Center of mass of a system of three particles of masses $1, 2, 3\, kg$ is at the point $(1\, m, 2\, m, 3\, m)$ and center of mass of another group of two particles of masses $2 \,kg$ and $3\, kg$ is at point $(-1 \,m, 3\, m, -2\, m)$ . Where a $5\, kg$ particle should be placed, so that center of mass of the system of all these six particles shifts to center of mass of the first system?

A circular plate of uniform thickness of diameter $56\, cm$, whose center is at origin. A circular part of diameter $42\, cm$ is removed from one edge. What is the distance of the centre of mass of the remaining part ........ $cm.$

From a circle of radius $a,$ an isosceles right angled triangle with the hypotenuse as the diameter of the circle is removed. The distance of the centre of gravity of the remaining position from the centre of the circle is

Two masses $m_1$ and $m_2$ ($m_1$ > $m_2$) are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of center of mass 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 $6 \mathrm{~m}$. 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

Three identical spheres each of mass $M$ are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $3\,m$ each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be $\sqrt{x} m$. The value of $x$ is