MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4b Fluid Flow Continuity Bernoulli

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4b Fluid Flow Continuity BernoulliQuestion 1 of 20

A researcher measures steady, incompressible flow through a horizontal tube and reports $A_1 v_1 = A_2 v_2$ for two points along the tube. Which experimental observation would most directly violate the use of the continuity equation in this form for the segment between points 1 and 2?

A side port between points 1 and 2 withdraws fluid at a constant rate

The tube diameter decreases smoothly between points 1 and 2

The tube remains horizontal between points 1 and 2

The fluid density is approximately constant along the segment

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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4b Fluid Flow Continuity Bernoulli

Practice 4b Fluid Flow Continuity Bernoulli in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on 4b Fluid Flow Continuity Bernoulli, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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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.

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Question 1

A side port between points 1 and 2 withdraws fluid at a constant rate (correct answer)
The tube diameter decreases smoothly between points 1 and 2
The tube remains horizontal between points 1 and 2
The fluid density is approximately constant along the segment

Explanation: This question tests understanding of when the continuity equation applies in its standard form. The continuity equation A₁v₁ = A₂v₂ assumes no fluid is added or removed between the two points—it's based on mass conservation for a closed system. Choice A describes fluid withdrawal between points, which violates this assumption and makes the standard continuity equation invalid, so A is correct. Choice B (diameter change) is exactly when continuity applies. Choice C (horizontal orientation) doesn't affect continuity. Choice D (constant density) is an assumption already built into the incompressible flow model. When applying continuity, always verify that the flow is truly continuous with no sources or sinks between measurement points.

Question 2

A horizontal pipe carries an incompressible fluid at steady flow. At point 1, the speed is $v_1$ and static pressure is $P_1$ . At point 2 downstream, the cross-sectional area is smaller and the speed is measured to be $v_2 > v_1$ . Neglecting viscosity, which statement is best explained by Bernoulli's equation?

$P_2 > P_1$ because higher speed requires higher static pressure
$P_2 < P_1$ because some pressure energy converts to kinetic energy (correct answer)
$P_2 = P_1$ because pressure depends only on depth in a fluid
$P_2 < P_1$ because the fluid's density decreases in the constriction

Explanation: This question tests conceptual understanding of Bernoulli's equation and energy conservation in fluid flow. Bernoulli's equation represents conservation of mechanical energy: pressure energy plus kinetic energy remains constant (neglecting potential energy for horizontal flow). When velocity increases from v₁ to v₂, kinetic energy increases, so pressure energy must decrease, meaning P₂ < P₁, making choice B correct. Choice A contradicts energy conservation by suggesting both forms of energy increase. Choice C incorrectly applies hydrostatic pressure principles to a flowing fluid. Choice D incorrectly suggests density changes in an incompressible fluid. To understand Bernoulli conceptually, think of it as an energy trade-off: faster flow means more kinetic energy, which must come from somewhere—specifically, from pressure energy.

Question 3

A syringe pump drives water (density $\rho = 1000\,\text{kg/m}^3$ ) through a rigid, horizontal tube that narrows from radius $r_1 = 2.0\,\text{mm}$ to $r_2 = 1.0\,\text{mm}$ . At the wide section, the static pressure is $P_1 = 120\,\text{kPa}$ and the average speed is $v_1 = 0.50\,\text{m/s}$ . Neglecting viscosity and elevation change, which conclusion is most consistent with Bernoulli's equation for the static pressure $P_2$ in the narrow section?

$P_2 < P_1$ because velocity increases in the constriction (correct answer)
$P_2 = P_1$ because the tube is horizontal
$P_2 > P_1$ because continuity requires higher pressure to maintain flow
$P_2$ is lower only if the flow is turbulent; Bernoulli's equation requires turbulence

Explanation: This question tests application of Bernoulli's equation to relate pressure and velocity changes in fluid flow. Bernoulli's equation for horizontal flow states that P₁ + ½ρv₁² = P₂ + ½ρv₂², meaning that as velocity increases, static pressure must decrease to conserve total mechanical energy. Since the tube narrows from radius 2.0 mm to 1.0 mm, the area decreases by a factor of 4, so by continuity, v₂ = 4v₁ = 2.0 m/s. With higher velocity at point 2, Bernoulli's equation requires P₂ < P₁, making choice A correct. Choice C incorrectly suggests pressure increases to maintain flow, confusing the cause-effect relationship—it's the pressure difference that drives flow, not continuity that requires pressure. When applying Bernoulli's equation, always check that kinetic energy increase corresponds to pressure energy decrease.

