Fluids
Unit 8 of AP Physics 1, worth 10–15% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Internal structure and density, pressure, fluids and Newton's laws, buoyancy, conservation of mass flow rate, conservation of energy in fluid flow (Bernoulli).
How this unit is tested
Start every fluids problem by asking whether the fluid is at rest or moving. Static problems (pressure at a depth, buoyancy, hydraulic lifts) use the pressure-depth relation and Archimedes' principle. Dynamic problems (fluid flowing through pipes, holes, or changing pipe diameters) use the continuity equation and Bernoulli's equation together.
For static problems, sketch the fluid and mark the depth of interest. Remember pressure depends only on depth, fluid density, and what's pushing down on the surface — not on the shape or total volume of the container. For buoyancy, always compare the object's density to the fluid's density before computing forces: this tells you immediately whether the object floats (partially submerged, buoyant force equals weight) or sinks (fully submerged, buoyant force less than weight).
For dynamic problems, apply the continuity equation first to relate the fluid speeds at the two locations you care about, since Bernoulli's equation has two unknowns (pressure and speed) unless you already know how the speeds compare. Then write Bernoulli's equation between the same two points on the same streamline, being careful to measure both heights from the same reference level.
Keep every quantity in SI units (kg, m, m³, Pa) before plugging into a formula — mixing cm² and m² is the single most common source of wrong answers on this unit. Finally, expect fluids to combine with earlier units: a hole draining from a tank is a Bernoulli problem followed by a projectile-motion problem, and a floating object is a buoyancy problem followed by a Newton's-second-law equilibrium problem.
What you have to know
Density
Density is mass per unit volume: $\rho = m/V$, with SI units of kg/m³.
Hydrostatic (pressure–depth) relation
In a static fluid, absolute pressure increases linearly with depth: $P = P_0 + \rho g h$, where $P_0$ is the pressure at the surface and $h$ is the depth below it. This follows from applying Newton's second law (net force = 0) to a thin horizontal slab of fluid in equilibrium.
Pascal's principle
A pressure applied to an enclosed, incompressible fluid is transmitted undiminished to every point in the fluid, so in a hydraulic system $F_1/A_1 = F_2/A_2$.
Archimedes' principle
The buoyant force on an object equals the weight of the fluid it displaces: $F_b = \rho_{fluid} V_{displaced} g$.
Continuity equation
For an incompressible fluid, the volume flow rate is the same at every cross-section of a pipe: $A_1 v_1 = A_2 v_2$.
Bernoulli's equation
Along a streamline in steady, incompressible, non-viscous flow, mechanical energy per unit volume is conserved: $P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$.
14 practice questions
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A metal cube has edge length 2.0 cm and mass 21.6 g. Calculate its density in kg/m³ and state whether it is more likely aluminum (2700 kg/m³) or gold (19,300 kg/m³).
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Answer. About 2700 kg/m³, so aluminum.
Convert to SI units: V = (0.020 m)³ = 8.0×10⁻⁶ m³ and m = 0.0216 kg. Then ρ = m/V = 0.0216/8.0×10⁻⁶ = 2700 kg/m³, matching aluminum rather than gold. -
A 60 kg person stands with both feet on the floor, and each foot contacts an area of 150 cm². What pressure does the person exert on the floor? (g = 9.8 m/s²)
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Answer. About 2.0×10⁴ Pa (19,600 Pa).
Total contact area is 2×150 cm² = 300 cm² = 0.030 m². Weight is mg = 60×9.8 = 588 N. Pressure is P = F/A = 588/0.030 ≈ 19,600 Pa. -
Three containers with different shapes (narrow, wide, and irregular) are each filled with water to the same height. How do the pressures at the bottom of the three containers compare?
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Answer. All three pressures are equal.
Hydrostatic pressure depends only on depth, fluid density, and the pressure at the surface, P = P0 + ρgh, not on the container's shape or the total weight of water it holds. Since all three have the same height of water, the bottom pressure is identical in each. -
Find the absolute pressure at a depth of 12 m in a freshwater lake. (ρ_water = 1000 kg/m³, atmospheric pressure = 1.01×10⁵ Pa, g = 9.8 m/s²)
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Answer. About 2.19×10⁵ Pa.
Use P = P0 + ρgh = 1.01×10⁵ + (1000)(9.8)(12) = 1.01×10⁵ + 117,600 ≈ 2.19×10⁵ Pa. The atmospheric term must be included since the question asks for absolute, not gauge, pressure. -
A hydraulic lift has an input piston of area 5.0 cm² and an output piston of area 250 cm². If a force of 200 N is applied to the input piston, what output force is produced, and does this violate conservation of energy?
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Answer. 10,000 N; no violation of energy conservation.
By Pascal's principle, F1/A1 = F2/A2, so F2 = F1(A2/A1) = 200×(250/5) = 10,000 N. Energy is still conserved because the output piston moves a proportionally shorter distance than the input piston, so work in equals work out. -
Using Newton's second law applied to a thin horizontal slab of fluid within a static tank, explain why pressure must increase with depth.
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Answer. Because the slab is in equilibrium, the upward pressure force on its bottom face must exceed the downward pressure force on its top face by exactly the weight of the slab, so pressure at the bottom is greater than at the top.
