CIE AS & A Level Physics 4.3 Density and pressure Study Notes- 2025-2027 Syllabus
CIE AS & A Level Physics 4.3 Density and pressure Study Notes – New Syllabus
CIE AS & A Level Physics 4.3 Density and pressure Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics Study Notes syllabus with Candidates should be able to:
- define and use density
- define and use pressure
- derive, from the definitions of pressure and density, the equation for hydrostatic pressure Δp = ρgΔh
- use the equation Δp = ρgΔh
- understand that the upthrust acting on an object in a fluid is due to a difference in hydrostatic pressure
- calculate the upthrust acting on an object in a fluid using the equation F = ρgV (Archimedes’ principle)
Definition and Use of Density
The density of a substance is defined as its mass per unit volume.
\( \mathrm{\rho = \dfrac{m}{V}} \)
- \( \mathrm{\rho} \): density (kg·m⁻³)
- \( \mathrm{m} \): mass (kg)
- \( \mathrm{V} \): volume (m³)
Explanation:
- Density measures how much mass is contained in a given volume.
- It depends on the nature of the material and its state (solid, liquid, or gas).
- For uniform materials, density is constant throughout; for non-uniform materials, density varies with position.
Typical Densities:
| Substance | Approximate Density (kg·m⁻³) |
|---|---|
| Air | 1.2 |
| Water | 1000 |
| Aluminium | 2700 |
| Iron | 7900 |
Example
A metal block has a mass of \( \mathrm{540\,g} \) and a volume of \( \mathrm{200\,cm^3.} \) Calculate its density in \( \mathrm{kg·m^{-3}.} \)
▶️ Answer / Explanation
Convert to SI units:
\( \mathrm{m = 0.540\,kg, \ V = 200\times10^{-6}\,m^3 = 2.0\times10^{-4}\,m^3.} \)
Using \( \mathrm{\rho = \dfrac{m}{V}} \):
\( \mathrm{\rho = \dfrac{0.540}{2.0\times10^{-4}} = 2700\,kg·m^{-3}.} \)
The density is \( \mathrm{2700\,kg·m^{-3},} \) typical of aluminium.
Definition and Use of Pressure
Pressure is defined as the force acting per unit area.
\( \mathrm{p = \dfrac{F}{A}} \)
- \( \mathrm{p} \): pressure (Pa or N·m⁻²)
- \( \mathrm{F} \): normal (perpendicular) force (N)
- \( \mathrm{A} \): area over which the force acts (m²)
Explanation:
- Pressure quantifies how concentrated a force is over a surface area.
- A smaller area or larger force results in higher pressure.
- Units: \( \mathrm{1\,Pa = 1\,N·m^{-2}.} \)
Applications of Pressure:
- Hydraulic systems — pressure transmission through liquids.
- Atmospheric pressure measurement (barometer).
- Design of snowshoes or tyres — increasing area reduces pressure.
Example
A force of \( \mathrm{400\,N} \) acts uniformly on an area of \( \mathrm{0.25\,m^2.} \) Find the pressure exerted.
▶️ Answer / Explanation
Using \( \mathrm{p = \dfrac{F}{A}} \):
\( \mathrm{p = \dfrac{400}{0.25} = 1600\,Pa.} \)
The pressure exerted is \( \mathrm{1.6\times10^3\,Pa.} \)
Derivation of Hydrostatic Pressure Equation
Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid above a given point.
Derivation:
Consider a vertical column of fluid of:
- cross-sectional area = \( \mathrm{A} \)
- height = \( \mathrm{\Delta h} \)
- density = \( \mathrm{\rho} \)
The mass of the column is:
\( \mathrm{m = \rho V = \rho A\Delta h} \)
Its weight is:
\( \mathrm{W = mg = \rho A\Delta h g} \)
The pressure difference between the bottom and top of the column is due to this weight acting over area \( \mathrm{A} \):
\( \mathrm{\Delta p = \dfrac{W}{A} = \dfrac{\rho A\Delta h g}{A} = \rho g \Delta h.} \)
Hydrostatic Pressure Equation:
\( \mathrm{\Delta p = \rho g \Delta h} \)
- \( \mathrm{\Delta p} \): pressure difference between two points (Pa)
- \( \mathrm{\rho} \): density of the fluid (kg·m⁻³)
- \( \mathrm{g} \): acceleration due to gravity (m·s⁻²)
- \( \mathrm{\Delta h} \): depth difference (m)
Key Features:
- Pressure increases linearly with depth in a fluid.