Question 4

An IV infusion line carries saline (incompressible) at a volumetric flow rate of $Q = 1.5\,\text{mL/s}$ . The line transitions from an internal radius of $r_1 = 1.0\,\text{mm}$ to $r_2 = 0.50\,\text{mm}$ without leakage. Based on conservation of mass (continuity), what is the best prediction for the relationship between average speeds $v_1$ and $v_2$ in the two sections?

$v_2 = 2 v_1$ because radius halves
$v_2 = 4 v_1$ because area scales as $r^2$ (correct answer)
$v_2 = v_1/4$ because smaller radius increases resistance
$v_2 = v_1$ because $Q$ is given in mL/s

Explanation: This question tests understanding of how the continuity equation relates to circular pipe geometry. For a circular pipe, area A = πr², so when radius changes from r₁ to r₂ = 0.50r₁, the area changes by a factor of (r₂/r₁)² = 0.25. By the continuity equation Q = A₁v₁ = A₂v₂, if A₂ = 0.25A₁, then v₂ = 4v₁, making choice B correct. Choice A incorrectly assumes velocity scales linearly with radius rather than inversely with area. Choice C reverses the relationship, suggesting smaller pipes have lower velocity, which contradicts continuity for steady flow. To avoid confusion, always remember that area scales as radius squared, and velocity is inversely proportional to area for constant flow rate.

Question 5

Blood (assume incompressible, $\rho = 1060\ \text{kg/m}^3$ ) flows steadily through a horizontal artery segment that has a localized stenosis. Upstream area is $A_1 = 4.0\ \text{mm}^2$ and at the stenosis is $A_2 = 1.0\ \text{mm}^2$ . If the upstream speed is $v_1 = 0.25\ \text{m/s}$ and viscous losses are neglected, what would be expected for the pressure at the stenosis compared with upstream, based on continuity and Bernoulli?

Higher pressure at the stenosis because smaller area increases resistance
Lower pressure at the stenosis because speed is higher in the smaller area (correct answer)
Same pressure at the stenosis because the artery is horizontal ( $\Delta h=0$ )
Lower pressure upstream because higher pressure is required to maintain constant volumetric flow

Explanation: This question tests understanding of pressure changes in arterial stenosis using Bernoulli's principle. The continuity equation shows that when area decreases from 4.0 mm² to 1.0 mm² (factor of 4), velocity increases from 0.25 m/s to 1.0 m/s (factor of 4). According to Bernoulli's equation for horizontal flow, this velocity increase requires a pressure decrease to conserve total mechanical energy. The pressure at the stenosis must be lower than upstream pressure because the fluid's kinetic energy has increased. Answer choice A incorrectly associates smaller area with higher resistance and pressure, confusing steady-state flow principles with viscous effects. In medical contexts, this pressure drop at stenoses can affect blood flow patterns and is why severe stenoses can compromise tissue perfusion despite maintaining flow continuity.

Question 6

A clinician models blood flow through a carotid artery segment as steady, incompressible flow along a streamline. At point 1 (lower elevation), $v_1 = 0.8\ \text{m/s}$ and $P_1 = 13\ \text{kPa}$ (gauge). At point 2, the vessel is at a higher elevation by $\Delta h = 0.20\ \text{m}$ and the speed is $v_2 = 0.8\ \text{m/s}$ . Take $\rho = 1060\ \text{kg/m}^3$ and $g = 9.8\ \text{m/s}^2$ , neglect viscosity. Based on Bernoulli's Equation, what is most consistent for $P_2$ relative to $P_1$ ?

