The centre of mass of a thin rectangular plate (as shown in the figure) with sides of length $a$ and $b$,whose mass per unit area $(\sigma)$ varies as $\sigma = \frac{\sigma_0 x}{a b}$ (where $\sigma_0$ is a constant),would be . . . . . .

The velocities of three particles of masses $20 \ g$,$30 \ g$ and $50 \ g$ are $10 \hat{i}$,$10 \hat{j}$ and $10 \hat{k}$,respectively. The velocity of the centre of mass of the three particles is

Two objects of mass $10\,kg$ and $20\,kg$ respectively are connected to the two ends of a rigid rod of length $10\,m$ with negligible mass. The distance of the center of mass of the system from the $10\,kg$ mass is:

$A$ rod of length $5L$ is bent at a right angle,keeping one side length as $2L$. Find the position of the center of mass of the system. (Consider $L = 10 \ cm$)

$A$ thin bar of length $L$ has a mass per unit length $\lambda$ that increases linearly with distance $x$ from one end. If its total mass is $M$ and its mass per unit length at the lighter end $(x=0)$ is $\lambda_0$,then the distance of the centre of mass from the lighter end is:

Center of mass of a system of three particles of masses $1 \, kg, 2 \, kg, 3 \, kg$ is at the point $(1 \, m, 2 \, m, 3 \, m)$ and center of mass of another group of two particles of masses $2 \, kg$ and $3 \, kg$ is at point $(-1 \, m, 3 \, m, -2 \, m)$. Where should a $5 \, kg$ particle be placed so that the center of mass of the system of all these six particles shifts to the center of mass of the first system?