MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4b Viscosity Poiseuille Flow

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4b Viscosity Poiseuille FlowQuestion 1 of 20

A lab compares two Newtonian fluids flowing through the same rigid cylindrical capillary (same $r$ and $L$ ) under the same pressure drop $\Delta P$ . Fluid X has viscosity $\eta_X = 1.0\ \text{mPa}\cdot\text{s}$ and Fluid Y has viscosity $\eta_Y = 4.0\ \text{mPa}\cdot\text{s}$ at the measurement temperature. Flow is laminar for both. Based on Poiseuille flow, what is the expected ratio $Q_X/Q_Y$ ?

Q_X/Q_Y = 1/4

because higher viscosity increases flow resistance.

Q_X/Q_Y = 4

because

Q \propto 1/\eta

with other variables fixed.

Q_X/Q_Y = 16

because

Q \propto 1/\eta^2

Q_X/Q_Y = 1

because both experience the same

\Delta P

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MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4b Viscosity Poiseuille Flow

Practice 4b Viscosity Poiseuille Flow 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.

What this quiz covers

This quiz focuses on 4b Viscosity Poiseuille Flow, 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

$Q_X/Q_Y = 1/4$ because higher viscosity increases flow resistance.
$Q_X/Q_Y = 4$ because $Q \propto 1/\eta$ with other variables fixed. (correct answer)
$Q_X/Q_Y = 16$ because $Q \propto 1/\eta^2$ .
$Q_X/Q_Y = 1$ because both experience the same $\Delta P$ .

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law demonstrates that Q = (πr⁴ΔP)/(8ηL), establishing an inverse proportionality between flow rate and viscosity. In this comparison, Fluid Y has four times the viscosity of Fluid X (4.0 vs 1.0 mPa·s) while using the same capillary and pressure drop, resulting in Fluid Y having one-fourth the flow rate. Choice B is correct because Q_X/Q_Y = η_Y/η_X = 4.0/1.0 = 4, accurately reflecting the Q ∝ 1/η relationship. Choice A reverses the ratio, choice C incorrectly squares the viscosity effect, and choice D ignores viscosity's crucial role. When comparing fluids in Poiseuille flow, the flow rate ratio equals the inverse of the viscosity ratio.

Question 2

A researcher evaluates whether a capillary-flow assay is sensitive to small manufacturing variation in tube radius. Two nominally identical rigid cylindrical capillaries have the same length $L$ and are used with the same Newtonian fluid at the same temperature and the same applied pressure drop $\Delta P$ . Capillary A has radius $r$ , while Capillary B has radius $0.90r$ . Assuming laminar Poiseuille flow, which outcome is most consistent with the expected change in volumetric flow rate?

$Q_B \approx 0.90Q_A$ because flow rate is directly proportional to radius.
$Q_B \approx 0.81Q_A$ because flow rate is proportional to cross-sectional area.
$Q_B \approx 0.66Q_A$ because flow rate scales as $r^4$ . (correct answer)
$Q_B \approx 1.11Q_A$ because a smaller radius increases speed (Bernoulli principle).

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law reveals that Q = (πr⁴ΔP)/(8ηL), demonstrating the critical r⁴ dependence that makes flow extremely sensitive to radius variations. When Capillary B has radius 0.90r compared to Capillary A's radius r, the flow rate ratio is Q_B/Q_A = (0.90)⁴ = 0.6561 ≈ 0.66. Choice C is correct because it accurately calculates that Q_B ≈ 0.66Q_A based on the fourth-power radius relationship in Poiseuille flow. Choices A and B underestimate the effect by assuming linear or quadratic relationships, while choice D contradicts physics with incorrect Bernoulli reasoning. This extreme sensitivity to radius (10% decrease causes 34% flow reduction) explains why precise manufacturing tolerances are crucial for microfluidic devices.

Question 3

To compare two capillaries, a student drives the same Newtonian fluid (viscosity $\eta$ ) through each under identical pressure drop $\Delta P = 1.0\ \text{kPa}$ . Capillary 1 has length $L$ and radius $r$ . Capillary 2 has length $2L$ and radius $r$ . The flow is laminar in both. Based on Poiseuille's equation, which outcome is consistent with these conditions?

