Fluids · quiz worked solutions

Tried the quiz first? Good. Here’s how to work through each answer. Follow the link beside a question to jump back to its original spot. Questions marked Extra exploration go beyond this AP Physics 1 course; Book controls questions explain the interface.

Bernoulli’s Equation · question 1

Back to the question · AP Topic 8.4

Question: A fluid flows through the inlet of a pipe with diameter 1 m and velocity 4 m/s . The fluid exits at the outlet of the pipe which has a diameter of 1.5 m . What is the velocity at this point?

Answer: 1.78 m/s.

1. Use A₁v₁ = A₂v₂ for steady incompressible filled-pipe flow. Why: The same volume per second crosses both sections.

2. Since area scales as diameter squared, v₂ = 4(1/1.5)² = 1.7778 m/s. Why: Using the unsquared diameter ratio would overestimate the outlet speed.

Bernoulli’s Equation · question 2

Back to the question · AP Topic 8.4

Question: Given that the final velocity in a pipe system is 12 m/s , what must be the ratio of A 2 : A 1 in order to achieve an initial velocity of 3 m/s ?

Answer: 1:4.

1. Write A₂/A₁ = v₁/v₂ = 3/12 = 1/4. Why: Continuity links areas and mean speeds inversely.

2. Check that the final section has one quarter of the initial area. Why: Its speed is four times the initial speed while volume flow rate stays constant.

Bernoulli’s Equation · question 3

Back to the question · AP Topic 8.4

Question: In steady incompressible lossless flow along a horizontal streamline, if fluid speed increases, its static pressure must:

Answer: Static pressure decreases.

1. Use P + ½ρv² = constant along the horizontal ideal streamline. Why: The elevation term is unchanged and no energy is lost.

2. If v increases, the kinetic-energy-density term increases, so P must decrease. Why: The conclusion depends on the stated ideal assumptions, not on all possible real pipe systems.

Bernoulli’s Equation · question 4

Back to the question · AP Topic 8.4

Question: For steady incompressible flow through the pictured sections, rank the mean speeds from greatest to least (I–IV).

Answer: Speeds: D > C > B > A; targets I–IV match that order.

1. Read the widths as decreasing from A to D. Why: The same flow rate passes every filled section.

2. Apply v = Q/A: smallest area D has largest speed, followed by C, B and A. Why: The rank labels refer to greatest through least, not the alphabetical pipe labels.

Bernoulli’s Equation · question 5

Back to the question · AP Topic 8.4

Question: Assume the pictured pipe is horizontal and the flow is steady, incompressible, and lossless. Rank the static pressures from greatest to least (I–IV).

Answer: Pressures: A > B > C > D; targets I–IV match that order.

1. From continuity, speed rises as area decreases from A to D. Why: The horizontal lossless-flow conditions permit Bernoulli comparison.

2. Use P = constant − ½ρv², so the widest, slowest part has the highest pressure. Why: Pressure and speed rank oppositely only under the assumptions supplied here.

Buoyancy · question 1

Back to the question · AP Topic 8.3

Question: A cube with side length 2 m and mass 50 kg floats with a horizontal face in water of density 1000 kg/m³. What is its submerged depth?

Answer: 12.5 mm.

1. For floating equilibrium, ρ_waterV_sub g = mg, so V_sub = 50/1000 = 0.050 m³. Why: Buoyancy balances weight, not the object’s total volume times water density.

2. Divide by the horizontal face area 2² = 4 m² to get depth 0.0125 m = 12.5 mm. Why: Only a small fraction of this large, low-density cube is submerged.

Buoyancy · question 2

Back to the question · AP Topic 8.3

Question: A floating boat in a small pond releases a dense anchor, which sinks to the bottom and is no longer supported by the boat. Does the pond water level rise or fall?

Answer: The water level falls.

1. On the boat, the anchor’s added weight requires displaced volume m_anchor/ρ_water. Why: The boat supports the anchor through buoyancy.

2. On the bottom, the dense anchor displaces only its physical volume m_anchor/ρ_anchor, which is smaller. Why: Less water displacement means a lower pond level; this assumes the sunk anchor is no longer pulling down on the boat.

Buoyancy · question 3

Back to the question · AP Topic 8.3

Question: A cylinder is 100 mm long, 250 mm in radius, and has mass 100 kg. It is released upright in water. Will a hole halfway up its side become submerged?

Answer: Yes; the cylinder sinks, so the hole becomes submerged.

1. Find full volume π(0.250)²(0.100) ≈ 0.019635 m³. Why: All given dimensions must use metres.

2. Even completely submerged it displaces only about 19.635 kg of water, much less than its 100 kg mass. Why: The maximum buoyant force is smaller than its weight; this is sinking rather than a half-submerged floating equilibrium.

Buoyancy · question 4

Back to the question · AP Topic 8.3

Question: An ice cube with 5 mm sides is floating in water. If ice has a density of 0.92 g/cm 3 , how far below the water’s surface will the bottom of the ice cube be?

Answer: 4.6 mm.