- It depends only on the height of the fluid column, not on the shape or volume of the container.
- It acts equally in all directions at a given depth (Pascal’s principle).
Example
Calculate the increase in pressure between two points in water separated by a vertical depth of \( \mathrm{5.0\,m.} \) Take \( \mathrm{\rho = 1000\,kg·m^{-3}} \) and \( \mathrm{g = 9.8\,m·s^{-2}.} \)
▶️ Answer / Explanation
Using \( \mathrm{\Delta p = \rho g \Delta h} \):
\( \mathrm{\Delta p = (1000)(9.8)(5.0) = 4.9\times10^4\,Pa.} \)
The pressure increases by \( \mathrm{49\,kPa} \) between the two points.
Applications of the Hydrostatic Pressure Equation
- Design of dams and submarines — pressure increases with depth, so structures must withstand greater forces at lower levels.
- Hydraulic systems — pressure transmission through incompressible fluids (Pascal’s law).
- Manometers and barometers — measuring fluid pressure differences.
| Quantity | Definition | Formula | SI Unit |
|---|---|---|---|
| Density | Mass per unit volume | \( \mathrm{\rho = \dfrac{m}{V}} \) | kg·m⁻³ |
| Pressure | Force per unit area | \( \mathrm{p = \dfrac{F}{A}} \) | Pa (N·m⁻²) |
| Hydrostatic Pressure | Pressure due to a column of fluid | \( \mathrm{\Delta p = \rho g \Delta h} \) | Pa |
Key Takeaways:
- Density links mass and volume; pressure links force and area.
- Hydrostatic pressure increases with fluid density, depth, and gravitational acceleration.
- At a given depth, pressure is the same in all directions and acts normally to surfaces.
Example
A diver is swimming at a depth of \( \mathrm{25\,m} \) below the surface of the sea. The density of seawater is \( \mathrm{1025\,kg·m^{-3}} \). Calculate the total pressure on the diver, given that atmospheric pressure at the surface is \( \mathrm{1.01\times10^5\,Pa.} \) Take \( \mathrm{g = 9.8\,m·s^{-2}.} \)
▶️ Answer / Explanation
Step 1: Calculate the pressure due to the water column using \( \mathrm{\Delta p = \rho g \Delta h} \).
\( \mathrm{\Delta p = (1025)(9.8)(25)} \)
\( \mathrm{\Delta p = 2.51\times10^5\,Pa.} \)
Step 2: Add atmospheric pressure to find total pressure at the diver’s depth:
\( \mathrm{p_{total} = p_{atm} + \Delta p} \)
\( \mathrm{p_{total} = 1.01\times10^5 + 2.51\times10^5 = 3.52\times10^5\,Pa.} \)
Step 3: State the result.
The total pressure on the diver at a depth of \( \mathrm{25\,m} \) is \( \mathrm{3.52\times10^5\,Pa} \) (approximately \( \mathrm{3.5\,atm.} \))
Understanding Upthrust in a Fluid
Upthrust (also called the buoyant force) is the upward force exerted by a fluid on a body immersed in it, partially or completely.
Origin of Upthrust — Difference in Hydrostatic Pressure:
- In a fluid, pressure increases with depth according to \( \mathrm{p = \rho g h.} \)
- When an object is immersed in a fluid, the pressure acting on its bottom surface is greater than that on its top surface.
- This difference in pressure produces a net upward force — the upthrust.
Reasoning:
- Top surface experiences downward force \( \mathrm{F_{top} = p_{top}A.} \)
- Bottom surface experiences upward force \( \mathrm{F_{bottom} = p_{bottom}A.} \)
- Since \( \mathrm{p_{bottom} > p_{top},} \) the resultant force acts upward.
