The center of mass of three particles of masses $10 \ kg, 20 \ kg$,and $30 \ kg$ is at $(0, 0, 0)$. Where should a fourth particle of mass $40 \ kg$ be placed so that the center of mass of the system shifts to $(3, 3, 3)$?

Look at the figure drawn with ink of uniform linear density. $A$ mass $m$ of ink is used to draw each of the two inner circles and each of the two line segments. $A$ mass $6m$ of ink is used to draw the outer circle. The coordinates of the centers of the different parts are: outer circle $(0, 0)$,left inner circle $(-a, a)$,right inner circle $(a, a)$,and the horizontal line segment $(0, -a)$. Find the $y$-coordinate of the center of mass of the ink in the figure.

Masses $m \left(\frac{1}{3}\right)^N \frac{1}{N}$ are placed at $x=N$,where $N=2, 3, 4, \ldots \infty$. If the total mass of the system is $M$,then the centre of mass is

The variation of density of a solid cylindrical rod of cross-sectional area $\alpha$ and length $L$ is given by $\rho = \rho_0 \frac{x^2}{L^2}$, where $x$ is the distance from one end of the rod. The position of its centre of mass from that end $(x=0)$ 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:

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: