1. Condition for pure rolling

For a wheel rolling without slipping:

• speed of centre of mass $= v$
• tangential speed of any rim particle relative to centre $= v$

So, the velocity of any point on the rim is the vector sum of:

• translational velocity of the centre: $v$ (forward)
• rotational velocity about the centre: magnitude $v$

2. Use the highest point

At the highest point of the wheel, the rotational velocity is also forward, so the two velocities add:

$v_{\text{top}} = v + v = 2v$

Given:

$v_{\text{top}} = 8\ \text{m/s}$

Therefore,

$2v = 8 \Rightarrow v = 4\ \text{m/s}$

3. Point at the same level as the centre

A point on the rim at the same level as the centre is either the frontmost point or the rearmost point.

For such a point:

• translational velocity = $4\ \text{m/s}$ forward
• rotational velocity = $4\ \text{m/s}$ vertically upward or downward

These two velocities are perpendicular.

Hence the resultant speed is

$v_{\text{side}} = \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2}$

Substitute $v=4$:

$v_{\text{side}} = 4\sqrt{2}\ \text{m/s}$

4. Check options

• A: $4\sqrt{2}\ \text{m/s}$ ✅
• B: $8\ \text{m/s}$ ❌
• C: $4\ \text{m/s}$ ❌
• D: $8\sqrt{2}\ \text{m/s}$ ❌

Therefore, the correct answer is A.