$A$ point object of mass $m$ is kept at $(a, 0)$ along the $x$-axis. What mass should be kept at $(-3a, 0)$ so that the center of mass lies at the origin?

The position of the center of mass of three masses $2 \text{ kg}$, $3 \text{ kg}$, and $15 \text{ kg}$ placed with respect to the midpoint $(p)$ of the base, as shown in the figure, is . . . . . . .

Two particles of masses $1\,kg$ and $3\,kg$ have position vectors $2\hat{i} + 3\hat{j} + 4\hat{k}$ and $-2\hat{i} + 3\hat{j} - 4\hat{k}$ respectively. The position vector of the centre of mass is:

The mass per unit length of a rod of length $l$ is given by $\lambda = \frac{M_0 x}{l}$,where $M_0$ is a constant and $x$ is the distance from one end of the rod. The position of the center of mass of the rod is:

$A$ uniform rod of length $L$ and mass $M$ is placed along the $x$-axis with one end at the origin. Find the $x$-coordinate of the centre of mass of the rod.

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