$P_2 > P_1$ because pressure increases with height in flowing fluids
$P_2 < P_1$ by approximately $\rho g\Delta h$ (correct answer)
$P_2 = P_1$ because speeds are equal
$P_2 < P_1$ by approximately $\tfrac{1}{2}\rho(v_2^2-v_1^2)$

Explanation: This question tests application of Bernoulli's equation including gravitational potential energy changes. The complete Bernoulli equation is P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. Since v₁ = v₂ = 0.8 m/s, the kinetic energy terms cancel, leaving P₁ + ρgh₁ = P₂ + ρgh₂. Rearranging: P₂ = P₁ - ρg(h₂ - h₁) = P₁ - ρgΔh. With Δh = 0.20 m positive (point 2 is higher), P₂ < P₁ by approximately ρgΔh = (1060 kg/m³)(9.8 m/s²)(0.20 m) ≈ 2080 Pa ≈ 2.1 kPa. Answer choice A incorrectly suggests pressure increases with height in flowing fluids, confusing static fluid behavior with dynamic flow. In medical contexts, this hydrostatic pressure difference explains why blood pressure varies with measurement height relative to the heart.

Question 7

A saline solution (density $\rho = 1000\ \text{kg/m}^3$ ) flows steadily through a horizontal catheter that narrows from cross-sectional area $A_1 = 4.0\ \text{mm}^2$ to $A_2 = 1.0\ \text{mm}^2$ . The average speed in the wide section is measured as $v_1 = 0.50\ \text{m/s}$ . Assuming incompressible flow and no leakage, what would be expected for the average speed $v_2$ in the narrow section?

$v_2 = 0.125\ \text{m/s}$
$v_2 = 2.0\ \text{m/s}$ (correct answer)
$v_2 = 0.50\ \text{m/s}$
$v_2 = 1.0\ \text{m/s}$

Explanation: This question tests understanding of the continuity equation for incompressible fluid flow. The continuity equation states that for steady flow of an incompressible fluid, the product of cross-sectional area and velocity must remain constant: A₁v₁ = A₂v₂. In this catheter scenario, we have A₁ = 4.0 mm², v₁ = 0.50 m/s, and A₂ = 1.0 mm². Solving for v₂: v₂ = (A₁/A₂)v₁ = (4.0/1.0) × 0.50 = 2.0 m/s. Choice A incorrectly divides velocities instead of multiplying, a common error when students confuse the inverse relationship between area and velocity. To verify continuity problems, always check that flow rate Q = Av remains constant at both points.

Question 8

A contrast agent (assume incompressible) flows through a bifurcating vessel where the parent vessel has cross-sectional area $A_0 = 6\ \text{mm}^2$ and average speed $v_0 = 0.40\ \text{m/s}$ . It splits into two daughter branches with areas $A_1 = 2\ \text{mm}^2$ and $A_2 = 4\ \text{mm}^2$ . If flow divides such that average speeds in both branches are equal ( $v_1 = v_2$ ), what is the expected common branch speed?

$0.20\ \text{m/s}$
$0.40\ \text{m/s}$ (correct answer)
$0.60\ \text{m/s}$
$0.80\ \text{m/s}$

Explanation: This question tests continuity equation application in bifurcating vessels. The continuity equation requires that flow rate entering equals flow rate exiting: Q₀ = Q₁ + Q₂, or A₀v₀ = A₁v₁ + A₂v₂. Given A₀ = 6 mm², v₀ = 0.40 m/s, A₁ = 2 mm², A₂ = 4 mm², and v₁ = v₂ = v, we solve: 6(0.40) = 2v + 4v = 6v, giving v = 0.40 m/s. Choice C incorrectly assumes flow splits proportionally to area ratios without considering the constraint of equal velocities. When branches have equal velocities, the parent velocity equals the branch velocity only when total branch area equals parent area, which is true here (2 + 4 = 6 mm²).

Question 9

A lab measures volumetric flow rate through an IV line as $Q = 2.0\ \text{mL/s}$ . The tubing has an inner radius $r = 0.50\ \text{mm}$ at a straight section. Assuming steady incompressible flow, what is the best estimate of the average fluid speed in this section? (Use $1\ \text{mL} = 1\ \text{cm}^3$ .)

