Density and pressure

1. Overview

Density and pressure are intrinsic macroscopic properties used to characterize the state and behavior of matter. Density describes the distribution of mass within a given volume, serving as a measure of how tightly matter is packed. Pressure describes the distribution of force over a surface area, specifically focusing on the component of force acting perpendicular to that surface. In the context of fluids (liquids and gases), these properties combine to explain hydrostatics. The increase of pressure with depth in a fluid leads directly to the phenomenon of upthrust, a resultant upward force that governs whether objects sink or float, as formalized by Archimedes’ Principle.

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2. Key Definitions

• Density ($\rho$): The mass per unit volume of a substance. It is a scalar quantity and an intrinsic property of a material, independent of the amount of substance present.
• Pressure ($p$): The normal force acting per unit cross-sectional area. "Normal" implies that only the component of force acting at $90^\circ$ to the surface contributes to pressure.
• Hydrostatic Pressure: The pressure exerted by a fluid at rest at a specific point, caused by the weight of the fluid column acting vertically above that point.
• Upthrust ($U$): The resultant upward force exerted by a fluid on a body submerged or floating in it. This force arises specifically due to the difference in hydrostatic pressure between the upper and lower surfaces of the body.
• Archimedes’ Principle: The upthrust acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

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3. Content

3.1 Density ($\rho$)

Density provides a way to compare different materials. For a uniform substance, density is constant regardless of the object's shape.

The Density Equation: $$\mathbf{\rho = \frac{m}{V}}$$ (This equation must be memorised)

Variables and Units:

• $\rho$ = density (SI unit: $\text{kg m}^{-3}$)
• $m$ = mass (SI unit: $\text{kg}$)
• $V$ = volume (SI unit: $\text{m}^3$)

Unit Conversions (Critical for Exams): In A-Level problems, units are often given in $\text{g cm}^{-3}$. You must convert these to SI units for consistency in calculations involving $g$ ($9.81 \text{ m s}^{-2}$).

• To convert $\text{g cm}^{-3}$ to $\text{kg m}^{-3}$: Multiply by $1000$.
  • Reasoning: $1 \text{ g} = 10^{-3} \text{ kg}$ and $1 \text{ cm}^3 = (10^{-2} \text{ m})^3 = 10^{-6} \text{ m}^3$.
  • $\frac{10^{-3} \text{ kg}}{10^{-6} \text{ m}^3} = 10^3 \text{ kg m}^{-3} = 1000 \text{ kg m}^{-3}$.
• To convert $\text{cm}^3$ to $\text{m}^3$: Multiply by $10^{-6}$.
• To convert $\text{mm}^3$ to $\text{m}^3$: Multiply by $10^{-9}$.

Density of Mixtures: If two substances are mixed, the density of the mixture ($\rho_{mix}$) is the total mass divided by the total volume: $$\rho_{mix} = \frac{m_1 + m_2}{V_1 + V_2}$$

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3.2 Pressure ($p$)

Pressure is a scalar quantity, even though it is derived from force (a vector). In fluids, pressure acts equally in all directions at a specific depth.

The Pressure Equation: $$\mathbf{p = \frac{F}{A}}$$ (This equation must be memorised)

Variables and Units:

• $p$ = pressure (SI unit: $\text{Pascal (Pa)}$ or $\text{N m}^{-2}$)
• $F$ = Normal force ($\text{N}$)
• $A$ = Cross-sectional area ($\text{m}^2$)

The "Normal Force" Requirement: If a force $F$ acts at an angle $\theta$ to the normal (the perpendicular line to the surface), the pressure is: $$p = \frac{F \cos \theta}{A}$$ If the angle $\theta$ is given relative to the surface itself, the pressure is: $$p = \frac{F \sin \theta}{A}$$ Always identify the component of the force that is perpendicular to the area.

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3.3 Derivation of Hydrostatic Pressure ($\Delta p = \rho g \Delta h$)

The syllabus requires you to derive the expression for the pressure exerted by a vertical column of liquid.

