A uniform rod of length 1 metre is bent at its mid-point to make 90° angle. The distance of the centre of mass from the centre of the rod is 

Rod POQ of length I= 100 cm is bent at its mid-point O so that $\angle$POQ = 90° (see Fig.). The mass of part PO of length l/2 can be taken to be concentrated at its mid-point A whose coordinates are (0, l/4) and of part OQ of length l/2 at its mid-point B whose coordinates are (l/4, 0). The centre of mass of these two equal masses is at mid-point C between A and B. The coordinates of C are (l/8, l/8). 

$$\begin{gathered} \therefore \mathrm{OC}=\sqrt{{\left(\mathrm{OE}\right)}^2+{\left(\mathrm{CE}\right)}^2}=\sqrt{{\left(\frac{\mathrm{l}}{8}\right)}^2+{\left(\frac{\mathrm{l}}{8}\right)}^2} \\ =\frac{\mathrm{l}}{\sqrt{32}}=\frac{100 \mathrm{cm}}{\sqrt{32}}=17.7 \mathrm{cm} \end{gathered}$$