Three particles each of mass `100g` are placed at the vertices of an equilateral triangle of side length `10cm`. Find the moment of inertia of the system about an axis passing through the centroid of the triangle and perpendicular to its plane.

Explanation

`m=100g=100xx10^(-3)kg`, side `a=10cm=0.1m`
Moment of inertia `I=3m^(2)`
`=3xx100xx10^(-3)((10)/(sqrt(3)xx10^(-2))^(2)=10^(-3)kkgm^(2)`