A thin uniform wire is bent to form the two equal sides $AB$ and $AC$ of triangle $ABC$, where $AB = AC = 5\,cm.$ The third side $BC$, of length $6\,cm,$ is made from  uniform wire of twice the density of the first. The distance of centre of mass from $A$ is

The position vector of three particles of masses $1\, kg, 2\, kg$ and $3\, kg$ are $\overrightarrow {{r_1}}  = (\widehat i + 4\widehat j + \widehat k)\,m,\overrightarrow {{r_2}}  = (\widehat i + \widehat j + \widehat k)\,m$ and $\overrightarrow {{r_3}}  = (2\widehat i - \widehat j - 2\widehat k)\,m$  respectively. The position  vector of their centre of mass is

A carpenter has constructed a toy as shown in the adjoining figure. If the density of the material of the sphere is $12$ times that of cone, the position of the centre of mass of the toy is given by

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

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

From a circular disc of radius $R$ a triangular portion is cut (see fig.). The distance of  $COM$ of the remaining disc from centre of disc $O$ is:-