For a fluid slab at rest, the net force is zero: (P_bottom)(A) − (P_top)(A) − mg = 0. Solving gives P_bottom − P_top = ρgΔh, which is positive, showing pressure must increase with depth in any static fluid. -
A solid ball of volume 2.0×10⁻³ m³ is fully submerged in water (ρ = 1000 kg/m³, g = 9.8 m/s²). Find the buoyant force on the ball.
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Answer. 19.6 N.
By Archimedes' principle, F_b = ρ_fluid V_displaced g = (1000)(2.0×10⁻³)(9.8) = 19.6 N. Since the ball is fully submerged, the displaced volume equals the ball's entire volume. -
An object weighs 50 N in air and 30 N when fully submerged in water (ρ_water = 1000 kg/m³, g = 9.8 m/s²). Find the buoyant force and the object's volume.
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Answer. Buoyant force = 20 N; volume ≈ 2.0×10⁻³ m³.
The apparent weight loss equals the buoyant force: F_b = 50 − 30 = 20 N. Since F_b = ρ_water V g, the volume is V = F_b/(ρg) = 20/(1000×9.8) ≈ 2.04×10⁻³ m³. -
A block of wood with density 600 kg/m³ floats in water (density 1000 kg/m³). What fraction of the block's volume is submerged?
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Answer. 60% of the block is submerged.
For a floating object, the buoyant force equals its weight: ρ_water V_sub g = ρ_object V_total g. Solving gives V_sub/V_total = ρ_object/ρ_water = 600/1000 = 0.60, or 60%. -
A steel ship floats on water even though steel is much denser than water. Which statement best explains why?
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Answer. The ship's hull shape displaces enough water that the weight of the displaced water equals the ship's total weight.
Buoyant force depends on the volume of fluid displaced, not on the material's density alone. Because the hollow hull displaces a large volume of water, the weight of that displaced water can equal the ship's weight even though solid steel itself is denser than water. -
Water flows through a horizontal pipe that narrows from a cross-sectional area of 8.0 cm² to 2.0 cm². If the speed in the wide section is 1.5 m/s, what is the speed in the narrow section?
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Answer. 6.0 m/s.
By the continuity equation, A1v1 = A2v2, so v2 = A1v1/A2 = (8.0×1.5)/2.0 = 6.0 m/s. The speed increases because the pipe narrows and volume flow rate must stay constant. -
A hose delivers water at a volume flow rate of 3.0×10⁻⁴ m³/s through a nozzle opening of area 5.0×10⁻⁵ m². What is the exit speed of the water?
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Answer. 6.0 m/s.
Volume flow rate is Q = Av, so v = Q/A = (3.0×10⁻⁴)/(5.0×10⁻⁵) = 6.0 m/s. -
Water flows horizontally through a pipe, speeding up from 2.0 m/s in a wide section to 8.0 m/s in a narrow section. If the pressure in the wide section is 1.5×10⁵ Pa, what is the pressure in the narrow section? (ρ = 1000 kg/m³)
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Answer. 1.2×10⁵ Pa.
Since the pipe is horizontal, the height terms in Bernoulli's equation cancel: P1 + ½ρv1² = P2 + ½ρv2². Solving, P2 = P1 + ½ρ(v1² − v2²) = 1.5×10⁵ + 0.5(1000)(4 − 64) = 1.5×10⁵ − 30,000 = 1.2×10⁵ Pa. Pressure drops where speed increases, as expected from the Venturi effect. -
Air moves faster over the curved top surface of an airplane wing than underneath it. According to Bernoulli's principle, what does this cause?
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Answer. Lower pressure above the wing than below it, producing a net upward force on the wing.
Bernoulli's equation shows that where fluid speed is higher, pressure is lower, provided height differences are small. Because air moves faster over the top of the wing, the pressure there is lower than beneath the wing, creating a net upward (lift) force.
What people get wrong
- Forgetting the atmospheric pressure term when a problem asks for absolute pressure at a depth. Instead, always check whether the question wants gauge pressure (just $\rho g h$) or absolute pressure (must add $P_0$).
- Using mass instead of weight in buoyancy problems, e.g. computing $F_b=\rho V$ and forgetting the factor of $g$. Instead, always write $F_b=\rho_{fluid}V_{displaced}g$ and check units come out in newtons.
- Assuming buoyant force depends on the object's own density or weight. Instead, remember buoyant force depends only on the fluid's density and the volume displaced — the object's density only determines whether it floats or sinks.
- Mixing up which area is A1 and which is v1 in the continuity equation, leading to an inverted answer (getting a slower speed in the narrower section). Instead, remember flow speeds up where the pipe narrows: smaller area always pairs with larger speed.
- Applying Bernoulli's equation between two points that aren't on the same streamline, or forgetting the $\rho g h$ term when the two points are at different heights. Instead, write out all three terms at each point before canceling anything, so you don't accidentally drop a height difference.
- Not converting cm² or cm³ to SI units before plugging into density, pressure, or flow-rate formulas. Instead, convert every area to m² and volume to m³ first, since pascals and kg/m³ require SI units throughout.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Physics 1. Free, and no account needed to start.