Capillary 2 has half the flow rate of Capillary 1 because $Q \propto 1/L$ . (correct answer)
Capillary 2 has twice the flow rate because a longer tube sustains flow longer.
Both capillaries have the same flow rate because $\Delta P$ is the same.
Capillary 2 has one-fourth the flow rate because $Q \propto 1/L^2$ .

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows that Q = (πr⁴ΔP)/(8ηL), indicating flow rate is inversely proportional to tube length. In this comparison, Capillary 2 has twice the length (2L) of Capillary 1 while all other parameters remain constant, resulting in half the flow rate. Choice A is correct because it accurately states that Capillary 2 has half the flow rate of Capillary 1 due to the Q ∝ 1/L relationship. Choice B incorrectly suggests longer tubes increase flow, choice C ignores the length effect, and choice D incorrectly proposes a quadratic relationship. When analyzing Poiseuille flow, verify that each parameter's effect matches the law's predictions: linear relationships for ΔP, η, and L, but fourth power for r.

Question 4

In a microfluidics experiment, a Newtonian buffer is driven through a straight cylindrical glass capillary of length $L = 10\ \text{cm}$ and radius $r = 0.50\ \text{mm}$ under a constant pressure drop $\Delta P = 2.0\ \text{kPa}$ . The flow is verified to be laminar and fully developed. The same capillary is then used with a second buffer at the same temperature, identical except its dynamic viscosity is doubled (from $\eta$ to $2\eta$ ). Based on Poiseuille flow, what change in volumetric flow rate $Q$ is expected under the same $\Delta P$ , $L$ , and $r$ ?

$Q$ decreases to one-fourth of its original value because viscous resistance scales with $\eta^2$ .
$Q$ remains unchanged because the pressure drop is held constant.
$Q$ decreases to one-half of its original value because $Q \propto 1/\eta$ for laminar capillary flow. (correct answer)
$Q$ increases because higher viscosity increases the pressure transmitted to the fluid (Bernoulli effect).

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states that volumetric flow rate Q = (πr⁴ΔP)/(8ηL), showing that flow rate is inversely proportional to viscosity. In this scenario, doubling the viscosity from η to 2η while keeping all other parameters constant will halve the flow rate. Choice C is correct because it accurately reflects that Q ∝ 1/η for laminar capillary flow, resulting in Q decreasing to one-half its original value. Choice A incorrectly suggests Q ∝ 1/η², while choices B and D contradict the fundamental inverse relationship between flow rate and viscosity. When solving Poiseuille flow problems, always check that the relationship between Q and each variable matches the law's predictions.

Question 5

A physiologist approximates flow through a small arteriole as steady laminar flow in a rigid cylindrical tube. A vasodilator increases the arteriole radius by 10% (from $r$ to $1.10r$ ) without changing $\Delta P$ across the segment, its length $L$ , or blood viscosity $\eta$ . Based on Poiseuille's equation, which change in flow rate is most consistent with this model?

Flow rate increases by about 10% because $Q \propto r$ .
Flow rate increases by about 21% because $Q \propto r^2$ .
Flow rate increases by about 46% because $Q \propto r^4$ . (correct answer)
Flow rate decreases because a larger radius lowers velocity and thus lowers $Q$ (Bernoulli reasoning).

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law establishes that Q = (πr⁴ΔP)/(8ηL), showing flow rate scales with the fourth power of radius. When the arteriole radius increases by 10% (from r to 1.10r), the flow rate increases by (1.10)⁴ = 1.4641, representing a 46.41% increase. Choice C is correct because it accurately calculates the 46% increase resulting from the Q ∝ r⁴ relationship in Poiseuille flow. Choices A and B underestimate by assuming linear or quadratic relationships, while choice D contradicts physics by suggesting larger radii decrease flow. In physiological applications, this r⁴ dependence explains why small vessel diameter changes dramatically affect blood flow and why vasoregulation is so effective.

