Select the pair of correct statments

$\left(1\right)$ Centre of gravity of a body is the point at which the weight of the body acts.
$\left(2\right)$ Centre of mass coincides with the centre of gravity if the earth is assumed to have infinitely large radius
$\left(3\right)$ The entire mass of the body can be considered to be concentrated at its centre of gravity to evaluate the gravitational field intensity due to any body at an external point
$\left(4\right)$ The radius of gyration of any body rotating about an axis is the length of the perendicular dropped from the centre of gravity of the body to the axis

Three particles of masses $50 \mathrm{g}$, $100\mathrm{ }\mathrm{g}$ and $150\mathrm{ }\mathrm{g}$ are placed at the vertices of an equilateral triangle of side $1 \mathrm{m}$ (as shown in the figure). The $\left(\mathrm{x},\mathrm{y}\right)$ coordinates of the centre of mass will be:

Three point particles of masses $1.0 \mathrm{k}\mathrm{g}, 1.5 \mathrm{kg}$ and $2.5 \mathrm{k}\mathrm{g}$ are placed at three corners of a right angle triangle of sides $4.0 \mathrm{c}\mathrm{m}, 3.0 \mathrm{c}\mathrm{m}$ and $5.0 \mathrm{c}\mathrm{m}$ as shown in the figure. The centre of mass of the system is at a point:

Three identical spheres, each of mass $M,$ are placed at the corners of a right angle triangle with mutually perpendicular sides equal to $2 \mathrm{m}$ (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, find the position vector of centre of mass.

As shown in figure, when a spherical cavity (centred at $O$ ) of radius $1$ is cut out of a uniform sphere of radius $R$ (centred at $C$ ), the centre of mass of remaining (shaded part of sphere is at $G,$ i.e., on the surface of the cavity. $R$ can be determined by the equation:

In the figure shown $ABC$ is a uniform wire. If the center of mass of the wire lies vertically below point $A$, then $\frac{BC}{AB}$ is close to

Figure shows a rectangular copper plate with its centre of mass at the origin $O$ and side $AB=2BC=2 \mathrm{m}.$ If a quarter part of the plate (shown as shaded) is removed, the centre of mass of the remaining plate would lie at

A uniform rectangular thin sheet $ABCD$ of mass $M$ has length $a$ and breadth $b$, as shown in the figure. If the shaded portion $HBGO$ is cut-off, the coordinates of the centre of mass of the remaining portion will be: