Two blocks of masses 2kg and 1 kg respectively are tied to the ends of a string which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in 2 seconds is 

$\left(\mathrm{g}=10 \mathrm{m}/{\mathrm{s}}^2\right)$

Two blocks have masses of 2 kg and 1 kg , and the time interval is 2 sec.

The formula for acceleration is,

$\mathrm{a}=\frac{\left({\mathrm{m}}_2-{\mathrm{m}}_1\right)\mathrm{g}}{({\mathrm{m}}_2+{\mathrm{m}}_1)}=\frac{(2-1)\mathrm{g}}{2+1}=\frac{\mathrm{g}}{3}$

The centre of mass's acceleration is expressed as,

$$\begin{gathered} {\mathrm{a}}_{\mathrm{cm}}=\frac{({\mathrm{m}}_2-{\mathrm{m}}_1){\displaystyle \frac{\mathrm{g}}{3}}}{{\mathrm{m}}_2+{\mathrm{m}}_1} \\ {\mathrm{a}}_{\mathrm{cm}}=\frac{(2-1){\displaystyle \frac{\mathrm{g}}{3}}}{2+1} \\ {\mathrm{a}}_{\mathrm{cm}}=\frac{\mathrm{g}}{9} \end{gathered}$$

The centre of mass's travelled distance is given as,

$\mathrm{s}=\frac{1}{2}{\mathrm{at}}^2=\frac{1}{2}\times \frac{\mathrm{g}}{9}\times 4=2.22\mathrm{m}$

Hence the correct answer is 2.22m.