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MCAT · Chemical and Physical Foundations · Physics
Fluids is part of Physics in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include continuity and Bernoulli in a constriction, poiseuille flow and radius scaling, hydrostatic pressure, gauge and absolute and archimedes and floating objects.
Common traps: predicting higher pressure where the fluid is faster; scaling speed with radius instead of radius squared; ignoring the height term when the tube tilts.
Question 1
An incompressible fluid flows at 1 m/s through a pipe of cross-sectional area 6 cm². The pipe then changes to an area of 2 cm². What is the flow speed in the second section?
0.333 m/s
1 m/s
3 m/s
9 m/s
Answer: C. 3 m/s
The continuity equation A₁v₁ = A₂v₂ gives v₂ = (6)(1)/2 = 3 m/s. Speed changes inversely with area, not with its square root or square.
Question 2
Two alveoli with the same surface tension are connected through a common airway. One has a radius of 50 µm and the other 100 µm. According to the law of Laplace, P = 2γ/r, what tends to happen?
The larger alveolus has twice the pressure and empties into the smaller one.
Both have the same pressure, because they share an airway.
The smaller alveolus has half the pressure, so air flows into it.
The smaller alveolus has twice the pressure and tends to empty into the larger one.
Answer: D. The smaller alveolus has twice the pressure and tends to empty into the larger one.
Pressure is inversely proportional to radius, so halving r doubles P; air flows from high to low pressure, so the small alveolus empties into the large one, the instability that surfactant counters (D). (A) and (C) reverse the dependence on radius; (B) ignores the surface-tension pressure, which is what differs between the two.
Question 3
A dialysis-fluid reservoir, open to the air, drains through a small hole 45 cm below the fluid surface. How fast does the fluid leave the hole? (g = 10 m/s²)
2.1 m/s
3.0 m/s
4.5 m/s
9.0 m/s
Answer: B. 3.0 m/s
Torricelli's result is v = √(2gh) = √(2 × 10 × 0.45) = √9.0 = 3.0 m/s (B), independent of the fluid's density and the hole's size. 2.1 m/s (A) drops the factor of 2; 4.5 m/s (C) computes gh without the 2 or the square root; 9.0 m/s (D) is v² = 2gh reported as a speed.
Question 4
A technician mixes 300 mL of a contrast agent with a specific gravity of 1.40 and 100 mL of water. Assuming the volumes add, what is the density of the mixture?
1.1 g/mL
1.2 g/mL
1.3 g/mL
1.4 g/mL
Answer: C. 1.3 g/mL
Mass = 300 × 1.40 + 100 × 1.00 = 520 g in 400 mL, so ρ = 520/400 = 1.3 g/mL (C). 1.2 g/mL (B) averages the two densities without weighting by volume; 1.1 g/mL (A) swaps the weights, as if there were 300 mL of water; 1.4 g/mL (D) ignores the dilution.
Question 5
A dry bone sample weighs 3.0 N in air and 2.0 N when fully submerged in water. What is the density of the bone? (water: 1.0 × 10³ kg/m³)
670 kg/m³
1500 kg/m³
2000 kg/m³
3000 kg/m³
Answer: D. 3000 kg/m³
The buoyant force is 3.0 − 2.0 = 1.0 N, the weight of the displaced water, so the bone is 3.0/1.0 = 3 times as dense as water: 3000 kg/m³ (D). 1500 kg/m³ (B) divides the weight in air by the weight in water; 2000 kg/m³ (C) divides the submerged weight by the buoyant force; 670 kg/m³ (A) inverts the ratio of the two weights.
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