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(B) For a wheel rolling without slipping,the velocity of the centre of mass is $v_{0} = \omega R$,where $\omega$ is the angular velocity and $R$ is th

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$2$

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(B) For a wheel rolling without slipping,the velocity of the centre of mass is $v_{0} = \omega R$,where $\omega$ is the angular velocity and $R$ is the radius of the wheel.For a particle on the rim at the same level as the centre,the velocity has two components:$1$. The translational velocity $v_{0}$ directed horizontally.$2$. The tangential velocity $v_{t} = \omega R$ directed vertically (downwards or upwards depending on the side).Since $v_{0} = \omega R$,the magnitude of the resultant velocity $v$ is given by:$v = \sqrt{v_{0}^{2} + v_{t}^{2}} = \sqrt{v_{0}^{2} + (\omega R)^{2}}$$v = \sqrt{v_{0}^{2} + v_{0}^{2}} = \sqrt{2 v_{0}^{2}} = \sqrt{2} \, v_{0}$Comparing this with the given expression $\sqrt{x} \, v_{0}$,we get $\sqrt{x} = \sqrt{2}$,which implies $x = 2$.

The centre of a wheel rolling on a plane surface moves with a speed $v_{0}$. $A$ particle on the rim of the wheel at the same level as the centre will be moving at a speed $\sqrt{x} \, v_{0}$. Then the value of $x$ is ...... .

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