PhysicsCore24 min read

Mechanical Properties of Matter

Density, pressure and elastic behaviour

This topic appears in:

9th Class Physics · ch.61st Year Physics · ch.5O Level Physics (5054) · ch.1A Level Physics (9702) · ch.4
01

Density

Definition

Density — The mass per unit volume of a substance, ρ = m/V. Measured in kg m⁻³ or g cm⁻³.

Density is a property of the material rather than of the object. A steel pin and a steel bridge have identical densities; they differ only in how much steel there is. This is why density is useful for identifying a substance when you cannot tell by looking.

Water has a density of 1000 kg m⁻³, which is the same as 1.0 g cm⁻³. That number is worth knowing, because it is the reference point for whether something floats. Anything less dense than water floats in it; anything denser sinks.

A steel ship floats even though steel is eight times denser than water, because the relevant density is that of the whole ship — steel plus the large volume of air inside the hull. Fill the hull with water and the average density rises above that of water, and the ship sinks.

ρ = m / Vm = ρ Vwater: 1000 kg m⁻³ = 1.0 g cm⁻³; 1 g cm⁻³ = 1000 kg m⁻³
ρ
densitykg m⁻³
m
masskg
V
volumem³
Worked example 14 marks

A metal block measures 4.0 cm × 3.0 cm × 2.0 cm and has a mass of 216 g. Calculate its density in g cm⁻³ and state whether it would float in water.

  1. Volume = 4.0 × 3.0 × 2.0 = 24 cm³.
  2. ρ = m/V = 216 / 24.
  3. ρ = 9.0 g cm⁻³.
  4. Water is 1.0 g cm⁻³, so the block is nine times denser and sinks.Compare with water rather than guessing.

9.0 g cm⁻³ — it sinks

02

Pressure

Definition

Pressure — The force acting per unit area, at right angles to a surface, p = F/A. Measured in pascals, where one pascal is one newton per square metre.

The same force spread over a different area produces a very different pressure, and that single idea explains a long list of everyday designs.

A drawing pin has a broad head and a sharp point. Your thumb pushes with a modest force over the large area of the head, so the pressure on your thumb is small and it does not hurt. That same force acts through the tiny area of the point, giving an enormous pressure that pushes into the wood.

The reverse trick is used to reduce pressure. Skis, snowshoes and the wide tracks of a tractor all spread the same weight over a much larger area, so the pressure on soft ground is low enough that the vehicle does not sink. Camels have broad feet for the same reason.

Pressure acts at right angles to whatever surface it meets, and in a fluid it acts in all directions equally.

p = F / AF = p Aone pascal is one newton per square metre; atmospheric pressure ≈ 1.0 × 10⁵ Pa
p
pressurePa
F
force at right angles to the surfaceN
A
aream²

Convert areas before substituting

A square centimetre is 10⁻⁴ m², not 10⁻² m². The prefix gets squared along with the unit. Substituting cm² directly into p = F/A makes the pressure ten thousand times too small, and the answer still looks like a number.

03

Pressure in a liquid

Pressure in a liquid increases with depth, because the deeper you go the greater the weight of liquid above pressing down. It does not depend on the shape of the container or on how much liquid there is in total — only on the depth, the density and the gravitational field strength.

That is why a dam is built much thicker at the bottom than at the top, and why a diver feels increasing pressure on the ears as they descend. It is also why water squirts furthest from the lowest hole in a punctured can.

At a given depth the pressure acts equally in all directions, not just downwards. This is what makes hydraulic systems possible: pressure applied at one point in an enclosed liquid is transmitted undiminished throughout it, so a small force on a small piston produces a large force on a large one.

p = ρ g hdepth h below the surface — the pressure from the liquid alone, before adding atmospheric
p
pressurePa
ρ
density of the liquidkg m⁻³
g
gravitational field strengthN kg⁻¹
h
depthm
Worked example 24 marks

A diver is 25 m below the surface of the sea, where the water has density 1030 kg m⁻³. Calculate the pressure due to the water. Take g = 9.8 N kg⁻¹.

  1. Use p = ρgh.All three quantities are given.
  2. p = 1030 × 9.8 × 25.
  3. p = 252 350 Pa.
  4. ≈ 2.5 × 10⁵ Pa, about two and a half atmospheres.Adding atmospheric pressure would give the total pressure on the diver.

2.5 × 10⁵ Pa from the water alone

Move the depth slider and watch the pressure climb in a straight line — p = ρgh. Switch to mercury and the line tilts sharply: 13.6 times the density means 13.6 times the pressure at the same depth. Notice the arrows at the marker: pressure acts equally in every direction.

04

Atmospheric pressure and the manometer

The atmosphere is a layer of air several kilometres deep, and its weight presses on everything at the surface at about 1.0 × 10⁵ Pa. We do not notice it because it acts equally in all directions, including from inside our bodies outwards.

Atmospheric pressure falls with altitude, because there is less air above you. This is why aircraft cabins are pressurised and why water boils at a lower temperature on a mountain.

A manometer measures the pressure of a gas supply by connecting it to a U-tube containing liquid. The gas pushes the liquid down on one side and up on the other, and the difference in the two levels gives the pressure difference directly through p = ρgh. If the levels are equal, the gas is at exactly atmospheric pressure.

