A uniform rod of length l=100 cm is bent at its mid-point to make 90∘ angle. The distance of the centre of mass from the centre of the rod is
17.7 cm
Let the rod POQ of length l be bent at its mid-point O so that ∠POQ=90∘ as shown in the figure. The mass of part PO of length l2 can be taken to be concentrated at its mid-point A whose co-ordinates are (0,l4) and of part OQ of length l2 at its mid-point B whose coordinates are (l4,0). The centre of mass of these two equal masses is at mid-point C between A and B. The coordinates of C are (l8),(l8).
∴OC=√(OE)2+(CE)2=√(l8)2+(l8)2=l√32=100 cm√32=17.7 cm
Hence, the correct choice is (c).