Question 6

A device uses laminar flow through a cylindrical capillary to deliver a drug solution. The designer can change only one parameter while keeping the others constant: pressure drop $\Delta P$ , tube length $L$ , and fluid viscosity $\eta$ remain fixed. Which modification is most effective for increasing the volumetric flow rate $Q$ by approximately an order of magnitude (about $10\times$ ) while remaining within the Poiseuille-flow model?

Increase the tube radius by a factor of 2 (since $Q \propto r^4$ ). (correct answer)
Decrease the tube length by a factor of 2 (since $Q \propto 1/L$ ).
Increase the tube radius by a factor of 1.8 (since $Q \propto r^2$ ).
Increase the pressure drop by a factor of 2 (since $Q \propto 1/\Delta P$ ).

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows Q = (πr⁴ΔP)/(8ηL), revealing that flow rate depends on r⁴, making radius changes most effective for large flow increases. To achieve a 10-fold increase in Q, the radius must increase by ⁴√10 ≈ 1.78, so doubling the radius yields 2⁴ = 16-fold increase, exceeding the target. Choice A is correct because increasing radius by factor of 2 produces the desired order-of-magnitude increase through the r⁴ dependence. Choice B only doubles flow rate, choice C incorrectly assumes r² dependence, and choice D incorrectly inverts the pressure relationship. For optimizing Poiseuille flow systems, radius adjustments provide the most dramatic effects due to the fourth-power relationship.

Question 7

A physiology lab models blood flow through a small artery as steady, laminar Poiseuille flow in a rigid cylindrical vessel. During a cold-pressor test, sympathetic activation causes the artery radius to decrease from $r$ to $0.90r$ while mean arterial pressure and vessel length remain approximately constant over the short interval. Viscosity is assumed unchanged. Based on Poiseuille's relationship $Q \propto r^4$ , what is the expected change in flow rate through that artery?

(You may use: $(0.90)^4 \approx 0.66$ .)

Flow decreases to about $0.66Q$ because of the $r^4$ dependence. (correct answer)
Flow decreases to about $0.90Q$ because flow is directly proportional to radius.
Flow increases because constriction increases fluid speed by Bernoulli's principle.
Flow is unchanged because pressure is the only determinant of flow in a tube.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law shows that flow rate depends on the fourth power of radius: Q ∝ r⁴ when other parameters are constant. In this scenario, the artery radius decreases from r to 0.90r during sympathetic activation, which dramatically reduces flow due to the r⁴ dependence. Choice A is correct because (0.90)⁴ ≈ 0.66, meaning flow decreases to about 66% of its original value. Choice B fails because it assumes a linear relationship between radius and flow, missing the critical fourth-power dependence. In similar questions, always remember the strong r⁴ dependence and calculate the effect of radius changes by raising the ratio to the fourth power.

Question 8

In a capillary viscometry setup, a constant pressure source applies $\Delta P$ across a rigid tube of radius $r$ and length $L$ . The fluid is Newtonian and the flow is verified to be laminar. The investigator accidentally records the tube diameter $d$ as if it were the radius when predicting the effect of changing tube size. If the true radius is doubled (from $r$ to $2r$ ) while $\Delta P$ , $L$ , and $\eta$ are held constant, which prediction for the change in $Q$ is consistent with Poiseuille flow?

$Q$ increases by a factor of 2 because $Q \propto r$ .
$Q$ increases by a factor of 4 because $Q \propto r^2$ .
$Q$ increases by a factor of 8 because $Q \propto r^3$ .
$Q$ increases by a factor of 16 because $Q \propto r^4$ . (correct answer)

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law clearly states that volumetric flow rate depends on the fourth power of radius: Q ∝ r⁴ when other parameters are constant. In this scenario, doubling the radius from r to 2r increases flow by a factor of 2⁴ = 16, regardless of any confusion about diameter versus radius in the setup description. Choice D is correct because it accurately reflects the r⁴ dependence that is fundamental to laminar flow in tubes. Choices A, B, and C fail because they suggest incorrect power relationships (r¹, r², and r³ respectively) that do not match Poiseuille's law. In similar questions, always remember the strong fourth-power radius dependence, which makes radius changes the most dramatic factor affecting flow rate.