$v \approx 0.025\ \text{m/s}$
$v \approx 0.25\ \text{m/s}$
$v \approx 2.5\ \text{m/s}$ (correct answer)
$v \approx 25\ \text{m/s}$

Explanation: This question tests calculation of flow velocity from volumetric flow rate using the continuity equation. The relationship between volumetric flow rate Q and average velocity v is Q = Av, where A is cross-sectional area. Given Q = 2.0 mL/s = 2.0 cm³/s and r = 0.50 mm = 0.050 cm, the area is A = πr² = π(0.050)² = 0.00785 cm². Therefore, v = Q/A = 2.0/0.00785 = 255 cm/s = 2.55 m/s ≈ 2.5 m/s. Choice B incorrectly uses diameter instead of radius or makes a unit conversion error, a common mistake when working with small medical tubing. Always verify units: when Q is in cm³/s and A in cm², velocity comes out in cm/s, which must be converted to m/s.

Question 10

A fluid with density $\rho = 1000\ \text{kg/m}^3$ flows steadily through a vertical pipe from point 1 (lower) to point 2 (higher). The pipe diameter is constant, so $v_1 = v_2$ . The height increase is $\Delta h = h_2 - h_1 = 0.50\ \text{m}$ . Neglecting viscosity, what pressure change $P_2 - P_1$ is most consistent with Bernoulli's equation? (Use $g = 9.8\ \text{m/s}^2$ .)

$P_2 - P_1 \approx +4.9\ \text{kPa}$
$P_2 - P_1 \approx -4.9\ \text{kPa}$ (correct answer)
$P_2 - P_1 \approx +0.20\ \text{kPa}$
$P_2 - P_1 \approx -0.20\ \text{kPa}$

Explanation: This question tests Bernoulli's equation with gravitational potential energy changes. The complete Bernoulli equation is P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂. Since the pipe has constant diameter, v₁ = v₂, so the kinetic energy terms cancel. This gives P₂ - P₁ = ρg(h₁ - h₂) = -ρgΔh = -1000(9.8)(0.50) = -4900 Pa = -4.9 kPa. The negative sign indicates pressure decreases with height, as expected since the fluid must overcome gravity. Choice C might result from using incorrect units or forgetting the density term. Remember: when fluid flows upward against gravity at constant velocity, pressure must decrease by ρgh to provide the driving force.

Question 11

A catheter measures blood flow through a horizontal artery segment that narrows due to plaque. At point 1 (before narrowing), $A_1 = 5.0\times 10^{-6}\,\text{m}^2$ , $v_1 = 0.30\,\text{m/s}$ , and $P_1 = 13.0\,\text{kPa}$ . At point 2 (in the narrowed region), $A_2 = 2.5\times 10^{-6}\,\text{m}^2$ . Assume blood is incompressible with density $\rho = 1060\,\text{kg/m}^3$ , and neglect viscosity and height change. What would be expected for the pressure change $P_2 - P_1$ ?

$P_2 - P_1 > 0$ because pressure must rise to accelerate the fluid
$P_2 - P_1 = 0$ because continuity conserves pressure
$P_2 - P_1 < 0$ because velocity increases where area decreases (correct answer)
$P_2 - P_1 < 0$ only if point 2 is at higher elevation

Explanation: This question tests combined application of continuity and Bernoulli's equations to predict pressure changes. First, continuity gives us v₂: since A₂ = 0.5A₁, we have v₂ = 2v₁ = 0.60 m/s. Then, applying Bernoulli's equation for horizontal flow: P₁ + ½ρv₁² = P₂ + ½ρv₂². Rearranging: P₂ - P₁ = ½ρ(v₁² - v₂²) = ½(1060)(0.30² - 0.60²) = ½(1060)(-0.27) = -143 Pa. Since this is negative, P₂ < P₁, making choice C correct. Choice A incorrectly suggests pressure must rise to accelerate fluid, misunderstanding that it's the pressure gradient that causes acceleration, not the reverse. When solving Bernoulli problems, always calculate the kinetic energy change first, then determine the corresponding pressure change.

