Calculate the velocity of the centre of mass of the system of particles shown in figure.

The correct option is B

0.2 m/s

As we already know,
$V-x{)}_{c}om=\frac{\sum m_i(v_i{)}_x}{\sum m_i};(V_y{)}_{com}=\frac{\sum m_i(v_i{)}_y}{\sum m_i}and(v_z{)}_{com}=\frac{\sum m_i(v_i{)}_z}{\sum m_i}$
As there is no particle in the given system where velocity has a component in Z direction, the component wise pictorial representation of all 5 particles is depicted as shown (assuming horizontal & vertical direction as x & y axes respectively).

Recall
cos ${37}^{\circ }$ = $\frac{4}{5}$
sin ${37}^{\circ }$ = $\frac{3}{5}$
Now ,
$(V_x{)}_{COM}$ = $\frac{-1.5\times \frac{4}{5}-(1\times \frac{4}{5}\times 1.5)+(0.5\times 3)+(1\times 2\times \frac{4}{5})}{1+1.2+1.5+0.5+1}$
= 0.135 m/s
Similarly ,
$(V_y{)}_{COM}$ = $\frac{-1.5\times \frac{3}{5}+(1.2\times 0.4)+(1.5\times \frac{3}{5})+(2\times 1\times \frac{3}{5})}{5.2}(V_y{)}_{COM}$ = 0.14 m/s
${\overrightarrow{V}}_{net}$ = $\sqrt{{(V_x{)}_{com}}^2+{(V_y{)}_{com}}^2}$
= $\sqrt{{\left(0.135\right)}^2+{\left(0.14\right)}^2}$
= 0.2 m/s