Question 9

A lab uses the same capillary (fixed $r$ and $L$ ) to measure flow of a Newtonian fluid at two temperatures. At $20^\circ\text{C}$ , the viscosity is $\eta_{20}$ and the measured flow rate is $Q_{20}$ . At $37^\circ\text{C}$ , the viscosity decreases to $0.75\eta_{20}$ while the applied pressure drop $\Delta P$ is kept constant and the flow remains laminar. Based on Poiseuille flow, what is the expected $Q_{37}$ relative to $Q_{20}$ ?

$Q_{37} = 0.75Q_{20}$ because flow is proportional to viscosity.
$Q_{37} = 1.33Q_{20}$ because flow is inversely proportional to viscosity. (correct answer)
$Q_{37} = 1.75Q_{20}$ because warmer fluids always flow faster by Bernoulli's principle.
$Q_{37} = Q_{20}$ because viscosity changes do not affect laminar flow.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states that flow rate is inversely proportional to viscosity: Q ∝ 1/η when other parameters remain constant. In this scenario, warming the fluid from 20°C to 37°C reduces viscosity to 0.75η₂₀, which increases flow rate by the reciprocal factor. Choice B is correct because Q₃₇ = Q₂₀ × (η₂₀/0.75η₂₀) = Q₂₀ × (1/0.75) = 1.33Q₂₀, properly applying the inverse viscosity relationship. Choice A fails because it incorrectly assumes flow is proportional to viscosity rather than inversely proportional. In similar questions, remember that lower viscosity means easier flow, and calculate the flow ratio as the inverse of the viscosity ratio.

Question 10

In a microfluidics study, researchers drive an incompressible Newtonian fluid through a straight glass capillary under steady, laminar conditions. The capillary length is held constant at $L = 10\ \text{cm}$ and the pressure drop is fixed at $\Delta P = 20\ \text{kPa}$ . The temperature is kept constant so viscosity does not change during a run. The team replaces Fluid 1 (viscosity $\eta_1 = 1.0\ \text{mPa}\cdot\text{s}$ ) with Fluid 2 (viscosity $\eta_2 = 2.0\ \text{mPa}\cdot\text{s}$ ) while keeping the same capillary radius. Based on Poiseuille flow, what change in volumetric flow rate $Q$ is expected?

(Use Poiseuille proportionality: $Q \propto \frac{\Delta P\, r^4}{\eta L}$ .)

$Q$ doubles because higher viscosity increases the pressure-driven flux.
$Q$ is unchanged because $\Delta P$ and $L$ are unchanged.
$Q$ decreases by a factor of 2 because $Q \propto 1/\eta$ . (correct answer)
$Q$ decreases by a factor of 16 because $Q \propto 1/r^4$ and viscosity increased.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states that volumetric flow rate is inversely proportional to viscosity: Q ∝ ΔPr⁴/(ηL). In this scenario, doubling the viscosity from 1.0 to 2.0 mPa·s while keeping all other parameters constant will halve the flow rate. Choice C is correct because it accurately reflects this inverse relationship: when viscosity doubles, flow rate decreases by a factor of 2. Choice A fails because it incorrectly suggests that higher viscosity increases flow, which contradicts the physics of viscous resistance. In similar questions, always check that flow rate decreases with increasing viscosity and verify the proportionality relationships are applied correctly.

Question 11

A student uses a syringe pump to generate steady laminar flow of a Newtonian fluid through a capillary. For a given trial, $\Delta P$ , $r$ , and $\eta$ are held constant while the student compares two capillary lengths: $L_1 = 4\ \text{cm}$ and $L_2 = 12\ \text{cm}$ . Which statement is consistent with Poiseuille flow for the ratio of flow rates $Q_1/Q_2$ ?

