Two blocks of masses $3\text{kg}$ and $2\text{kg}$ respectively are tied to the ends of a string which passes over a light pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance moved by $COM$ in $5$ seconds is

The correct option is B $5\text{m}$ downward
FBD of blocks:-


Let the acceleration of the blocks be $a$.
Therefore, for $3\text{kg}$ block
$mg-T=ma\Rightarrow 3g-T=3a--\left(i\right)$
For $2\text{kg}$ block
$T-mg=ma\Rightarrow T-2g=2a--\left(ii\right)$

From eqn (i) and (ii)
$3g-3a=2g+2a\Rightarrow 5a=g$
$\Rightarrow a=g/5=2{\text{m/s}}^2$

Both blocks start from rest. So, distance travelled by the block in 5 sec is
$s_1=s_2=ut+\frac{1}{2}a{t}^2=0\times t+\frac{1}{2}\times 2\times 5^2=25\text{m}$.

Now, the distance travelled by $COM$ is
$S_{COM}=\frac{M_1{s}_1+M_2{s}_2}{M_1+M_2}=\frac{2\times 25+3(-25)}{2+3}=-5\text{m}$
Thus $COM$ will move $5\text{m}$ downwards in $5\text{s}$