1. For floating ice, V_sub/V_total = ρ_ice/ρ_water = 0.92. Why: Equating buoyancy and weight cancels g.

2. For the upright cube, depth = 0.92(5 mm) = 4.6 mm. Why: The remaining 0.4 mm sits above the water surface.

Buoyancy · question 5

Back to the question · AP Topic 8.3

Question: A rectangular box 1 m long with a 0.2 m square end floats with its long axis horizontal in water of density 1000 kg/m³. Find its mass if the submerged depth is 0.02 m.

Answer: 4 kg.

1. Use submerged volume = (1 m)(0.2 m)(0.02 m) = 0.004 m³. Why: The long axis is horizontal and the submerged depth is smaller than the 0.2 m side.

2. Floating equilibrium gives m = ρ_waterV_sub = (1000)(0.004) = 4 kg. Why: The full box volume is not the displaced volume.

Ideal Gas Law · question 1

Back to the question · Extra exploration

Question: For this problem, use the reference conditions 0 °C and 1 atm. What is the ideal-gas volume of one mole at those conditions?

Answer: 22.414 L.

1. Use PV = nRT with T = 273.15 K and P = 1 atm. Why: The problem specifies the reference conditions rather than leaving STP convention ambiguous.

2. Find V = (1)(0.0820574)(273.15)/1 ≈ 22.41397 L. Why: The gas constant here uses litres and atmospheres.

Ideal Gas Law · question 2

Back to the question · Extra exploration

Question: If a gas mixture is made up of 4 L of Nitrogen Gas, and 2 L of Oxygen gas, and the gas is at 20 ∘ C and 1 atm , what is the mass of the gas mixture?

Answer: 7.32 g.

1. At the same P and T, find n_N₂ = 4/(0.0820574·293.15) and n_O₂ = 2/(0.0820574·293.15). Why: The separate supplied volumes are proportional to each gas’s amount.

2. Use molar masses 28 g/mol and 32 g/mol: m = [4(28) + 2(32)]/(0.0820574·293.15) ≈ 7.317 g. Why: Add masses, not molar masses without amount weighting.

Ideal Gas Law · question 3

Back to the question · Extra exploration

Question: How many moles are inside a Helium gas container, if the container has a volume of 5 L , is kept at 25 ∘ C , and is pressurized to 20 atm ?

Answer: 4.09 mol.

1. Convert 25°C to 298.15 K. Why: Ideal-gas temperature must be absolute.

2. Find n = PV/(RT) = (20)(5)/(0.0820574·298.15) ≈ 4.0874 mol. Why: The given pressure must be treated as absolute in this model.

Ideal Gas Law · question 4

Back to the question · Extra exploration

Question: If you have a balloon filled with 0.04 mol of air,what will its volume be if you bring it to the top of Mt. Everest, where the pressure is 0.341 atm , and the temperature is - 20 ∘ C ?

Answer: 2.44 L.

1. Convert −20°C to 253.15 K. Why: Negative Celsius temperature is still positive absolute temperature.

2. Use V = nRT/P = (0.04)(0.0820574)(253.15)/0.341 ≈ 2.4367 L. Why: The lower ambient pressure permits a larger ideal balloon volume.

Ideal Gas Law · question 5

Back to the question · Extra exploration

Question: What is the ideal-gas density of air at 0 °C and 1 atm, given a molar mass of 29 g/mol?

Answer: 1.29 g/L.

1. One ideal mole occupies RT/P ≈ 22.414 L at 273.15 K and 1 atm. Why: The reference conditions match the earlier molar-volume calculation.

2. Divide molar mass by molar volume: 29/22.414 ≈ 1.2939 g/L. Why: Density is mass per volume, not the molar mass alone.

Ideal Gas Law · question 1

Back to the question · Extra exploration

Question: If a balloon has a volume of 4 L at 1 atm , and the balloon is taken to the bottom of a pool, where the pressure is 1.3 atm , what will its new volume be? Assume no change in temperature.

Answer: 3.08 L.

1. With constant amount and temperature, P₁V₁ = P₂V₂. Why: This is Boyle’s law using absolute pressures.

2. Find V₂ = (1)(4)/1.3 ≈ 3.0769 L. Why: The larger external pressure compresses the ideal balloon.

Ideal Gas Law · question 2

Back to the question · Extra exploration

Question: A hot air balloon contains 10 m 3 of air at 80 ∘ C . If the balloon rises from sea level to a height where the pressure is 0.7 atm , what is the new volume of the balloon? Assume the fire keeps the temperature constant.

Answer: 14.3 m³.

1. Use P₁V₁ = P₂V₂ with temperature held constant. Why: The simplified problem also treats the amount of gas as fixed.

2. Find V₂ = (1)(10)/0.7 ≈ 14.286 m³. Why: A real open hot-air balloon exchanges gas, so this fixed-amount model is the stipulated classroom approximation.

Ideal Gas Law · question 3

Back to the question · Extra exploration

Question: A 10 L helium tank is initially at −20 °C and absolute pressure 50 atm. If all of its gas is transferred, how many full 0.4 L balloons can it supply at 1 atm and 20 °C? Assume ideal gas behavior and disregard gas left in the tank.

Answer: 1447 full balloons.

1. Convert initial and final temperatures to 253.15 K and 293.15 K. Why: The ideal-gas amount is unchanged during the complete transfer.

2. Find total balloon-state volume V₂ = (50)(10)(293.15)/(1·253.15) ≈ 579.005 L. Why: Pressure and temperature both affect the volume available.

3. Divide by 0.4 L and take the floor: 1447 full balloons. Why: A fractional remaining balloon does not count as full, and the prompt excludes gas left in the tank.

Ideal Gas Law · question 4

Back to the question · Extra exploration

Question: What pressure will gas have if 10 L of gas at STP are forced into a 2 L tank at 50 ∘ C ?

Answer: 5.92 atm.

1. Use reference T₁ = 273.15 K and T₂ = 323.15 K. Why: The retained section uses 0°C and 1 atm for this STP calculation.

2. Find P₂ = P₁(V₁/V₂)(T₂/T₁) = 1(10/2)(323.15/273.15) ≈ 5.915 atm. Why: Compressing and heating both increase the absolute pressure.

Ideal Gas Law · question 5

Back to the question · Extra exploration

Question: A child blows up a balloon to 0.25 L and ties it to his mailbox, before running inside. About an hour later, a hurricane comes through, dropping the pressure from 1 atm to 0.888 atm , and cooling of the day from 30 ∘ C to 25 ∘ C . What is the new volume of the balloon, assuming it remains steadfastly tied to the mailbox?

Answer: 0.277 L.

1. Convert T₁ = 303.15 K and T₂ = 298.15 K. Why: The tied balloon’s amount is fixed in this ideal model.

2. Find V₂ = 0.25(1/0.888)(298.15/303.15) ≈ 0.2769 L. Why: Cooling slightly shrinks it, but the larger effect of reduced pressure makes it expand overall.

FFR - More on Fluids · question 1

Back to the question · Extra exploration

Question: What do the non-ideal gas laws take into account that the ideal gas laws did not?

Answer: Molecular volume and intermolecular forces: A and C.

1. The ideal model treats particles as point-like and neglects interactions between collisions. Why: Those are its relevant simplifying assumptions.

2. The nonideal correction accounts for finite occupied volume and intermolecular attraction or repulsion. Why: Ordinary collisions already appear in the ideal-gas picture, so all of the above is too broad.

FFR - More on Fluids · question 2

Back to the question · Extra exploration

Question: What types of head loss occur in a pipe system?

Answer: Both frictional and minor losses.

1. Distinguish wall friction distributed along a pipe from local disturbances at fittings. Why: They produce two kinds of irreversible mechanical-energy loss.

2. Select both categories. Why: Minor means localized, not necessarily small in magnitude.

FFR - More on Fluids · question 3

Back to the question · Extra exploration

Question: Which pictured route contains the most approximately right-angle bends? Use this as a qualitative illustration of local-loss sources; the images alone do not determine numerical head loss.

Answer: Option 2 has the most approximately right-angle bends.

1. Count bends in each pictured route. Why: The question asks for a visual count rather than a quantitative hydraulic calculation.

2. Choose the route with the greatest number of those bends. Why: Head loss also depends on flow, size and fittings, so the drawings alone cannot rank actual numerical losses.

FFR - More on Fluids · question 4

Back to the question · Extra exploration

Question: Indicate on the image where friction and minor losses occur in both Sections 1 & 2.

Answer: Targets 1–4: friction 1, minor 1, minor 2, friction 2.

1. Assign distributed friction to the lengths of straight pipe in each section. Why: Wall shear acts along the wetted pipe.

2. Assign minor losses to the depicted bends and fittings. Why: Local geometry changes disturb flow; these are additional sources to the distributed wall losses.

FFR - More on Fluids · question 5

Back to the question · Extra exploration

Question: Using the van der Waals model, find the temperature in °C of 8 mol of ozone in a 30 L tank at absolute pressure 20 atm. Use a = 3.520 atm·L²/mol², b = 0.049 L/mol, and R = 0.0820574 L·atm/(mol·K).

Answer: About 640°C.

1. Use T = (P + a(n/V)²) × (V − nb)/(nR) in matching litre–atmosphere units. Why: The van der Waals equation corrects pressure and usable volume.

2. Insert n = 8, V = 30, P = 20, a = 3.520, b = 0.049 and R = 0.0820574 to get about 913.36 K. Why: The nb correction is 0.392 L and the attraction correction is about 0.2503 atm.

3. Subtract 273.15 to get about 640.2°C. Why: Only the final absolute temperature is converted back to Celsius.

Model references

These original worked steps use the stated question models. Physics references: Buoyancy, Continuity, Ideal gases. AP topic labels follow the supplied-edition College Board CED, Effective Fall 2024, ©2026, pinned in the course audit.