Density and Pressure in Fluids — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Density definition and calculation, hydrostatic pressure variation with depth, gauge vs absolute pressure, manometer analysis, and Pascal’s principle for hydraulic systems, with core problem-solving techniques for AP exam questions.

You should already know: Basic definitions of mass, volume, and SI units; proportional reasoning for algebraic relationships; definitions of force and pressure for solid objects.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Density and Pressure in Fluids?

Fluids are substances that flow (liquids and gases), and this topic covers the fundamental properties of static (non-moving) fluids that underpin all other fluid concepts in AP Physics 1. Density describes how much mass is packed into a given volume of fluid (or solid), while pressure describes the perpendicular force per unit area exerted by a fluid on any surface it contacts. According to the AP Physics 1 CED, this topic makes up roughly 30% of Unit 8 (Fluids and Thermal Physics), translating to 2-4% of total exam score. Concepts from this topic appear in both multiple-choice (MCQ) and as a foundational component of longer free-response questions (FRQ), often combined with force balance or buoyancy. AP Physics 1 does not require analysis of dynamic fluid pressure, so all problems in this topic focus on static fluids at equilibrium.

2. Density

Density is defined as the ratio of an object’s mass to its total volume, with the standard symbol $\rho$ (Greek rho) and SI units of kilograms per cubic meter (${\text{kg/m}}^3$). The core formula is: $$\rho =\frac{m}{V}$$ Common reference values you should memorize for AP problems: pure water has a density of $1000{\text{kg/m}}^3$, which is an easy number for proportional reasoning problems. For composite objects (like a hollow sphere or mixed material), we calculate average density, which is total mass of the object divided by total volume (including any hollow space or different materials). Density is an intensive property, meaning it does not depend on how much of the material you have: a small chip of aluminum has the same density as a large block of aluminum. AP Physics 1 frequently tests proportional reasoning for density: for example, if two objects have the same mass, density is inversely proportional to volume; if they have the same volume, density is directly proportional to mass.

Worked Example

A hollow aluminum sphere has an outer radius of $0.10\text{m}$ and a total mass of $2.7\text{kg}$. Solid aluminum has a density of $2700{\text{kg/m}}^3$. What is the radius of the hollow cavity inside the sphere?

1. First find the volume of aluminum used to make the sphere, rearranging the density formula: $V_{\text{Al}}=\frac{m}{{\rho }_{\text{Al}}}=\frac{2.7\text{kg}}{2700{\text{kg/m}}^3}=0.001{\text{m}}^3$.
2. Calculate the total outer volume of the sphere: $V_{\text{outer}}=\frac{4}{3}\pi R_{\text{outer}}^3=\frac{4}{3}\pi (0.10\text{m}{)}^3\approx 0.00419{\text{m}}^3$.
3. The volume of the hollow cavity is the difference between total outer volume and aluminum volume: $V_{\text{cavity}}=V_{\text{outer}}-V_{\text{Al}}=0.00419-0.001=0.00319{\text{m}}^3$.
4. Solve for the cavity radius $r$: $r={\left(\frac{3V_{\text{cavity}}}{4\pi }\right)}^{1/3}={\left(\frac{3(0.00319)}{4\pi }\right)}^{1/3}\approx 0.092\text{m}$.

Exam tip: On proportional reasoning MCQs, cancel all constants before plugging in numbers. For example, if you are asked for the ratio of densities of two objects, you can cancel $\pi$, $g$, and any common constants to simplify the calculation and save time.

3. Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a static fluid at a given depth, caused by the weight of the fluid above the point of interest. We can derive the core formula by considering a horizontal area $A$ at depth $h$ below the fluid surface. The mass of fluid above this area is $m=\rho V=\rho Ah$, so the weight (force exerted on the area) is $F=mg=\rho Ahg$. Pressure is force per unit area, so the pressure change from the surface is: $$\Delta P=\frac{F}{A}=\rho gh$$ A key result here is that hydrostatic pressure only depends on depth, fluid density, and $g$ — it does not depend on the shape of the container (the "hydrostatic paradox"). This is because a wider container has more total weight of fluid, but that weight is distributed over a larger area, so pressure per unit area stays the same at equal depth. We distinguish between two types of pressure: gauge pressure is pressure relative to atmospheric pressure ($P_{\text{gauge}}=\rho gh$), and absolute pressure is total pressure including atmospheric pressure at the surface ($P_{\text{abs}}=P_{\text{atm}}+P_{\text{gauge}}$). Pressure is equal at the same horizontal depth in a static fluid, which is the core principle for solving manometer problems.

