density of seawater is about 1.03*10^3 kg/m^2
bulk modulus of tanker hull is about 0.9
call draft = h
so mass of water displaced = 1.03 *10^3 * 0.9 * 400 * 60 * d
that must equal the mass of the fully loaded tanker
2.3 * 10^8 = 2.22*10*7 h
h = 10.4 meters
Math people are likely to not know about bulk modulus (ratio of underwater volume to rectangular box) so if this is a math subject and not a ship design subject divide answer by 0.9
A supertanker can carry 2.2x10 m of oil with a density of 0.88g/cm. When fully loaded its mass is 2.3 x 108 kg, and the dimensions of its hull are approximately 400 m long, 60 m wide, and 38 m high. Given that the density of seawater is 1.03 g/cm , how deeply is the hull submerged in the water?
3 answers
Typo corrections:
density of seawater is about 1.03*10^3 kg/m^2
bulk modulus of tanker hull is about 0.9
call draft = d
so mass of water displaced = 1.03 *10^3 * 0.9 * 400 * 60 * d
that must equal the mass of the fully loaded tanker
2.3 * 10^8 = 2.22*10*7 d
d = 10.4 meters
Math people are likely to not know about bulk modulus (ratio of underwater volume to rectangular box) so if this is a math subject and not a ship design subject multiply d by 0.9
density of seawater is about 1.03*10^3 kg/m^2
bulk modulus of tanker hull is about 0.9
call draft = d
so mass of water displaced = 1.03 *10^3 * 0.9 * 400 * 60 * d
that must equal the mass of the fully loaded tanker
2.3 * 10^8 = 2.22*10*7 d
d = 10.4 meters
Math people are likely to not know about bulk modulus (ratio of underwater volume to rectangular box) so if this is a math subject and not a ship design subject multiply d by 0.9
By the way, it is unlikely that the question is worded correctly. The ship would float lower in the water if fully loaded. However that is what it says and we do not know the cargo volume anyway.