The position of the centre of mass of a cube of uniform density will be at

Two particles of masses $2$ g and $4$ g are situated at the opposite ends, $A$ and $B$ of a wooden bar respectively. Let $l(AB) = 9$ cm. The center of mass of the system will be

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

To find the centre of mass of a rigid body,why is it not possible to know $\sum m_i \vec{r}_i$ for all the particles?

Particles of masses $8 \, kg, 2 \, kg, 4 \, kg$,and $2 \, kg$ are placed at the vertices $A, B, C$,and $D$ of a square respectively. The distance of the center of mass of the system from vertex $A$ is $....... \, cm$. The length of the diagonal of the square is $80 \, cm$.

The centre of mass of a triangle shown in the figure has coordinates