sides by 2 gives:
\[ 4r = gt^2 \]
Rearranging this gives:
\[ t^2 = \frac{4r}{g} \]
Taking the square root of both sides, we find:
\[ t = 2\sqrt{\frac{r}{g}} \]

### Step 3: Calculate the horizontal distance traveled (s)
The horizontal distance traveled by the blob while it is falling can be calculated using the formula:
\[ s = vt \]
where:
- \( v = 2v \) (the horizontal velocity of the blob),
- \( t = 2\sqrt{\frac{r}{g}} \) (the time of flight).

Substituting these values into the equation gives:
\[ s = 2v \cdot 2\sqrt{\frac{r}{g}} \]
This simplifies to:
\[ s = 4v\sqrt{\frac{r}{g}} \]

### Step 4: Conclusion
The horizontal distance from point A to point B where the blob of mud lands is:
\[ AB = 4v\sqrt{\frac{r}{g}} \]

### Final Answer
The distance AB is \( 4v\sqrt{\frac{r}{g}} \).