A mercury barometer works on the same principle to measure atmospheric pressure itself. Atmospheric pressure supports a column of mercury about 760 mm tall — and mercury is used rather than water precisely because it is so dense that the column is a manageable height. A water barometer would need to be over ten metres tall.

Key points

  1. Density is mass per unit volume and identifies the material.
  2. Pressure is force per unit area — a sharp point concentrates it, a broad foot spreads it.
  3. Pressure in a liquid depends on depth and density, not on the shape of the container.
  4. At a given depth, pressure acts equally in all directions.
  5. Convert cm² to m² by 10⁻⁴ and cm³ to m³ by 10⁻⁶.

Practice questions

7 questions · 23 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define density and state its SI unit.
Model answer

Mass per unit volume, ρ = m/V. SI unit: kg m⁻³.

Examiner tip. One mark for the definition, one for the unit. Never leave the unit off a definition question.

SQ2[2 marks]
Why does a camel have broad feet?
Model answer

Broad feet spread the camel's weight over a larger area, so the pressure on the sand is smaller and it does not sink.

Examiner tip. The mark scheme wants "larger area" and "smaller pressure". The weight itself is unchanged — saying the feet reduce the weight loses both marks.

SQ3[2 marks]
State Hooke's law.
Model answer

The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.

Examiner tip. The proviso is worth a mark. A statement of proportionality without it is incomplete.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A tank contains oil of density 800 kg m⁻³ to a depth of 1.5 m. Calculate the pressure at the base due to the oil, and the force this exerts on a base of area 2.0 m². Take g = 10 N kg⁻¹.

Given. ρ = 800 kg m⁻³, h = 1.5 m, A = 2.0 m², g = 10 N kg⁻¹

Full working
  1. Uses p = ρgh[1]
  2. p = 800 × 10 × 1.5 = 12 000 Paunit required[1]
  3. Rearranges p = F/A to F = pA[1]
  4. F = 12 000 × 2.0 = 24 000 N[1]

p = 1.2 × 10⁴ Pa, F = 2.4 × 10⁴ N

Exam questions

3 · 13 marks

Multi-part questions with a full mark scheme.

Q1[5 marks]
A rectangular block of wood has dimensions 20 cm × 10 cm × 5.0 cm and a mass of 0.80 kg.
  1. Calculate the density of the wood in kg m⁻³.
  2. The block is placed on a table on its largest face. Calculate the pressure it exerts. Take g = 10 N kg⁻¹.
Mark scheme
  1. Volume = 0.20 × 0.10 × 0.050 = 1.0 × 10⁻³ m³converting every length to metres first[1]
  2. Density = 0.80 / 1.0 × 10⁻³ = 800 kg m⁻³[1]
  3. Weight = mg = 0.80 × 10 = 8.0 Npressure needs force, and the force here is the weight[1]
  4. Largest face area = 0.20 × 0.10 = 0.020 m²largest face gives the lowest pressure[1]
  5. Pressure = 8.0 / 0.020 = 400 Pa[1]

(a) 800 kg m⁻³ (b) 400 Pa

Examiner tip. Convert centimetres to metres at the very start, not halfway through. A volume in cm³ divided into a mass in kg gives a number that is wrong by a factor of a million, and the error is invisible once written down.

Q2[4 marks]
A diver is 12 m below the surface of a lake. The density of the water is 1000 kg m⁻³ and g = 10 N kg⁻¹.
  1. Calculate the pressure on the diver due to the water alone.
  2. The lake narrows sharply near the bottom. State and explain the effect of this on the pressure at 12 m depth.
Mark scheme
  1. Uses p = ρgh[1]
  2. p = 1000 × 10 × 12 = 1.2 × 10⁵ Paunit required[1]
  3. No effect / the pressure is unchanged[1]
  4. Pressure in a liquid depends only on depth, density and g — not on the shape or width of the containerthe reasoning mark; the statement alone scores 1 of 2[1]

(a) 1.2 × 10⁵ Pa (b) no change — pressure depends only on depth

Examiner tip. The question says "state and explain", so a bare "no effect" earns half the marks available. Every "explain" is a separate mark waiting to be collected.

Q3[4 marks]
A spring of natural length 8.0 cm extends to 12.0 cm when a load of 5.0 N is hung from it.
  1. Calculate the spring constant.
  2. Calculate the length of the spring when a load of 8.0 N is applied, assuming the limit of proportionality is not exceeded.
Mark scheme
  1. Extension = 12.0 − 8.0 = 4.0 cm = 0.040 mextension, not total length — the mark most often lost on this topic[1]
  2. k = F/x = 5.0 / 0.040 = 125 N m⁻¹[1]
  3. New extension = 8.0 / 125 = 0.064 m = 6.4 cm[1]
  4. New length = 8.0 + 6.4 = 14.4 cmadding the natural length back on[1]

(a) 125 N m⁻¹ (b) 14.4 cm

Examiner tip. Two traps in one question: x is the extension going in, and the answer wants a total length coming out. Read what is asked for, then check your final number is that thing.