Fluid Properties and Hydrostatics (4B)
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MCAT Chemical and Physical Foundations of Biological Systems › Fluid Properties and Hydrostatics (4B)
A rigid object is suspended from a force sensor and lowered into a beaker of ethanol ($\rho = 0.79,\text{g/mL}$) until it is fully submerged. The object’s volume is $V = 60,\text{mL}$. Take $g = 9.8,\text{m/s}^2$. Which prediction aligns with buoyancy regarding the force sensor reading after submersion (relative to in air)?
The reading decreases only if the object’s density is less than ethanol.
The reading is unchanged because buoyant force exists only for floating objects.
The reading decreases by $\rho g V$ because the fluid exerts an upward buoyant force.
The reading increases by $\rho g V$ because the fluid adds downward pressure.
Explanation
This question tests how buoyancy affects force measurements. When the object is submerged, it experiences an upward buoyant force F_B = ρ_fluid × V × g = (0.79 g/mL)(60 mL)(9.8 m/s²) = (0.79 × 10³ kg/m³)(60 × 10⁻⁶ m³)(9.8 m/s²) = 0.465 N. This buoyant force reduces the apparent weight, so the sensor reading decreases by this amount. Choice C incorrectly claims buoyancy exists only for floating objects, but all submerged objects experience buoyant force regardless of whether they sink or float. To find the change in force sensor reading, calculate the buoyant force and subtract it from the object's weight in air.
A rigid cube of side length $3.0,\text{cm}$ is fully submerged in a tank containing a uniform fluid with density $\rho = 1200,\text{kg/m}^3$. Take $g = 9.8,\text{m/s}^2$. Which expression is most consistent with the buoyant force on the cube?
$F_B = \rho g (3.0,\text{cm})^3$
$F_B = (\rho g)/(3.0,\text{cm})^3$
$F_B = \rho g (3.0,\text{cm})$
$F_B = \rho g (3.0,\text{cm})^2$
Explanation
This question tests the mathematical expression for buoyant force. The buoyant force equals the weight of displaced fluid: F_B = ρ_fluid × V_displaced × g. For a cube with 3.0 cm sides, the volume is V = (3.0 cm)³ = 27 cm³ = 27 × 10⁻⁶ m³. Therefore, F_B = ρg(3.0 cm)³, where proper unit conversion is implied. Choice A incorrectly uses linear dimension instead of volume, while choice B uses area instead of volume. When calculating buoyant force, always use the three-dimensional volume of the displaced fluid, not linear or area measurements.
A diver carries a rigid, sealed instrument case with a small trapped air volume. The diver descends in seawater. Assume the case does not change volume. Which prediction aligns with hydrostatics for the net buoyant force on the case as depth increases (uniform seawater density)?
It remains approximately constant because displaced volume and fluid density are unchanged.
It increases because hydrostatic pressure increases with depth.
It becomes zero at sufficient depth because pressure equalizes on all sides.
It decreases because hydrostatic pressure compresses the surrounding water.
Explanation
This question tests buoyant force on rigid objects at varying depths. For a rigid, sealed case that maintains constant volume, the buoyant force F_B = ρ_fluid × V_case × g remains constant as long as the fluid density is uniform. The displaced volume doesn't change because the case is rigid, and seawater density is approximately uniform over moderate depth ranges. Choice A incorrectly suggests buoyant force increases with the increasing pressure at depth, but pressure affects only compressible objects' volumes. For incompressible objects in uniform fluids, buoyant force is independent of depth—only the surrounding pressure increases.
In a benchtop demonstration, a small rubber balloon is fully submerged in a tank of water and held at depth. The balloon material is flexible, and the gas inside can be compressed. Which prediction aligns with hydrostatics for the balloon’s volume when moved to a greater depth (assume temperature constant)?
The balloon volume decreases because external pressure increases with depth.
The balloon volume increases because buoyant force increases with depth.
The balloon volume decreases only if the water density decreases with depth.
The balloon volume is unchanged because water is incompressible.
Explanation
This question tests the effect of hydrostatic pressure on compressible objects. As depth increases, external pressure on the balloon increases according to P = P₀ + ρgh. This increased external pressure compresses the gas inside the balloon, reducing its volume according to Boyle's law (PV = constant at constant temperature). The balloon shrinks until internal pressure matches the higher external pressure. Choice B incorrectly suggests volume increases with depth, which would require internal pressure to somehow decrease. For flexible containers with compressible contents, increasing depth always causes volume reduction due to the higher surrounding pressure.
A sealed syringe filled with saline is connected to a second syringe by rigid tubing, forming a closed hydraulic system. Syringe 1 has plunger area $A_1 = 2.0,\text{cm}^2$ and Syringe 2 has plunger area $A_2 = 0.5,\text{cm}^2$. A force is applied to Syringe 1 to generate a static pressure increase of $\Delta P = 40,\text{kPa}$ in the fluid. Based on Pascal’s principle, which outcome is most likely for Syringe 2?
Syringe 2 experiences a larger $\Delta P$ because pressure concentrates in narrower plungers.
Syringe 2 experiences the same $\Delta P$ and thus a force $F_2 = \Delta P,A_2$.
Syringe 2 experiences no pressure change because the tubing is rigid.
Syringe 2 experiences a smaller $\Delta P$ because pressure decreases with smaller area.
Explanation
This question tests Pascal's principle in a closed hydraulic system. Pascal's principle states that pressure changes are transmitted equally throughout an incompressible fluid. The pressure increase of 40 kPa in Syringe 1 is transmitted unchanged to Syringe 2. The force on Syringe 2's plunger is F₂ = ΔP × A₂ = (40 × 10³ Pa)(0.5 × 10⁻⁴ m²) = 2.0 N. Choice C incorrectly suggests pressure concentrates in narrower plungers, but pressure is a scalar quantity that doesn't depend on geometry in static fluids. In hydraulic systems, pressure is constant throughout, but forces vary proportionally with piston areas, creating mechanical advantage.
A lab models pulmonary edema by measuring pressure at two heights in a static column of fluid in a transparent tube. The tube contains a uniform fluid ($\rho = 1000,\text{kg/m}^3$). Point A is $15,\text{cm}$ above Point B. Take $g = 9.8,\text{m/s}^2$. Which statement best describes the behavior of the system under the given conditions?
Pressure at A is lower than at B by $\rho g(0.15,\text{m})$.
Pressure at A is higher than at B by $\rho g(0.15,\text{m})$.
Pressure at A equals pressure at B because the tube is narrow.
Pressure difference depends on the tube’s cross-sectional area, not height.
Explanation
This question tests hydrostatic pressure variation in a vertical fluid column. In a static fluid, pressure decreases with height according to P = P₀ - ρgh when moving upward. Since Point A is 15 cm (0.15 m) above Point B, the pressure at A is lower by ΔP = ρgh = (1000 kg/m³)(9.8 m/s²)(0.15 m) = 1,470 Pa. This occurs because there is less fluid weight above Point A compared to Point B. Choice C incorrectly states that tube narrowness affects pressure distribution, but in static fluids, pressure depends only on depth, not on container geometry. When comparing pressures at different heights, always identify which point is higher and apply the rule that pressure decreases with elevation.
A tissue-engineering bioreactor uses a closed reservoir connected to a vertical viewing tube. The fluid is static and the tube is open to the atmosphere at its top. When the reservoir pressure is increased, the fluid level in the tube rises by $\Delta h$. Which statement best describes the behavior of the system under the given conditions, based on hydrostatics?
The reservoir gauge pressure increase equals $\rho g\Delta h$.
The fluid rises because buoyant force increases when reservoir pressure increases.
The rise depends only on tube diameter because narrower tubes amplify pressure.
The reservoir absolute pressure increase equals $\rho g\Delta h$ regardless of atmospheric pressure.
