Given: A wheel rolls without slipping on a plane surface.

• Speed of centre of wheel = $v_0$
• We need the speed of a particle on the rim which is at the same level as the centre.

Rolling without slipping: For pure rolling, $v_0 = \omega R$ where $\omega$ is angular speed and $R$ is radius of the wheel.

Velocity of the required point: A point on the rim at the same height as the centre is the side point of the wheel.

Its velocity is the vector sum of:

• translational velocity of centre: $v_0$ (horizontal)
• rotational velocity about centre: $\omega R = v_0$ (vertical, for the side point)

These two velocities are perpendicular.

Resultant speed: $v = \sqrt{v_0^2 + v_0^2} = \sqrt{2v_0^2} = \sqrt{2}\,v_0$

Comparing with the form $\sqrt{x}\,v_0$, $\sqrt{x}\,v_0 = \sqrt{2}\,v_0$ so, $x = 2$

Comparison with stored answer: Stored correct answer = $02$

Our derived answer is $2$, which matches the stored answer.