$Q_1/Q_2 = 3$ because $Q \propto 1/L$ . (correct answer)
$Q_1/Q_2 = 1/3$ because longer tubes increase velocity via reduced wall interactions.
$Q_1/Q_2 = 9$ because $Q \propto 1/L^2$ .
$Q_1/Q_2 = 1$ because viscosity sets the flow rate, not length.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law demonstrates that Q = (πr⁴ΔP)/(8ηL), establishing an inverse linear relationship between flow rate and tube length. Comparing capillaries with L₁ = 4 cm and L₂ = 12 cm while keeping other parameters constant, the flow rate ratio is Q₁/Q₂ = L₂/L₁ = 12/4 = 3. Choice A is correct because it accurately identifies that Q ∝ 1/L, resulting in the shorter capillary having three times the flow rate of the longer one. Choice B reverses the relationship, choice C incorrectly squares the length effect, and choice D ignores length's influence entirely. For Poiseuille flow comparisons, remember that doubling length halves flow rate, tripling length reduces flow to one-third, and so on.

Question 12

An investigator models blood flow through a straight artery segment as laminar Poiseuille flow. During mild hypothermia, plasma viscosity increases by 25% (from $\eta$ to $1.25\eta$ ) while arterial radius, length, and the pressure drop across the segment are assumed unchanged. Under these assumptions, which statement best describes the influence of viscosity in this system?

Volumetric flow rate decreases to $0.80$ of baseline because $Q \propto 1/\eta$ . (correct answer)
Volumetric flow rate increases to $1.25$ of baseline because $Q \propto \eta$ .
Volumetric flow rate is unchanged because viscosity only affects turbulent flow.
Volumetric flow rate decreases to $0.64$ of baseline because $Q \propto 1/\eta^2$ .

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law establishes that Q = (πr⁴ΔP)/(8ηL), showing flow rate is inversely proportional to viscosity. In this hypothermia scenario, increasing viscosity by 25% (from η to 1.25η) while keeping radius, length, and pressure drop constant will reduce flow rate to Q/1.25 = 0.80Q. Choice A is correct because it accurately calculates that flow rate decreases to 0.80 of baseline, reflecting the Q ∝ 1/η relationship. Choice B incorrectly suggests a direct proportionality, choice C wrongly limits viscosity effects to turbulent flow, and choice D proposes an incorrect quadratic relationship. For medical applications of Poiseuille flow, remember that even modest viscosity changes can significantly impact perfusion.

Question 13

In an experiment, laminar flow of a Newtonian fluid is established through a cylindrical capillary. The operator accidentally increases the pressure drop from $\Delta P$ to $2\Delta P$ while simultaneously switching to a fluid with twice the viscosity (from $\eta$ to $2\eta$ ). The tube radius $r$ and length $L$ are unchanged. Under Poiseuille flow, what is the expected net effect on the volumetric flow rate $Q$ ?

$Q$ doubles because pressure dominates viscosity in laminar flow.
$Q$ is unchanged because the factor of 2 in $\Delta P$ cancels the factor of 2 in $\eta$ . (correct answer)
$Q$ is halved because $Q \propto \eta/\Delta P$ .
$Q$ increases by a factor of 4 because $Q \propto (\Delta P)^2$ .

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law states Q = (πr⁴ΔP)/(8ηL), showing flow rate is directly proportional to pressure drop and inversely proportional to viscosity. When pressure doubles (ΔP to 2ΔP) and viscosity doubles (η to 2η) simultaneously, these effects exactly cancel: Q_new = (πr⁴·2ΔP)/(8·2η·L) = Q_original. Choice B is correct because it recognizes that the factor of 2 increase in pressure drop perfectly compensates for the factor of 2 increase in viscosity, leaving flow rate unchanged. Choices A and C incorrectly weight one parameter over the other, while choice D proposes an incorrect quadratic pressure dependence. When multiple parameters change in Poiseuille flow, multiply their individual effects to find the net result.

