AP Physics 1 / Topic

Fluids

Unit 8 covers fluids at rest and in motion: density and pressure, hydrostatic pressure with depth, Pascal's principle, buoyancy and Archimedes' principle, the continuity equation, and Bernoulli's equation.

43 practice questions3 FRQs8 skill areas

What the exam asks

  • Compute absolute and gauge pressure at depth in a fluid
  • Apply Pascal's principle to hydraulic lifts
  • Use Archimedes' principle for floating and submerged objects and apparent weight
  • Apply the continuity equation to flow through changing pipe cross-sections
  • Use Bernoulli's equation (and Torricelli's theorem) to relate pressure, speed, and height

Key formulas and rules

Density and pressure

ρ = m/V; P = F/A

Hydrostatic pressure

P = P0 + ρgh (depends on depth, not container shape)

Buoyant force

F(b) = ρ(fluid)·V(displaced)·g; floating: F(b) = weight

Continuity

A1v1 = A2v2 - narrower pipe, faster flow

Bernoulli

P + ½ρv² + ρgh = constant; faster flow ⇒ lower pressure

Torricelli

efflux speed v = √(2gh)

Question bank breakdown

Multiple choice

43

Free response

3

Difficulty mix

11 easy · 26 medium · 6 hard

Skills covered

Pressure and Pascal's PrincipleBuoyancy and Archimedes' PrincipleFluid Continuity EquationBernoulli's EquationApplications of Fluid DynamicsDensityPressure and DepthPascal's Principle

Every question in the bank comes with a worked explanation, and most add a per-choice breakdown so you learn why each wrong answer is wrong - not just the key.

Sample question

From the free tier of the ScoreMint AP Physics 1 bank - try it, then check the answer.

A swimming pool is filled with water (ρ=1000 kg/m3\rho = 1000 \text{ kg/m}^3). What is the gauge pressure at a depth of 3.0 m3.0 \text{ m} below the surface? Use g=10 m/s2g = 10 \text{ m/s}^2.

  • A. 3,000 Pa3,000 \text{ Pa}
  • B. 30,000 Pa30,000 \text{ Pa}
  • C. 131,000 Pa131,000 \text{ Pa}
  • D. 300 Pa300 \text{ Pa}
Show answer and explanation

Answer: B. Gauge pressure at depth hh in a fluid of density ρ\rho is Pgauge=ρghP_{gauge} = \rho g h. This represents the pressure due to the fluid column above that point, not including atmospheric pressure. Pgauge=(1000)(10)(3.0)=30,000P_{gauge} = (1000)(10)(3.0) = 30{,}000 Pa. The absolute pressure at this depth would be Pabs=Patm+ρgh101,000+30,000=131,000P_{abs} = P_{atm} + \rho g h \approx 101{,}000 + 30{,}000 = 131{,}000 Pa.

Study tip

The buoyant force uses the fluid's density and the displaced volume - never the object's density. A floating object displaces its weight in fluid; a sunken one displaces its volume.