Question 12

A microfluidic chip delivers buffer through two serial sections of equal length: section 1 has cross-sectional area $A_1$ , section 2 has cross-sectional area $A_2 = 3A_1$ . The fluid is incompressible and flow is steady with no branching. Which statement is most consistent with the continuity equation regarding the volumetric flow rates $Q_1$ and $Q_2$ and average speeds $v_1$ and $v_2$ ?

$Q_2 = 3Q_1$ and $v_2 = v_1$ because area is larger
$Q_2 = Q_1$ and $v_2 = v_1/3$ because $Q = Av$ (correct answer)
$Q_2 = Q_1/3$ and $v_2 = v_1$ because flow slows in wider channels
$Q_2 = Q_1$ and $v_2 = 3v_1$ because velocity is proportional to area

Explanation: This question tests understanding of volumetric flow rate conservation in series flow. For incompressible flow with no branching, the volumetric flow rate Q must be constant throughout: Q₁ = Q₂. Since Q = Av and A₂ = 3A₁, we must have v₂ = v₁/3 to maintain constant Q, making choice B correct. Choice A incorrectly suggests flow rate increases in the wider section, violating mass conservation. Choice D reverses the velocity relationship, suggesting velocity increases with area. When analyzing series flow, remember that volumetric flow rate (not velocity) remains constant, and velocity varies inversely with cross-sectional area.

Question 13

A respiratory therapist models airflow in an idealized, rigid airway as incompressible flow for a short segment. The airway narrows from $A_1$ to $A_2 = 0.5A_1$ while remaining approximately horizontal. If the static pressure at the wide section is $P_1$ , which statement best reflects the expected change at the narrow section according to continuity and Bernoulli (neglecting viscosity)?

Speed decreases and static pressure increases in the narrow section
Speed increases and static pressure decreases in the narrow section (correct answer)
Speed increases and static pressure increases in the narrow section
Speed and static pressure are unchanged because the segment is horizontal

Explanation: This question tests conceptual understanding of how continuity and Bernoulli's principles work together. When area decreases to A₂ = 0.5A₁, continuity requires velocity to double: v₂ = 2v₁. With increased velocity, Bernoulli's equation dictates that static pressure must decrease to conserve total mechanical energy, making choice B correct. Choice A reverses both effects, while choice C incorrectly suggests both speed and pressure increase, violating energy conservation. Choice D ignores the area change entirely, misunderstanding that horizontal orientation only eliminates gravitational potential energy changes, not the velocity-pressure relationship. To analyze constrictions, always apply continuity first to find velocity change, then use Bernoulli to determine the corresponding pressure change.

Question 14

A closed-loop perfusion system circulates a saline solution (density $\rho = 1000\,\text{kg/m}^3$ ) through a horizontal constriction. At point 1, the speed is $v_1 = 1.0\,\text{m/s}$ and pressure is $P_1 = 200\,\text{kPa}$ . At point 2 in the constriction, the speed is measured as $v_2 = 3.0\,\text{m/s}$ . Neglecting viscosity and elevation change, what is the best prediction for $P_2$ ?

$P_2 = 200\,\text{kPa}$ because pressure is conserved in closed loops
$P_2 = 204\,\text{kPa}$ because higher speed increases static pressure by $\tfrac{1}{2}\rho v^2$
$P_2 = 196\,\text{kPa}$ because pressure decreases as kinetic energy increases (correct answer)
$P_2 = 182\,\text{kPa}$ because pressure drop scales with $v$ (not $v^2$ ) in Bernoulli flow

Explanation: This question tests direct application of Bernoulli's equation when velocities are given. For horizontal flow, Bernoulli's equation states P₁ + ½ρv₁² = P₂ + ½ρv₂². Solving for P₂: P₂ = P₁ + ½ρ(v₁² - v₂²) = 200,000 + ½(1000)(1² - 3²) = 200,000 + 500(-8) = 196,000 Pa = 196 kPa, making choice C correct. Choice B incorrectly adds the kinetic energy term instead of recognizing the trade-off between pressure and kinetic energy. Choice D suggests a linear relationship with velocity rather than the correct quadratic relationship. When using Bernoulli's equation, always remember that kinetic energy terms involve velocity squared, and increased velocity means decreased pressure.

