Oil flows steadily and incompressibly through a pipe that widens from area to . If the speed in the narrow section is , what is the speed in the wide section?
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AP Physics 1 Quiz
Practice Fluids And Conservation Laws in AP Physics 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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Oil flows steadily and incompressibly through a pipe that widens from area A to 3A. If the speed in the narrow section is 6 m/s, what is the speed in the wide section?
This quiz focuses on Fluids And Conservation Laws, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 1.
Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.
Oil flows steadily and incompressibly through a pipe that widens from area A to 3A. If the speed in the narrow section is 6 m/s, what is the speed in the wide section?
Explanation: This question assesses understanding of the continuity equation in incompressible flow, based on mass conservation. For incompressible fluids, the volume flow rate remains constant because density doesn't change, so A v is the same everywhere. When the pipe widens, the fluid slows down to keep the flow rate steady. This ensures the same amount of fluid volume passes through wider and narrower sections in the same time. Choice A, 18 m/s, could be a distractor if someone multiplies instead of dividing the area ratio. A useful strategy for these problems is to identify the ratio of areas and inversely apply it to the velocities while conserving A v.
Water flows steadily and incompressibly through a pipe with A1=3 cm2 and v1=4 m/s. If A2=6 cm2, what is v2?
Explanation: This question examines the continuity equation for incompressible fluids, stemming from the conservation of mass. For incompressible flow, density is uniform, so mass conservation requires the volumetric flow rate to be constant throughout the pipe. When the cross-sectional area increases, the velocity decreases proportionally to maintain this flow rate. This qualitative understanding helps predict that doubling the area halves the speed. Choice A, 8 m/s, might be selected if someone doubles the speed instead of halving it due to confusion with area ratios. To tackle similar questions, consistently use A1 v1 = A2 v2 and verify units for consistency.
A liquid flows steadily and is incompressible through a pipe that expands from area A to area 4A. The speed in the smaller section is 12 m/s. For steady, incompressible flow, what is the speed in the larger section?
Explanation: This problem applies the continuity equation to find speed in an expanding pipe section. For incompressible flow, A₁v₁ = A₂v₂ ensures constant volume flow rate. The pipe expands from area A to area 4A, with initial speed 12 m/s. Applying continuity: (A)(12 m/s) = (4A)(v₂), which gives 12A m/s = 4Av₂, so v₂ = 3 m/s. Choice A (48 m/s) incorrectly multiplies instead of dividing, misunderstanding that larger areas require slower speeds. Remember that area and speed are inversely proportional in incompressible flow: when area quadruples, speed becomes one-fourth.
A fluid flows steadily and is incompressible through a pipe. At section 1, A1=A and v1=5 m/s. At section 2, the speed is v2=1 m/s. For steady, incompressible flow, what is A2?
Explanation: This problem requires finding the cross-sectional area when speeds are known at two pipe sections. The continuity equation A₁v₁ = A₂v₂ applies for incompressible flow. Given A₁ = A, v₁ = 5 m/s, and v₂ = 1 m/s, we solve for A₂. Substituting: (A)(5 m/s) = (A₂)(1 m/s), which gives 5A m/s = A₂ m/s, so A₂ = 5A. Choice A (A/5) incorrectly inverts the relationship, assuming area decreases when speed decreases. For incompressible flow, remember that area and speed are inversely related: when speed decreases by a factor of 5, area must increase by the same factor.
Water flows steadily and is incompressible through a pipe. At section 1, the radius is r. At section 2, the radius is 2r. The speed at section 1 is v. For steady, incompressible flow, what is the speed at section 2?
Explanation: This problem involves the continuity equation with circular pipe cross-sections of different radii. For incompressible flow, A₁v₁ = A₂v₂, where area A = πr² for circular pipes. At section 1, radius is r and speed is v; at section 2, radius is 2r. The areas are A₁ = πr² and A₂ = π(2r)² = 4πr². Applying continuity: (πr²)(v) = (4πr²)(v₂), which simplifies to v = 4v₂, giving v₂ = v/4. Choice C (2v) incorrectly assumes speed doubles when radius doubles, ignoring that area depends on radius squared. When radius doubles, area quadruples, so speed becomes one-fourth to maintain constant flow rate.
Oil flows steadily and is incompressible through a pipe that narrows from area 3A to area A. The speed in the wider section is 2 m/s. For steady, incompressible flow, what is the speed in the narrow section?
Explanation: This problem applies the continuity equation for incompressible fluid flow through a narrowing pipe. For steady, incompressible flow, the product of cross-sectional area and speed remains constant: A₁v₁ = A₂v₂. The pipe narrows from area 3A to area A, and the initial speed is 2 m/s. Applying continuity: (3A)(2 m/s) = (A)(v₂), which gives 6A m/s = Av₂, so v₂ = 6 m/s. Choice A (⅔ m/s) incorrectly divides instead of multiplying, misunderstanding the inverse relationship. To solve continuity problems, set up the equation A₁v₁ = A₂v₂ and solve for the unknown quantity.