Question 14

A researcher measures laminar flow of a glycerol–water mixture through a rigid cylindrical capillary at constant temperature. The capillary length is held fixed at $L = 5.0\ \text{cm}$ and the applied pressure drop is held fixed at $\Delta P = 1.5\ \text{kPa}$ . The capillary radius is increased from $r$ to $2r$ by switching to a wider tube of the same length. Assuming Poiseuille flow applies, what change in volumetric flow rate $Q$ is expected?

$Q$ increases by a factor of 2 because $Q \propto r$ .
$Q$ increases by a factor of 4 because $Q \propto r^2$ .
$Q$ decreases by a factor of 2 because wider tubes reduce shear rate and slow the flow.
$Q$ increases by a factor of 16 because $Q \propto r^4$ . (correct answer)

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law reveals that volumetric flow rate Q = (πr⁴ΔP)/(8ηL), demonstrating that flow rate is proportional to the fourth power of radius. In this experiment, doubling the radius from r to 2r while maintaining constant pressure drop, length, and viscosity will increase flow rate by a factor of 2⁴ = 16. Choice D is correct because it accurately identifies that Q ∝ r⁴, resulting in a 16-fold increase in flow rate. Choices A and B incorrectly assume linear or quadratic relationships, while choice C contradicts the physics by suggesting wider tubes decrease flow. For Poiseuille flow problems, remember that the r⁴ dependence makes radius changes the most dramatic factor affecting flow rate.

Question 15

In a microfluidics experiment, a Newtonian buffer is driven through a straight glass capillary (length $L = 10\ \text{cm}$ , radius $r = 0.50\ \text{mm}$ ) under a constant pressure drop $\Delta P = 2.0\ \text{kPa}$ . Flow is confirmed laminar (Re $< 200$ ). The buffer is then replaced with a glycerol-water mixture whose dynamic viscosity is 4 times higher, while $\Delta P$ , $L$ , and $r$ are held constant. Based on Poiseuille flow, what change in volumetric flow rate $Q$ is expected?

$Q$ increases by a factor of 4 because higher viscosity increases shear-driven flow.
$Q$ decreases by a factor of 4 because $Q \propto 1/\eta$ for laminar capillary flow. (correct answer)
$Q$ is unchanged because laminar flow rate depends only on $\Delta P$ and $r$ .
$Q$ decreases slightly because higher viscosity makes the flow turbulent, reducing throughput.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q proportional to ΔP r^4 / (η L). In this scenario, increasing the viscosity by a factor of 4 while keeping ΔP, L, and r constant directly impacts the flow rate. Choice B is correct because it accurately reflects the inverse relationship between flow rate and viscosity as per Poiseuille's law, leading to a decrease by a factor of 4. Choice A fails because higher viscosity decreases shear-driven flow, not increases it. In similar questions, always verify the flow regime is laminar and check the inverse proportionality to viscosity for consistency. Remember that Poiseuille's law applies only to Newtonian fluids in laminar flow.

Question 16

A capillary viscometer uses a fixed glass tube (radius $r$ , length $L$ ) and measures flow rate $Q$ under a known pressure drop $\Delta P$ . A student accidentally uses a tube of the same radius but half the length ( $L/2$ ), keeping $\Delta P$ and fluid viscosity $\eta$ the same. Under laminar conditions, what outcome is consistent with Poiseuille flow?

$Q$ is halved because shorter length reduces time for acceleration.
$Q$ doubles because hydraulic resistance is proportional to $L$ . (correct answer)
$Q$ increases by a factor of 4 because $Q \propto 1/L^2$ .
$Q$ is unchanged because only radius determines capillary flow.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q inversely proportional to L. In this scenario, halving the tube length while keeping other parameters constant affects the hydraulic resistance. Choice B is correct because it accurately reflects that shorter length halves resistance, doubling Q. Choice C fails because Q is proportional to 1/L, not 1/L^2. In similar questions, always isolate the changed variable and check its proportionality in Poiseuille's equation for consistency.