Explanation
This question tests the relationship between reservoir pressure and manometer readings. When reservoir gauge pressure increases, the fluid in the open tube must rise to balance this pressure increase. The height change Δh creates a hydrostatic pressure ρgΔh that exactly equals the reservoir pressure increase. This is because at equilibrium, pressure at any horizontal level must be the same throughout the connected system. Choice C incorrectly suggests the effect depends on tube diameter, but hydrostatic pressure depends only on height and fluid density, not cross-sectional area. Open-tube manometers provide direct visual indication of pressure changes through height changes.
A field test estimates lake depth using a pressure probe. At the surface, the probe reads atmospheric pressure. At a depth where the gauge pressure is $49,\text{kPa}$ in freshwater ($\rho = 1000,\text{kg/m}^3$), take $g = 9.8,\text{m/s}^2$. Which depth is most consistent with the measurement (ignore small density changes)?
$\approx 50,\text{m}$
$\approx 5.0,\text{m}$
$\approx 500,\text{m}$
$\approx 0.50,\text{m}$
Explanation
This question tests depth calculation from pressure measurements. Using the hydrostatic pressure formula P = ρgh, we solve for depth: h = P/(ρg) = 49,000 Pa / [(1000 kg/m³)(9.8 m/s²)] = 49,000 / 9,800 = 5.0 m. This depth corresponds to the gauge pressure reading, measuring pressure relative to atmospheric at the surface. Choice C (50 m) would produce a pressure of 490 kPa, ten times larger than measured. When calculating depth from pressure, ensure you're using gauge pressure (not absolute) and check that the calculated pressure matches the measurement order of magnitude.
A static fluid-filled chamber contains a narrow vertical capillary connected to a wide reservoir; both contain the same fluid and are open to air. Surface tension effects are neglected. Which statement best describes the behavior of the system under the given conditions, based on hydrostatics?
The fluid levels match because pressure at equal depths must be equal in connected static fluids.
The fluid is higher in the reservoir because larger area increases hydrostatic pressure.
The fluid rises higher in the capillary because pressure is greater in narrower tubes.
The levels differ because Pascal’s principle applies only to gases.
Explanation
This question tests hydrostatic equilibrium in connected vessels. In static, connected fluids open to the same atmospheric pressure, the fluid level must be the same in all parts regardless of vessel shape or size. This occurs because pressure at any given horizontal level must be uniform throughout the connected system. If levels differed, there would be a pressure gradient causing flow until equilibrium is reached. Choice B incorrectly suggests pressure is greater in narrower tubes, confusing this with capillary action (which requires surface tension). In the absence of surface tension effects, hydrostatic principles alone determine that connected static fluids reach the same level.
A vertical column contains two immiscible layers at rest: oil on top ($\rho_o = 800,\text{kg/m}^3$) and water below ($\rho_w = 1000,\text{kg/m}^3$). The oil layer thickness is $0.20,\text{m}$ and the water layer thickness is $0.30,\text{m}$. Take $g = 9.8,\text{m/s}^2$. Which prediction aligns with hydrostatics for the gauge pressure at the bottom relative to the top surface (open to air)?
It is independent of density and equals $g(0.50,\text{m})$.
It equals $\rho_o g(0.30,\text{m}) + \rho_w g(0.20,\text{m})$ because layers swap contributions.
It equals $\rho_w g(0.50,\text{m})$ because only the denser fluid matters.
It equals $\rho_o g(0.20,\text{m}) + \rho_w g(0.30,\text{m})$.
Explanation
This question tests pressure calculation through multiple fluid layers. In stratified fluids, total pressure accumulates by summing contributions from each layer: P = P₀ + ρ₁gh₁ + ρ₂gh₂ + ... Starting from the top (atmospheric pressure), we add the oil layer contribution ρ_o × g × (0.20 m) and the water layer contribution ρ_w × g × (0.30 m). The total gauge pressure at the bottom is (800)(9.8)(0.20) + (1000)(9.8)(0.30) = 1,568 + 2,940 = 4,508 Pa. Choice C incorrectly swaps the layer thicknesses, which would give a different result. When working with layered fluids, carefully track which density corresponds to which layer thickness.