Two point masses M and 3M are placed at a distance r. The moment of inertia of the system about an Axis passing through the centre of mass of system and perpendicular to the line joining the points masses is

Dear Student ,
As mass of both of the object are not equal so the centre of mass will be more towards 3m mass than mass m .
Let the distance of mass m be x and that from 3m be (r - x) .
Now , mx =3m ( r -x )
=> x = 3r /4 .
and r -x = r /4 .
Now the moment of inertia of the system is ,
$$\begin{gathered} I=3m{\left(r-x\right)}^2+m{x}^2 \\ =\frac{3m{r}^2}{16}+\frac{9m{r}^2}{16} \\ =\frac{3}{4}m{r}^2 \end{gathered}$$
Regards