Question 17

A lab measures laminar flow of a Newtonian fluid through a capillary at constant $\Delta P$ , $L$ , and $r$ . Temperature is increased from $20^\circ\text{C}$ to $40^\circ\text{C}$ , and the fluid's dynamic viscosity decreases by 30% (to $0.70\eta_0$ ). All other parameters are unchanged. What change in $Q$ is expected from Poiseuille flow?

$Q$ decreases to $0.70Q_0$ because viscosity and flow rate are directly proportional.
$Q$ increases to about $1.43Q_0$ because $Q \propto 1/\eta$ . (correct answer)
$Q$ is unchanged because viscosity only affects turbulent flow.
$Q$ increases slightly because Bernoulli predicts higher temperature increases pressure.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q inversely proportional to η. In this scenario, decreasing viscosity to 0.70 η0 through temperature increase alters the flow rate. Choice B is correct because it accurately reflects Q ∝ 1/η, leading to Q ≈ 1.43 Q0. Choice A fails because viscosity and flow rate are inversely, not directly, proportional. In similar questions, always recall the inverse relationship with viscosity and verify if other parameters remain constant.

Question 18

In an in vitro model of an arteriole, researchers keep the same blood-mimicking fluid (constant $\eta$ ) and the same tube radius $r$ , but double the tube length from $L$ to $2L$ while maintaining the same pressure drop $\Delta P$ . Flow remains laminar. What change in $Q$ is expected?

$Q$ doubles because the fluid experiences the pressure drop over a longer distance.
$Q$ halves because resistance is proportional to length. (correct answer)
$Q$ decreases by a factor of 4 because $Q \propto 1/L^2$ .
$Q$ is unchanged because viscosity dominates over length in laminar flow.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, with Q inversely proportional to L. In this scenario, doubling the length increases the resistance to flow. Choice B is correct because it accurately reflects the relationship Q ∝ 1/L, halving the flow rate. Choice A fails because longer distance increases, not decreases, resistance. In similar questions, always check the linear inverse dependence on length and ensure constant radius and pressure.

Question 19

A physiology lab compares two arterioles with the same radius and viscosity but different lengths: Vessel X has length $L$ , Vessel Y has length $3L$ . Both experience the same pressure drop and exhibit laminar flow. What is the expected ratio $Q_Y/Q_X$ ?

$Q_Y/Q_X = 3$ because longer vessels allow more time for fluid to accelerate.
$Q_Y/Q_X = \tfrac{1}{3}$ because $Q \propto 1/L$ . (correct answer)
$Q_Y/Q_X = \tfrac{1}{9}$ because $Q \propto 1/L^2$ .
$Q_Y/Q_X = 1$ because length does not affect flow if radius is constant.

Question 20

A device measures flow through a rigid capillary for two conditions. Condition 1: $\Delta P_1$ , viscosity $\eta$ , radius $r$ , length $L$ , flow $Q_1$ . Condition 2: pressure drop is reduced to $0.80\Delta P_1$ and radius is increased to $1.10r$ with $\eta$ and $L$ unchanged (laminar). Which prediction for $Q_2/Q_1$ is most consistent with Poiseuille flow?

$Q_2/Q_1 \approx 0.80 \times (1.10)^4 \approx 1.17$ . (correct answer)
$Q_2/Q_1 \approx 0.80 \times (1.10)^2 \approx 0.97$ .
$Q_2/Q_1 \approx 0.80/(1.10)^4 \approx 0.55$ .
$Q_2/Q_1 \approx 1.10/0.80 \approx 1.38$ because velocity increases with radius.

Explanation: This question tests understanding of viscosity and Poiseuille flow, a key concept in fluid dynamics. Poiseuille's law describes how flow rate is affected by parameters like viscosity, tube length, and radius, combining multiple factors. In this scenario, reducing ΔP to 0.80 and increasing r to 1.10 alters Q via Q ∝ ΔP r^4. Choice A is correct because it accurately calculates ≈ 0.80 * (1.10)^4 ≈ 1.17. Choice C fails because it inverts the radius factor incorrectly. In similar questions, compute the product of all factors and verify laminar conditions.