Question 15

In a lab demonstration, water flows steadily through a rigid pipe that widens from diameter $d_1 = 1.0\,\text{cm}$ to $d_2 = 2.0\,\text{cm}$ . The average speed in the narrow section is measured as $v_1 = 0.80\,\text{m/s}$ . Assuming incompressible flow and no leaks, what is the best estimate for $v_2$ in the wide section?

$v_2 = 0.40\,\text{m/s}$ because diameter doubles
$v_2 = 0.20\,\text{m/s}$ because area increases by a factor of 4 (correct answer)
$v_2 = 3.2\,\text{m/s}$ because velocity is inversely proportional to diameter
$v_2 = 0.80\,\text{m/s}$ because flow speed is unchanged by geometry

Explanation: This question tests understanding of how diameter changes affect flow velocity through the continuity equation. When diameter doubles from d₁ to d₂ = 2d₁, the area increases by a factor of 4 (since A ∝ d²). By continuity, A₁v₁ = A₂v₂, so if A₂ = 4A₁, then v₂ = v₁/4 = 0.80/4 = 0.20 m/s, making choice B correct. Choice A incorrectly assumes velocity is halved when diameter doubles, forgetting that area depends on diameter squared. Choice C reverses the relationship entirely, suggesting velocity increases in the wider section. To solve diameter-based problems efficiently, remember that doubling diameter quadruples area, which quarters the velocity for steady, incompressible flow.

Question 16

In a hemodynamics study, an incompressible fluid (modeled as blood) flows steadily through a horizontal vessel segment that narrows from cross-sectional area $A_1 = 6.0\,\text{mm}^2$ to $A_2 = 2.0\,\text{mm}^2$ . At the wider segment, the average speed is $v_1 = 0.40\,\text{m/s}$ . Assuming no leakage and steady flow, what would be expected for the average speed $v_2$ in the narrowed segment based on the continuity equation?

$v_2 = 0.13\,\text{m/s}$ because velocity decreases in a constriction
$v_2 = 1.2\,\text{m/s}$ because $A_1 v_1 = A_2 v_2$ (correct answer)
$v_2 = 0.40\,\text{m/s}$ because speed is constant in horizontal flow
$v_2 = 3.6\,\text{m/s}$ because pressure triples when area decreases by a factor of 3

Explanation: This question tests understanding of the continuity equation for incompressible fluid flow. The continuity equation states that for steady flow with no leaks, the product of cross-sectional area and velocity must remain constant: A₁v₁ = A₂v₂. Given A₁ = 6.0 mm², v₁ = 0.40 m/s, and A₂ = 2.0 mm², we can solve for v₂: v₂ = (A₁v₁)/A₂ = (6.0 × 0.40)/2.0 = 1.2 m/s. Choice B correctly applies this principle and arrives at the right answer. Choice A incorrectly assumes velocity decreases in a constriction, which contradicts the continuity equation—velocity actually increases when area decreases. To approach similar problems, remember that for incompressible flow, the volumetric flow rate Q = Av must remain constant throughout the system.

Question 17

In a horizontal flow chamber, incompressible fluid moves from point 1 to point 2. The measured static pressure decreases by $500\ \text{Pa}$ from 1 to 2. Neglect viscosity and assume constant height. Which observation is most consistent with Bernoulli's equation?

The flow speed is lower at point 2 than at point 1.
The flow speed is higher at point 2 than at point 1. (correct answer)
The density must be lower at point 2 than at point 1.
The volumetric flow rate must be zero at point 2.

