Distance Traversed by Centre of Mass of Pulley System in Given Time

To find the distance traversed by the centre of mass of the system in $t = 2\text{ s}$, we first determine the acceleration of each block.

Let the masses be $m_1 = 2\text{ kg}$ and $m_2 = 1\text{ kg}$.

The common acceleration $a$ of the system of blocks connected over a light, frictionless pulley is given by: $a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g$

Substitute the given values ($g = 10\text{ m/s}^2$): $a = \left(\frac{2 - 1}{2 + 1}\right) \times 10 = \frac{10}{3}\text{ m/s}^2$

Here, mass $m_1 = 2\text{ kg}$ moves vertically downwards with acceleration $a$, while mass $m_2 = 1\text{ kg}$ moves vertically upwards with acceleration $a$.

Taking the vertically downward direction as positive, the acceleration of the individual masses can be represented vectorially as: $\vec{a}_1 = +\frac{10}{3}\hat{j}\text{ m/s}^2$ $\vec{a}_2 = -\frac{10}{3}\hat{j}\text{ m/s}^2$

The acceleration of the centre of mass ($\vec{a}_{\text{cm}}$) is given by: $\vec{a}_{\text{cm}} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2}$

Substituting the values: $\vec{a}_{\text{cm}} = \frac{2 \left(+\frac{10}{3}\hat{j}\right) + 1 \left(-\frac{10}{3}\hat{j}\right)}{2 + 1} = \frac{\frac{20}{3}\hat{j} - \frac{10}{3}\hat{j}}{3} = \frac{10}{9}\hat{j}\text{ m/s}^2$

The magnitude of the acceleration of the centre of mass is: $a_{\text{cm}} = \frac{10}{9}\text{ m/s}^2$

Since the system starts from rest ($u_{\text{cm}} = 0$), the distance $s_{\text{cm}}$ traversed by the centre of mass in time $t = 2\text{ s}$ is: $s_{\text{cm}} = u_{\text{cm}} t + \frac{1}{2} a_{\text{cm}} t^2$ $s_{\text{cm}} = 0 + \frac{1}{2} \times \left(\frac{10}{9}\right) \times (2)^2 = \frac{20}{9}\text{ m} \approx 2.22\text{ m}$

Thus, the correct option is C.