Thursday, August 6, 2026

Density, Pressure, and Buoyancy UPCAT Reviewer: Lesson and Practice

TEACHER ABI UPCAT SCIENCE

Density, Pressure, and Buoyancy

Use mass and volume evidence, follow pressure through fluids, and compare weight with buoyant force to predict motion.

5-10 minute lesson27 original questionsAdaptive practiceSaves progress
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Density, Pressure, and Buoyancy

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Density predicts the tendency to float; force balance determines what actually happens

density: ρ = m/Vpressure: P = F/Aliquid gauge pressure: P = ρghbuoyant force: F_b = weight of displaced fluid = ρ_fluid gV_displaced

An object sinks when its weight exceeds the maximum buoyant force available, rises when buoyancy exceeds weight, and floats or remains suspended when the forces balance. For a floating object, the amount submerged adjusts until the displaced fluid weighs exactly as much as the object.

Pressure is force distributed over area. In a liquid at rest, pressure increases with depth and depends on fluid density—not on the container’s shape.

Separate density from force

Density is a material or average-object property. Floating is a force-balance condition. Shape can let a steel ship have a low average density and displace enough water.

Use apparent-weight loss

For a submerged object, buoyant force equals weight in air minus apparent weight in the fluid.

DO IT FAST

For floating questions, compare densities before calculating forces

ρobject < ρfluid: The object rises and can float partially submerged.

ρobject = ρfluid: A fully submerged object can remain suspended.

ρobject > ρfluid: The object sinks unless another force supports it.

Floating fraction:

fraction submerged = ρobject / ρfluid

Why it works

Density comparison immediately predicts the direction of motion, while the submerged-fraction relation explains how a floating object reaches equilibrium.

WORKED EXAMPLES

Five forms you should recognize

1. Density from dimensions

Problem: A 240 g block occupies 80 cm³.

ρ = 240/80 = 3.0 g/cm³
2. Volume by displacement

Problem: Water rises from 120 mL to 165 mL after a stone is submerged.

Vstone = 165 − 120 = 45 cm³
3. Liquid pressure

Problem: Find gauge pressure 2.0 m below water using ρ = 1000 kg/m³ and g = 10 m/s².

P = 1000(10)(2.0) = 20,000 Pa
4. Apparent weight

Problem: An object weighs 50 N in air and 35 N underwater.

Fb = 50 − 35 = 15 N
5. Hydraulic lift

Problem: A 50 N input acts on 0.010 m²; output area is 0.20 m².

F₂ = 50(0.20/0.010) = 1000 N

The force gain is paired with a shorter output distance.

COMMON TRAPS

Check before you commit

  • Using mass instead of density to predict floating without considering volume
  • Forgetting to subtract initial water level from final level
  • Assuming a floating object has no weight
  • Thinking pressure at equal depth depends on container width
  • Using total object volume instead of displaced volume for a partially floating object
  • Treating a hydraulic lift as creating energy
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QUICK ANSWERS

Density, Pressure, and Buoyancy FAQ

Does a floating object experience buoyant force?

Yes. At equilibrium, buoyant force equals its weight.

Why does pressure increase with depth?

A deeper point supports the weight of a taller column of fluid above it.

Can a dense material float?

Yes, if its shape and enclosed air make the whole object’s average density lower than the fluid’s.

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