Worked Example

A U-tube manometer has one open end exposed to the atmosphere, and the other end connected to a sealed tank of compressed gas. Mercury (density $13600{\text{kg/m}}^3$) rests $0.25\text{m}$ higher on the open end side than on the gas-tank side. Atmospheric pressure is $1.013\times 10^5\text{Pa}$. What is the absolute pressure of the gas in the tank?

1. For static fluids, pressure at the same horizontal level is equal. We take the horizontal level of the mercury surface on the gas side.
2. Pressure at this level from the gas side equals the gas pressure $P_{\text{gas}}$, and pressure from the open side equals atmospheric pressure plus the pressure from the $0.25\text{m}$ mercury column: $P_{\text{gas}}=P_{\text{atm}}+\rho gh$.
3. Calculate the gauge pressure from the mercury column: $\rho gh=(13600)(9.8)(0.25)\approx 3.33\times 10^4\text{Pa}$.
4. Add to atmospheric pressure to get absolute pressure: $P_{\text{gas}}=1.013\times 10^5+3.33\times 10^4\approx 1.35\times 10^5\text{Pa}$.

Exam tip: Always circle whether the question asks for gauge or absolute pressure. AP exam writers intentionally set traps where the wrong pressure type is a common incorrect answer choice.

4. Pascal's Principle

Pascal's principle states that any change in pressure applied to an enclosed, incompressible fluid is transmitted undiminished to every point in the fluid and to the walls of the container. This principle is the basis for hydraulic systems (like car lifts and brake lines), which are common AP problems. For a hydraulic system with two pistons, the pressure change is the same at both pistons, so: $$\frac{F_1}{A_1}=\frac{F_2}{A_2}$$ where $F_1,A_1$ are the force and area of the input piston, and $F_2,A_2$ are the force and area of the output piston. This means a small input force on a small piston creates a large output force on a large piston — it acts as a force multiplier, similar to a lever. Work is still conserved: the small piston moves a much larger distance than the large piston, so input work equals output work (ignoring friction). Pascal's principle only applies to incompressible fluids (liquids, generally), as gases compress and do not transmit pressure changes undiminished.

Worked Example

A hydraulic lift used to raise cars has a small input piston with cross-sectional area $0.005{\text{m}}^2$ and a large output piston with cross-sectional area $0.4{\text{m}}^2$. The output piston supports a $2000\text{kg}$ car. What is the minimum input force required to hold the car stationary?

1. Pascal's principle gives equal pressure at both pistons, so $\frac{F_1}{A_1}=\frac{F_2}{A_2}$.
2. The output force $F_2$ is the weight of the car: $F_2=mg=(2000)(9.8)=19600\text{N}$.
3. Rearrange to solve for input force: $F_1=F_2\cdot \frac{A_1}{A_2}=19600\cdot \frac{0.005}{0.4}=19600\cdot 0.0125=245\text{N}\approx 250\text{N}$.
4. Check for reasonableness: this small input force (equivalent to the weight of a 25 kg object) lifting a 2000 kg car matches how hydraulic lifts are designed to work.

Exam tip: If you are given diameters (or radii) instead of areas for hydraulic pistons, remember that area is proportional to the square of diameter, so the area ratio is $(d_2/d_1{)}^2$. Never use the diameter ratio directly.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the diameter ratio directly instead of squaring it to get the area ratio for Pascal's principle. Why: Students confuse linear and area proportionality, and forget that area depends on the square of linear dimensions. Correct move: Always write $A=\pi (d/2{)}^2$, so the $\pi$ and $1/4$ cancel, leaving $A_2/A_1=(d_2/d_1{)}^2$ — confirm this ratio before solving.
• Wrong move: Forgetting to add atmospheric pressure to gauge pressure when asked for absolute pressure. Why: Most problems default to gauge pressure for fluid columns, so students develop a habit of only calculating $P=\rho gh$ without checking the question. Correct move: Circle the words "gauge" or "absolute" in the question, and write $P=P_{\text{atm}}+\rho gh$ explicitly for absolute pressure before starting calculations.
• Wrong move: Using the height of a submerged object instead of the depth of the point of interest below the surface when calculating hydrostatic pressure. Why: Students mix up object height and depth, and assume the object's own dimension is the $h$ in $P=\rho gh$. Correct move: Label $h$ on your diagram as the distance from the fluid surface to the point you are analyzing, and confirm it is depth, not an object dimension.
• Wrong move: Calculating the density of a hollow object's material by dividing total mass by total outer volume including the hollow cavity. Why: Students confuse average density of the whole object with density of the material it is made from. Correct move: Check whether the question asks for material density or average density. For material density, only use the volume of the material itself, not the total outer volume of the object.
• Wrong move: Assuming pressure at the same depth is higher in a wider container because it holds more total mass of fluid. Why: Intuition about total weight overrides the definition of pressure as force per unit area. Correct move: Remember hydrostatic pressure only depends on depth, density, and $g$ — container shape has no effect, so always use $P=\rho gh$ regardless of container width.
• Wrong move: Using $g=10{\text{m/s}}^2$ when the problem expects $9.8{\text{m/s}}^2$, leading to an answer that does not match the MCQ options. Why: Students memorize one value for $g$ and do not check the problem's convention. Correct move: Always use $g=9.8{\text{m/s}}^2$ unless the problem explicitly states to use $10{\text{m/s}}^2$ for approximation.

6. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

Two vertical cylinders are filled with water to the same depth $h$. Cylinder A has twice the diameter of Cylinder B. What is the ratio of the total hydrostatic force on the bottom of Cylinder A to the total hydrostatic force on the bottom of Cylinder B? A) $1:1$ B) $2:1$ C) $4:1$ D) $8:1$

Worked Solution: Hydrostatic pressure at the bottom is $P=\rho gh$, which is identical for both cylinders because depth $h$, density $\rho$, and $g$ are the same. Total force on the bottom is $F=P\cdot A$, where $A$ is the area of the bottom. Area of a circular cylinder is proportional to the square of diameter, so $A_A=(2{)}^2{A}_B=4A_B$. Substituting, $F_A=P\cdot 4A_B=4F_B$, so the ratio is $4:1$. The common mistake is confusing pressure ratio and force ratio, leading to answer A. Correct answer: C.

Question 2 (Free Response)

A student measures an irregularly shaped chunk of unknown solid, finding its mass is 45 g. The student submerges the chunk in a graduated cylinder of water, and observes the water level rises from 25.0 mL to 41.5 mL. (a) Calculate the density of the solid. (b) The student claims the solid is pure gold, which has a density of $19300{\text{kg/m}}^3$. Is the student's claim supported? Justify your answer. (c) The solid is actually a composite of a low-density aluminum core surrounded by higher-density gold plating. Does the average density of the composite increase or decrease as the thickness of the gold plating increases? Justify your answer.

Worked Solution: (a) The volume of the solid equals the change in water volume: $V=41.5\text{mL}-25.0\text{mL}=16.5\text{mL}=16.5\times 10^{-6}{\text{m}}^3$. Mass $m=45\text{g}=0.045\text{kg}$. Density $\rho =\frac{m}{V}=\frac{0.045}{16.5\times 10^{-6}}\approx 2700{\text{kg/m}}^3$. (b) The calculated density of ~$2700{\text{kg/m}}^3$ is much lower than the density of pure gold ($19300{\text{kg/m}}^3$), so the student's claim is not supported. (c) The average density of the composite increases. Average density is total mass divided by total volume. Adding more high-density gold increases the total mass proportionally more than it increases total volume, so the average density of the composite approaches the density of gold as plating thickness increases.

Question 3 (Application / Real-World Style)

Atmospheric pressure decreases with increasing altitude. A rough approximation models near-surface air as a static fluid with constant density of $1.2{\text{kg/m}}^3$. Atmospheric pressure at sea level is $1.013\times 10^5\text{Pa}$. Use this model to estimate the atmospheric pressure at the top of Mount Elbrus, which is 5642 m above sea level. What does your result suggest about the constant density assumption for air?

Worked Solution: Pressure change with altitude is $\Delta P=-\rho gh$, where $h$ is height above sea level. Calculate $\Delta P=-(1.2{\text{kg/m}}^3)(9.8{\text{m/s}}^2)(5642\text{m})\approx -6.6\times 10^4\text{Pa}$. Pressure at the summit is $P=P_{\text{sea}}+\Delta P=1.013\times 10^5-6.6\times 10^4=3.5\times 10^4\text{Pa}$. The actual measured pressure at the summit is ~$5.0\times 10^4\text{Pa}$, so the model underestimates pressure. This shows the constant density assumption for air is incorrect: air density decreases with increasing altitude, so the actual pressure drop with height is smaller than the constant density model predicts.

7. Quick Reference Cheatsheet

8. What's Next

This chapter lays the foundation for all other fluid concepts in AP Physics 1 Unit 8. Next, you will apply density and pressure concepts to understand buoyancy and Archimedes' principle, which is one of the most heavily tested topics in the fluid unit. Without a solid grasp of hydrostatic pressure and density, you cannot correctly derive the buoyant force or solve force balance problems for submerged or floating objects, which frequently appear in both MCQ and FRQ sections. This topic also connects to thermal physics concepts later in the unit, where density changes due to thermal expansion explain convection currents and pressure changes in gases. It also reinforces force and Newton's laws from earlier in the course, as most static fluid problems rely on force balance analysis.

Buoyancy and Archimedes' Principle Thermal Expansion and Ideal Gases Static Equilibrium and Force Balance