A solid steel ball and a large steel cargo ship have the same total mass. The ball sinks; the ship floats. What is the correct explanation?
AThe ship is made of a different, lower-density alloy than the solid ball
BThe ship's large surface area increases drag, slowing its descent and keeping it afloat
CThe ship is hollow — the average density of the hull, enclosed air, and cargo together is less than water, so the ship displaces its weight in water before sinking
DThe buoyant force on the ship is greater because its bottom surface area is larger, amplifying the upward pressure
Buoyancy depends on the weight of fluid displaced, not on the object's material density in isolation. A solid steel ball has an average density (~7800 kg/m³) far above water, so it sinks before displacing its own weight. A steel ship is hollow: the enclosed volume of air brings the average density of the entire system (hull + air + cargo) below ~1000 kg/m³. As the ship settles, it displaces water until the displaced water's weight equals the ship's total weight — and it floats. Option D is wrong because buoyancy depends on displaced *volume*, not surface area in contact with fluid.
Question 2 Multiple Choice
A submarine is fully submerged at 50 m depth in seawater (treated as incompressible). It descends to 200 m without changing its volume. How does the buoyant force change?
AIt increases significantly — hydrostatic pressure is much higher at 200 m, increasing the upward force
BIt remains the same — for an incompressible fluid, buoyant force equals ρ_fluid × g × V_displaced and is independent of depth
CIt decreases — the greater water pressure at depth compresses the submarine slightly, reducing displaced volume
DIt increases because the water is slightly denser at depth, increasing the weight of the displaced fluid
Archimedes' principle: F_b = ρ_fluid × g × V_displaced. For an incompressible fluid, ρ_fluid and g are constant, and V_displaced is fixed (submarine volume unchanged). The buoyant force is therefore constant regardless of depth. While pressure increases with depth, the key insight is that pressure increases equally on all faces — the net upward force (bottom pressure minus top pressure) depends only on the depth *difference* across the object (equal to its height), not on the absolute depth. For an incompressible fluid this depth difference and ρ_fluid are constant. (Option D is technically true at very large depths, but negligible for this context.)
Question 3 True / False
An iceberg floats with approximately 89% of its volume submerged because the ratio V_submerged/V_total equals ρ_ice/ρ_seawater ≈ 917/1025.
TTrue
FFalse
Answer: True
For a floating object, weight equals buoyant force: ρ_obj × V_obj × g = ρ_fluid × V_submerged × g. Rearranging: V_submerged/V_obj = ρ_obj/ρ_fluid = 917/1025 ≈ 0.894, confirming that ~89% is submerged and ~11% is above water. This is a direct consequence of Archimedes' principle applied to partial submersion. The iceberg keeps sinking until the weight of displaced seawater exactly equals the weight of the entire iceberg.
Question 4 True / False
The buoyant force on a submerged object increases as the object sinks deeper, because the hydrostatic pressure surrounding the object increases with depth.
TTrue
FFalse
Answer: False
For an incompressible fluid, the buoyant force is F_b = ρ_fluid × g × V_displaced — it depends only on the displaced volume, not on depth. As an object descends, pressure increases on all surfaces, but the pressure on the bottom face increases by exactly the same amount as the pressure on the top face, so the net upward force (their difference integrated over the surfaces) remains constant. This is derivable from the pressure-depth relationship: the net upward force equals ρ_f × g × H × A = ρ_f × g × V, where H is the object's height — independent of the absolute depth to the top or bottom.
Question 5 Short Answer
Using the concept of displaced fluid weight, explain why a large steel cargo ship floats even though solid steel is about 8 times denser than water.
Think about your answer, then reveal below.
Model answer: The buoyant force equals the weight of fluid displaced — not the weight of the ship's steel, and not any property of the steel's density alone. A cargo ship is hollow: its hull encloses a large volume of air and cargo space. The *average* density of the entire system (steel hull + air + any cargo) can be less than the density of water. As the ship is lowered into water, it displaces an increasing volume of water. It stops sinking (reaches equilibrium) when the weight of displaced water equals the ship's total weight. For a steel ship, this happens before the deck goes underwater because the enclosed air volume is large enough that the average density of the ship falls below water density. The ship's maximum load capacity is determined by how deep it can settle before flooding — at maximum draft, buoyancy force still equals total weight.
The key insight is that Archimedes' principle applies to the displaced volume — the volume swept out by the submerged portion of the hull — not to the volume of steel material. A hollow object displaces far more water than a solid object of the same mass and material, which is why the shape matters: a steel sphere sinks, but the same steel formed into a bowl shape floats.