If the density of material of a square plate and a circular plate shown in is same, the centre of mass of the composite system will be

 

Mass of square plate ${\mathrm{M}}_1 = {\mathrm{l}}^2\mathrm{\sigma }$

Mass of circular disc ${\mathrm{M}}_2 = \left(\mathrm{\pi }\frac{{\mathrm{l}}^2}{4}\right)\mathrm{\sigma }$

${\mathrm{\alpha }}_{\mathrm{cm}} = \frac{{\mathrm{M}}_1\times 0+{\displaystyle \frac{{\mathrm{\pi l}}^2\mathrm{\sigma l}}{4}}}{{\mathrm{l}}^2\mathrm{\sigma }[1+{\displaystyle \frac{\mathrm{\pi }}{4}}] } \Rightarrow {\mathrm{x}}_{\mathrm{cm}} = \frac{\mathrm{\pi l}}{4{\displaystyle \frac{(\mathrm{\pi }+4)}{4}}} = \frac{\mathrm{\pi l}}{\mathrm{\pi }+4}$

${\mathrm{x}}_{\mathrm{cm}}<\frac{\mathrm{l}}{2}, \mathrm{i}.\mathrm{e}., \mathrm{C}.\mathrm{O}.\mathrm{M}. \mathrm{lies} \mathrm{in} \mathrm{the} \mathrm{square}$