What is the weight of the cargo taken off the steamer if its draft has decreased by 0.2 m?

What is the weight of the cargo taken off the steamer if its draft has decreased by 0.2 m? The horizontal section of the ship at water level is 4000 m2.

Given:
h = 0.2 meters – the amount by which the draft of the steamer decreased after the cargo was removed from it;
s = 4000 m2 (square meters) – the area of ​​the horizontal section of the steamer at water level;
ro = 1000 kg / m3 (kilogram per cubic meter) – water density;
g = 9.8 N / kg (Newton per kilogram) – acceleration of gravity.
It is required to determine P (Newton) – what weight was removed from the ship.
According to Archimedes’ law, the weight of the removed cargo will be numerically equal to the difference between the Archimedean force acting on the ship before the cargo is removed and after:
P = A1 – A2 = ro * g * V1 – ro * g * V2 = ro * g * s * h1 – ro * g * s * h2 = ro * g * s * (h1 – h2) = ro * g * s * h;
P = 1000 * 9.8 * 4000 * 0.2 = 9800 * 800 = 7840000 Newton = 7.84 MN.
Answer: a cargo weighing 7.84 MN was removed from the steamer.