A rod of length 3 m and its mass acting per unit length is directly proportional to distance x from one of its end, then its centre of gravity from that end will be at

Given that mass per unit length is proportional to distance x, i.e., p =k x, where k = constantNow mass of dx length at a distance x=p dx =k x dxLet the centre of mass is at a distance x0 from one end. Then mass on either side of the rod must be equal. Thus∫0x0 kxdx=∫x03 kxdxor x220x0=x22x03or x022=92−x022∴ x02=92  or  x0=32So, x0≈2m