The coordinates of the center of mass of the triangular lamina shown in the figure are ....... .

$A$ 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 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:

$A$ rigid body consists of a $3 \ kg$ mass and a $2 \ kg$ mass connected by a massless rod. The $3 \ kg$ mass is at $\vec{r}_1 = (2 \hat{i} + 5 \hat{j}) \ m$ and the $2 \ kg$ mass is at $\vec{r}_2 = (4 \hat{i} + 2 \hat{j}) \ m$. Find the length of the rod and the coordinates of the center of mass.

$A$ 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:

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