Explanation: This question applies Bernoulli's equation to infer speed from pressure changes. Bernoulli's equation at constant height gives P1 + (1/2) ρ v1² = P2 + (1/2) ρ v2²; if P2 < P1, then v2 > v1. With static pressure decreasing by 500 Pa from point 1 to 2, speed must increase. Choice B is correct because higher speed at point 2 accounts for the pressure drop via kinetic energy gain. Choice A is incorrect as it would require P2 > P1 for lower speed, opposing the observation. In such inferences, solve for v from the pressure difference using rearranged Bernoulli. Neglect viscosity only if stated and confirm horizontal setup.

Question 18

A researcher compares flow in two horizontal segments of the same rigid tube carrying incompressible saline. The speed is higher in segment 2 than segment 1, but the cross-sectional area is the same in both segments. Which explanation is most consistent with continuity and steady flow assumptions?

The fluid density must be lower in segment 2.
The flow cannot be steady and incompressible if $A_1=A_2$ yet $v_2\ne v_1$ . (correct answer)
Bernoulli's equation requires speed to increase along the flow direction.
Higher speed in segment 2 implies higher area in segment 2.

Explanation: This question tests implications of the continuity equation for steady incompressible flow. Continuity requires A1 v1 = A2 v2; if A1 = A2, then v1 must equal v2 for steady flow. With A1 = A2 but v2 > v1, this violates the assumptions. Choice B is correct because unequal speeds with equal areas imply non-steady or compressible flow, inconsistent with given conditions. Choice C is wrong as Bernoulli does not require speed increases without cause like area change. For diagnostics, check if A v is constant; discrepancies suggest violated assumptions. Examine if density or steadiness is truly constant.

Question 19

A horizontal pipe carries water ( $\rho=1000\ \text{kg/m}^3$ ) at steady flow. At point 1, $A_1=2A_2$ and $v_1=0.90\ \text{m/s}$ . Neglect viscosity. Based on Bernoulli's equation, which statement is most consistent about the static pressure difference $P_1-P_2$ ?

$P_1-P_2=0$ because continuity enforces equal pressure.
$P_1-P_2>0$ because $v_2>v_1$ in the narrower section. (correct answer)
$P_1-P_2<0$ because faster flow increases static pressure.
$P_1-P_2>0$ because the fluid gains gravitational potential energy.

Explanation: This question probes Bernoulli's equation for pressure differences in a narrowing pipe. Bernoulli's equation relates pressure, speed, and height: P + (1/2) ρ v² + ρ g h = constant, with viscosity neglected. In this horizontal pipe with A1 = 2 A2 and v1 = 0.90 m/s, continuity gives v2 = 2 v1 = 1.8 m/s, affecting pressure. Choice B is correct because P1 - P2 > 0, as higher v2 lowers P2 to conserve energy. Choice C fails by claiming faster flow increases pressure, confusing kinetic and potential energy terms. To approach such problems, use continuity to find speeds, then compare pressures via Bernoulli. Check for horizontal orientation to omit height terms.

Question 20

A horizontal tube carries an ideal incompressible fluid. At point 1, $A_1=5.0\ \text{mm}^2$ , $v_1=0.20\ \text{m/s}$ . At point 2, $A_2=2.0\ \text{mm}^2$ . Which value is most consistent with the volumetric flow rate $Q$ ?

$Q = 1.0\times 10^{-6}\ \text{m}^3/\text{s}$ (correct answer)
$Q = 1.0\times 10^{-3}\ \text{m}^3/\text{s}$
$Q = 2.5\times 10^{-6}\ \text{m}^3/\text{s}$
$Q = 1.0\times 10^{-5}\ \text{m}^3/\text{s}$

Explanation: This question tests calculation of volumetric flow rate using continuity. Q = A1 v1 = A2 v2 must be constant, allowing computation from any point. With A1 = 5.0 mm² = 5e-6 m², v1 = 0.20 m/s, Q = 1e-6 m³/s. Choice A is correct because it matches Q = 5e-6 * 0.20 = 1.0e-6 m³/s. Choice B is off by orders of magnitude, likely from unit conversion errors. For flow rate problems, convert areas to m² and speeds to m/s for consistent units. Verify by calculating Q at point 2 to confirm constancy.