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A. $0.4\pi \,{\text{m/}}{{\text{s}}^{\text{2}}}$

B.$0.4{\pi ^2}\,{\text{m/}}{{\text{s}}^{\text{2}}}$

C. $\dfrac{{{\pi ^2}}}{{0.4}}\,{\text{m/}}{{\text{s}}^{\text{2}}}$

D.$\dfrac{{0.4}}{{{\pi ^2}}}\,{\text{m/}}{{\text{s}}^{\text{2}}}$

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As we know that angular velocity is the rate of change of angular displacement. For an object rotating about an axis every point of the object will have the same angular velocity. It is denoted by $w$.

In circular motion angular acceleration occurs when the rate of angular velocity changes. It is denoted as $\alpha $.

We also know that when acceleration of a body is expressed as $a$ then it has a relationship with angular velocity. The relationship is expressed as

$ \Rightarrow a = {w^2}r$

Where $w$ is the angular velocity and $r$ is the radius of the circle.

Time period is known as the time taken for completing a complete revolution. It is denoted as $T$.

We have given the following data

Radius of cycle wheel $ = r = 0.4{\text{m}}$

Time for one revolution is $2$ seconds that is time period $T = 2$ seconds.

We know that relation between angular velocity and time period is expressed as

$ \Rightarrow w = \dfrac{{2\pi }}{T}$

So here we can get the value of angular velocity as,

$ \Rightarrow w = \dfrac{{2\pi }}{2} = \pi {\text{rad/s}}$.

Now we will calculate the acceleration of the wheel as

$ \Rightarrow a = {w^2}r$

Now, we substitute the values in the above equation as,

$ \Rightarrow a = {\pi ^2} \times 0.4{\text{m/}}{{\text{s}}^{\text{2}}}$

After simplification we get,

$\therefore a = 0.4{\pi ^2}\,{\text{m/}}{{\text{s}}^{\text{2}}}$

Thus, we get the acceleration of the cycle wheel as $0.4{\pi ^2}\,{\text{m/}}{{\text{s}}^{\text{2}}}$.