Therefore,
\( \mathrm{F_{upthrust} = F_{bottom} – F_{top}} \)
Key Point:
The upthrust depends only on the volume of fluid displaced, not on the depth of immersion or shape of the object.
Example
Explain why an object experiences an upward force when immersed in water, and why this force is greater for a larger object.
▶️ Answer / Explanation
Pressure in water increases with depth. The bottom of the object is deeper than the top, so it experiences a larger pressure and hence a larger upward force. The difference between the upward and downward forces produces a net upward upthrust. A larger object displaces more fluid, so the pressure difference (and therefore upthrust) is greater.
Example
A solid metal sphere has a radius of \( \mathrm{0.05\,m.} \) It is completely immersed in a liquid of density \( \mathrm{1200\,kg·m^{-3}.} \) Calculate the upthrust acting on the sphere. Take \( \mathrm{g = 9.8\,m·s^{-2}.} \)
▶️ Answer / Explanation
Step 1: Write the formula for upthrust (Archimedes’ principle)
\( \mathrm{F = \rho g V} \)
Step 2: Calculate the volume of the sphere
\( \mathrm{V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi (0.05)^3 = 5.24\times10^{-4}\,m^3.} \)
Step 3: Substitute values into the upthrust equation
\( \mathrm{F = (1200)(9.8)(5.24\times10^{-4})} \)
\( \mathrm{F = 6.16\,N.} \)
Step 4: State the result
The upthrust acting on the sphere is \( \mathrm{6.2\,N} \) (upward).
Archimedes’ Principle and the Equation for Upthrust
Archimedes’ Principle:
The upthrust acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.
Derivation of the Upthrust Equation:
Let the object displace a volume \( \mathrm{V} \) of fluid of density \( \mathrm{\rho.} \)
Mass of displaced fluid:
\( \mathrm{m = \rho V} \)
Weight of displaced fluid (which equals upthrust):
\( \mathrm{F = mg = \rho g V} \)
Hence,
\( \mathrm{F = \rho g V} \)
- \( \mathrm{F} \): upthrust (N)
- \( \mathrm{\rho} \): density of the fluid (kg·m⁻³)
- \( \mathrm{g} \): gravitational field strength (m·s⁻²)
- \( \mathrm{V} \): volume of fluid displaced (m³)
Key Features:
- Upthrust depends on the density of the fluid and the volume of fluid displaced.
- It acts vertically upward through the centre of gravity of the displaced fluid — the centre of buoyancy.
- If upthrust = weight → object floats (equilibrium).
- If upthrust < weight → object sinks.
Example
A metal cube of volume \( \mathrm{5.0\times10^{-4}\,m^3} \) is completely immersed in water of density \( \mathrm{1000\,kg·m^{-3}.} \) Calculate the upthrust acting on the cube. Take \( \mathrm{g = 9.8\,m·s^{-2}.} \)
▶️ Answer / Explanation
Using \( \mathrm{F = \rho g V} \):
\( \mathrm{F = (1000)(9.8)(5.0\times10^{-4})} \)
\( \mathrm{F = 4.9\,N.} \)
The upthrust on the cube is \( \mathrm{4.9\,N.} \)
Example
A solid block of density \( \mathrm{800\,kg·m^{-3}} \) is placed in water of density \( \mathrm{1000\,kg·m^{-3}.} \) Explain whether the block will float or sink and determine the fraction of its volume submerged when floating.
▶️ Answer / Explanation
The block will float because its density is less than that of water. In equilibrium, upthrust = weight.
\( \mathrm{\rho_{fluid} g V_{submerged} = \rho_{block} g V_{total}} \)
Therefore,
\( \mathrm{\dfrac{V_{submerged}}{V_{total}} = \dfrac{\rho_{block}}{\rho_{fluid}} = \dfrac{800}{1000} = 0.8.} \)
Thus, \( \mathrm{80\%} \) of the block’s volume is submerged while floating.