Masses of $8 \ kg, 2 \ kg, 4 \ kg,$ and $2 \ kg$ are placed at the corners $A, B, C,$ and $D$ respectively of a square $ABCD$ with a diagonal of $80 \ cm$. The distance of the center of mass from $A$ is ....... $cm$.

$A$ system consists of $3$ particles each of mass $m$ located at $(1, 1), (2, 2),$ and $(3, 3)$. The coordinates of the centre of mass are

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 masses and positions (in rectangular coordinates) of four particles are as follows: $1 \ kg$ at $(a, a)$,$2 \ kg$ at $(-a, a)$,$3 \ kg$ at $(-a, -a)$,and $4 \ kg$ at $(a, -a)$. The position vector of the centre of mass of the system of four particles is:

Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles,and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6m$. The coordinates of the centres of the different parts are: outer circle $(0,0)$,left inner circle $(-a, a)$,right inner circle $(a, a)$,vertical line $(0,0)$ and horizontal line $(0,-a)$. The $y$-coordinate of the centre of mass of the ink in this drawing 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 $L = 10 \ cm$)