Water flows steadily through a horizontal pipe and can be treated as incompressible. At section 1 the pipe radius is 2r and the average speed is v. Farther downstream at section 2 the pipe radius is r. The flow is steady, so the volume flow rate is conserved. What is the average speed of the water at section 2?
Explanation: This question assesses the application of the continuity equation in fluid dynamics for incompressible fluids. For incompressible fluids in steady flow, the mass flow rate is conserved, which means the volume flow rate is the same at every cross-section since density is constant. The volume flow rate is given by Q = A v, where A is the cross-sectional area and v is the average speed. Therefore, A1 v1 = A2 v2; here, A1 = π (2r)^2 = 4 π r², A2 = π r², so v2 = (A1 / A2) v1 = 4 v. A common distractor is B, 2v, which might result from incorrectly using the radius ratio instead of the area ratio, since radius halves, but area quarters. To solve similar problems, always remember to use the continuity equation Q1 = Q2 and calculate areas properly from given dimensions.
A steady, incompressible stream of water flows through a pipe of area A. The pipe then splits into two identical branches, each of area 2A. What is the speed in each branch compared to the original speed v?
Explanation: This question explores the continuity equation in branching pipes for incompressible flow, based on mass conservation. In steady flow, the mass flow rate into the split equals the sum out, and with constant density, volumetric rates conserve similarly. When splitting into identical branches, each carries half the flow, but speeds depend on areas. Qualitatively, same area halves would maintain speed if flow halves per branch. Choice C, 2v, could be chosen if someone assumes speed doubles when area halves without considering the split. A key strategy is to calculate total flow rate and divide appropriately among branches, then apply A v per branch.
In steady, incompressible flow, water moves through a pipe. At point 1, A1=2A and v1=3v. If point 2 has area A2=A, what is v2?
Explanation: This question evaluates the application of the continuity equation for steady, incompressible flow, rooted in mass conservation. Conservation of mass implies that the mass entering a section equals the mass leaving, and with constant density, this translates to equal volumetric flow rates. Thus, A1 v1 = A2 v2 holds, allowing us to relate speeds and areas at different points. Here, the area halves while the initial speed is given, leading to a proportional increase in speed at the narrower point. Choice D, v/6, might be chosen if someone inverts the velocity ratio incorrectly. Remember, for any incompressible flow problem, compute the flow rate at one point and set it equal at the other to find unknowns.
Air is treated as incompressible and flows steadily in a duct that narrows from 5A to A. If the speed in the narrow section is 10 m/s, what is the speed in the wide section?
Explanation: This question probes the continuity principle in incompressible flow, derived from mass conservation laws. In such flows, since the fluid can't be compressed, the volume flow rate must be identical across sections to conserve mass. This means that in a wider section, the fluid moves slower to match the flow in a narrower, faster section. The relationship A v = constant captures this qualitatively, with velocity inversely proportional to area. A distractor like choice C, 50 m/s, could result from multiplying areas instead of dividing properly. A transferable approach is to always calculate the flow rate using known values and equate it for the unknown section.
Water flows steadily through a horizontal pipe and is incompressible. At section 1 the cross-sectional area is 2A and the speed is v. At section 2 the area is A. For this steady, incompressible flow, what is the speed at section 2?
Explanation: This problem tests the continuity equation for incompressible fluids flowing through pipes with varying cross-sectional areas. For steady, incompressible flow, the volume flow rate must remain constant throughout the pipe, meaning A₁v₁ = A₂v₂. At section 1, we have area 2A and speed v, while at section 2, the area is A. Substituting into the continuity equation: (2A)(v) = (A)(v₂), which simplifies to 2Av = Av₂, giving v₂ = 2v. Choice C (½v) incorrectly assumes speed decreases when area decreases, reversing the relationship. When solving continuity problems, remember that speed and area are inversely proportional for incompressible flow.
Water flows steadily and is incompressible through a pipe with two sections. At section 1, A1=8 cm2 and v1=10 cm/s. At section 2, v2=40 cm/s. For steady, incompressible flow, what is A2?
Explanation: This problem tests the continuity equation with numerical values for areas and speeds. For incompressible flow, the volume flow rate A₁v₁ equals A₂v₂ at all pipe sections. Given A₁ = 8 cm², v₁ = 10 cm/s, and v₂ = 40 cm/s, we find A₂. Applying continuity: (8 cm²)(10 cm/s) = (A₂)(40 cm/s), which gives 80 cm³/s = 40A₂ cm/s, so A₂ = 2 cm². Choice B (32 cm²) incorrectly multiplies areas instead of recognizing the inverse relationship with speed. When speed quadruples in incompressible flow, area must become one-fourth to conserve mass flow rate.
Incompressible water flows steadily through a pipe. At section 1 the diameter is D and the average speed is 8 m/s. At section 2 the diameter is 2D. Assuming steady flow, what is the average speed at section 2?
