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.

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.

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.

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.

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.

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.

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.

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.

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.

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.