1. Consider a cylinder of liquid with a horizontal cross-sectional area $A$ and a vertical height $h$.
2. Volume ($V$) of the liquid column: $$V = A \times h$$
3. Mass ($m$) of the liquid column (using the density $\rho$ of the liquid): $$m = \rho \times V = \rho A h$$
4. Weight ($W$) of the liquid column: The weight is the force $F$ acting downwards on the base of the cylinder. $$F = W = m \times g = (\rho A h)g$$
5. Pressure ($p$) at the base: Substitute the force into the definition of pressure: $$p = \frac{F}{A} = \frac{\rho A h g}{A}$$
6. Simplify: The area $A$ cancels out, demonstrating that hydrostatic pressure is independent of the shape or area of the container. $$\mathbf{p = \rho g h}$$
7. Change in Pressure: For a change in depth $\Delta h$, the change in pressure $\Delta p$ is: $$\mathbf{\Delta p = \rho g \Delta h}$$ (This equation is provided on the Data Sheet)

Note on Total Pressure: If a liquid is exposed to the atmosphere, the total pressure at depth $h$ is the sum of the atmospheric pressure ($p_0 \approx 1.01 \times 10^5 \text{ Pa}$) and the hydrostatic pressure: $$p_{total} = p_0 + \rho g h$$

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3.4 Upthrust and the Origin of the Buoyant Force

Upthrust is not a "new" fundamental force; it is the result of the pressure gradient in a fluid.

The Mechanism:

1. Pressure in a fluid increases with depth ($p = \rho g h$).
2. Consider a submerged rectangular block. The bottom surface is at a greater depth than the top surface.
3. Therefore, the upward pressure on the bottom surface is greater than the downward pressure on the top surface.
4. Since $F = pA$, the upward force on the bottom is greater than the downward force on the top.
5. The resultant of these two forces is the Upthrust.

Deriving the Upthrust Equation ($F = \rho g V$):

1. Let a block of height $h$ and area $A$ be submerged in a fluid of density $\rho_f$.
2. Top surface depth = $h_{top}$; Bottom surface depth = $h_{bottom}$.
3. Pressure at top: $p_1 = \rho_f g h_{top}$.
4. Pressure at bottom: $p_2 = \rho_f g h_{bottom}$.
5. Force at top (downwards): $F_1 = p_1 A = \rho_f g h_{top} A$.
6. Force at bottom (upwards): $F_2 = p_2 A = \rho_f g h_{bottom} A$.
7. Upthrust ($U$) = $F_2 - F_1$: $$U = \rho_f g A (h_{bottom} - h_{top})$$
8. Since $(h_{bottom} - h_{top}) = h$ (the height of the block) and $A \times h = V$ (the volume of the block): $$\mathbf{U = \rho_f g V}$$

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3.5 Archimedes’ Principle

Archimedes' Principle states that the upthrust is equal to the weight of the displaced fluid.

The Equation: $$\mathbf{F = \rho g V}$$ (This equation must be memorised)

Crucial Distinctions:

• $\rho$ is the density of the fluid, not the object.
• $V$ is the volume of the fluid displaced.
  • If the object is fully submerged: $V_{displaced} = V_{object}$.
  • If the object is partially submerged (floating): $V_{displaced} = \text{Volume of the part of the object below the surface}$.

The Principle of Flotation: For an object to float in equilibrium: $$\text{Upthrust} = \text{Weight of the object}$$ $$\rho_{fluid} g V_{displaced} = \rho_{object} g V_{total}$$ This leads to the ratio: $$\frac{V_{displaced}}{V_{total}} = \frac{\rho_{object}}{\rho_{fluid}}$$

• If $\rho_{object} < \rho_{fluid}$, the object floats partially submerged.
• If $\rho_{object} = \rho_{fluid}$, the object is neutrally buoyant (floats fully submerged).
• If $\rho_{object} > \rho_{fluid}$, the object sinks (Weight > Upthrust).

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Worked Example 1 — Hydrostatic Pressure on a Diver

A diver is swimming in a lake (density of water = $1000 \text{ kg m}^{-3}$). The atmospheric pressure at the surface is $1.0 \times 10^5 \text{ Pa}$. Calculate the total pressure acting on the diver at a depth of $25 \text{ m}$.

Step 1: Identify the components of total pressure. $$p_{total} = p_{atm} + p_{hydrostatic}$$

Step 2: Calculate the hydrostatic pressure. $$p = \rho g h$$ $$p = 1000 \times 9.81 \times 25$$ $$p = 245,250 \text{ Pa}$$

Step 3: Add atmospheric pressure. $$p_{total} = (1.0 \times 10^5) + 245,250$$ $$p_{total} = 100,000 + 245,250 = 345,250 \text{ Pa}$$

Step 4: Final answer to appropriate significant figures. $$p_{total} = 3.5 \times 10^5 \text{ Pa} \text{ (to 2 s.f.)}$$

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Worked Example 2 — Upthrust and Tension

A solid metal sphere of radius $3.0 \text{ cm}$ and density $8000 \text{ kg m}^{-3}$ is suspended by a thin wire so that it is completely submerged in oil of density $850 \text{ kg m}^{-3}$. Calculate the tension in the wire.