Explanation: This question assesses conservation of volume flow rate using pipe diameters. For incompressible steady flow, mass conservation implies A1 v1 = A2 v2. Areas are proportional to diameter squared, so A2 / A1 = (2D/D)^2 = 4. Thus v2 = v1 (A1 / A2) = 8 / 4 = 2 m/s. Distractor A 16 m/s might come from incorrectly doubling the speed instead of quartering it. Always compute the area ratio using (d2/d1)^2 when diameters are given.
In steady incompressible flow, a pipe splits into branches X and Y. Branch X has flow rate 2 L/s and branch Y has 5 L/s. What is the flow rate in the main pipe before the split?
Explanation: This question assesses mass conservation in splitting pipes for steady incompressible flow, requiring total flow rate equality. Since the fluid is incompressible, the volume flow rate before the split must equal the sum after, preventing mass buildup. This conservation law applies at junctions, adding branch rates to get the main. Qualitatively, higher branch flows imply a proportionally larger main flow. Choice C, 10 L/s, might be a distractor if someone multiplies instead of adding the branches. For similar problems, always add or subtract flow rates at junctions while remembering total conservation.
Water flows steadily and incompressibly through a horizontal pipe that narrows from cross-sectional area 4A to A. If the speed in the wide section is v, what is the speed in the narrow section?
Explanation: This question tests the continuity equation for incompressible fluid flow, which arises from the conservation of mass. In incompressible flow, the fluid's density remains constant, ensuring that the mass flow rate is the same at every point in the pipe. Since mass flow rate equals density times volumetric flow rate, and density is constant, the volumetric flow rate A v must be conserved. Qualitatively, when the pipe narrows, the fluid speeds up to maintain the same volume passing through per unit time. A common distractor is choice D, v/4, which might be selected if someone incorrectly assumes velocity is proportional to area instead of inversely proportional. To solve similar problems, always apply the continuity equation A1 v1 = A2 v2 and solve for the unknown velocity.
A steady, incompressible fluid flows through a pipe that splits into two branches. The main pipe carries volumetric flow rate Q. One branch carries 41Q. What flow rate must the other branch carry?
Explanation: This question tests conservation of mass in branching pipes for incompressible flow, where total flow rate is preserved. In incompressible fluids, the volumetric flow rate into a junction equals the sum of flow rates out, as mass can't accumulate in steady flow. This means the main pipe's flow splits among branches without loss. Qualitatively, if one branch takes a fraction, the other takes the remainder to conserve the total. Choice A, 1/4 Q, is a distractor possibly chosen if someone subtracts incorrectly or confuses addition. A general strategy is to sum the branch flow rates to find the main or vice versa, ensuring conservation.
Water flows steadily and incompressibly through a pipe. At point 1, A1=A and v1=v. At point 2, A2=3A. What is v2?
Explanation: This question evaluates the continuity equation for incompressible flow, grounded in conservation of mass principles. For incompressible fluids, constant density means volumetric flow rate A v is conserved along the pipe. A decrease in area requires an increase in velocity to keep the flow rate steady. This inverse relationship ensures the same volume passes per second regardless of section. Choice A, v/3, is a distractor likely picked by inverting the area ratio incorrectly. Always solve by setting A1 v1 = A2 v2 and isolating the unknown, checking if the result makes physical sense.
Incompressible water flows steadily through a pipe that splits into two branches. The incoming volume flow rate is 9 L/s. One branch carries 4 L/s. Assuming steady flow, what volume flow rate must the other branch carry?
Explanation: This question focuses on conservation of volume flow rate in branching pipes for incompressible fluids. When a pipe splits, the total volume flow rate into the junction equals the sum of flow rates out of the branches. Since the fluid is incompressible and flow is steady, no accumulation occurs, so Qin = Qbranch1 + Qbranch2. Here, 9 L/s = 4 L/s + Q2, so Q2 = 5 L/s. A distractor like C, 13 L/s, could be from adding instead of subtracting. Always apply the continuity principle at junctions by summing the outgoing flows to match incoming.
Incompressible water flows steadily through a pipe that narrows from cross-sectional area 5A to A. The average speed in the wide section is 0.6 m/s. Using conservation of volume flow rate, what is the speed in the narrow section?
Explanation: This question applies conservation of volume flow rate in a narrowing pipe. Incompressible fluid in steady flow has constant Q = A v. As area decreases, speed increases proportionally. Here, area ratio 5A/A =5, v2=5*0.6=3.0 m/s. Distractor B 0.6 m/s might be chosen if forgetting the area change affects speed. Consistently use the continuity equation to relate speeds and areas in different sections.
A steady, incompressible fluid flows through a pipe that splits into two equal-area branches. The main pipe carries volume flow rate Q. Each branch has the same cross-sectional area. Assuming steady flow, what is the volume flow rate in each branch?
Explanation: This question examines flow rate conservation in a splitting pipe with equal branches. For steady incompressible flow, the incoming flow rate equals the sum of branch flow rates. If branches have equal area and are symmetric, the flow splits equally. Thus, each branch carries 2Q. Distractor C 2Q might be from misunderstanding and doubling instead. Always verify assumptions like equal splitting when branches are identical.