Volume V of the barge=75×6×4=1800m3. Fresh water density=1000kg/m3.
Displacement=180 tonnes=180000kg.
Density=mass/volume=180000/volume=1000, so volume=180m3=V/10.
Assuming that 4m is the height of the barge, then the base area is 6×75=450m2. The draft of the empty barge would be 180/450=0.4m. The total mass of the barge with a load of 360t of iron=360+180=540t=540000kg. This is the mass of freshwater displaced so the corresponding volume is: v=540000/1000=540m3. If we divide this by the area of the base we get 540/450=1.2m, giving us the draft. (3 times the weight (compared with the unladen barge) produces 3 times the draft.)