Step 1: Calculate the volume of the sphere. $$V = \frac{4}{3} \pi r^3$$ $$V = \frac{4}{3} \pi (0.030)^3 = 1.131 \times 10^{-4} \text{ m}^3$$

Step 2: Calculate the weight of the sphere ($W$). $$W = m_{sphere} g = (\rho_{sphere} V) g$$ $$W = 8000 \times (1.131 \times 10^{-4}) \times 9.81 = 8.877 \text{ N}$$

Step 3: Calculate the upthrust ($U$). Note: Use the density of the oil. $$U = \rho_{oil} g V$$ $$U = 850 \times 9.81 \times (1.131 \times 10^{-4}) = 0.943 \text{ N}$$

Step 4: Use the equilibrium condition. The forces acting on the sphere are Tension ($T$) upwards, Upthrust ($U$) upwards, and Weight ($W$) downwards. $$T + U = W$$ $$T = W - U$$ $$T = 8.877 - 0.943 = 7.934 \text{ N}$$

Step 5: Final answer. $$T = 7.9 \text{ N} \text{ (to 2 s.f.)}$$

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4. Key Equations

Equation	Symbols	SI Units	Status
$\rho = \frac{m}{V}$	$\rho$: density, $m$: mass, $V$: volume	$\text{kg m}^{-3}, \text{kg}, \text{m}^3$	Memorise
$p = \frac{F}{A}$	$p$: pressure, $F$: normal force, $A$: area	$\text{Pa}, \text{N}, \text{m}^2$	Memorise
$\Delta p = \rho g \Delta h$	$\Delta p$: pressure change, $g$: $9.81 \text{ m s}^{-2}$	$\text{Pa}$	Data Sheet
$F = \rho g V$	$F$: Upthrust, $V$: volume displaced	$\text{N}, \text{m}^3$	Memorise

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5. Common Mistakes to Avoid

• ❌ Wrong: Using the density of the object ($\rho_s$) to calculate upthrust.

• ✅ Right: Always use the density of the fluid ($\rho_f$) in the upthrust equation ($F = \rho_f g V$). Upthrust is about the fluid being pushed out of the way.

• ❌ Wrong: Forgetting that $1 \text{ cm}^2$ is $10^{-4} \text{ m}^2$.

• ✅ Right: Area is a squared dimension. $(10^{-2} \text{ m})^2 = 10^{-4} \text{ m}^2$. Similarly, $1 \text{ cm}^3 = 10^{-6} \text{ m}^3$.

• ❌ Wrong: Using the total height of an object in $p = \rho g h$ when calculating pressure at the top surface.

• ✅ Right: $h$ is the depth from the surface of the fluid to the point in question.

• ❌ Wrong: Assuming upthrust only acts on floating objects.

• ✅ Right: Upthrust acts on every object in a fluid, whether it is sinking, rising, or suspended. It only depends on the volume of fluid displaced.

• ❌ Wrong: Ignoring atmospheric pressure when asked for "total" or "absolute" pressure.

• ✅ Right: If the question mentions the surface is open to the air, add $1.01 \times 10^5 \text{ Pa}$ to your hydrostatic calculation.

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6. Exam Tips

1. The Derivation Question: If asked to derive $\Delta p = \rho g \Delta h$, you must show the intermediate steps: $V = Ah$, then $m = \rho Ah$, then $W = \rho Ahg$. Simply writing $p = F/A = \rho gh$ will lose marks for lack of detail.
2. Free-Body Diagrams: For upthrust problems, always draw the object and label the forces ($W$ down, $U$ up, and any $T$ or $F_{applied}$). This prevents sign errors in equilibrium equations.
3. Significant Figures: Cambridge 9702 usually provides data to 2 or 3 significant figures. Your final answer should match the precision of the data provided (usually 2 s.f. is safe, but 3 s.f. is often preferred if the data allows).
4. The Value of $g$: Always use $g = 9.81 \text{ m s}^{-2}$ as stated on the data sheet. Using $10 \text{ m s}^{-2}$ will result in a rounding error and loss of marks.
5. Check the "Normal": In pressure questions involving slopes or angled forces, double-check if the angle is with the horizontal or the vertical. Draw a small triangle to confirm you are using $\sin$ or $\cos$ correctly to find the perpendicular component.
6. Upthrust is a Force: Remember that Upthrust is a force measured in Newtons. If a question asks for the "mass of fluid displaced," you must divide the upthrust by $g$.