Density, ρ= m/V, and how to measure it for regular and irregular solids and for liquids

Upthrust as the weight of fluid displaced, and why it exists

Floating and sinking, and the fraction of a floating body below the surface

Hydrometers and hot-air balloons

Viscosity, laminar and turbulent flow, and how viscosity depends on temperature

Stokes' law, F = 6πηrv, and when it can be used

Terminal velocity of a sphere in a liquid: weight = upthrust + viscous drag

Core Practical 2: measuring viscosity by the falling-ball method

1. A fluid exerts two forces on a body inside it, upthrust (always up, equal to the weight of fluid displaced) and viscous drag (opposing motion), and with the weight these explain floating, balloons and terminal velocity in a liquid.

2. Density ρ= m/V, in kg m⁻³; 1 g cm⁻³ = 1000 kg m⁻³.

3. Upthrust = weight of fluid displaced = ρ_fluidV_displaced g. It exists because fluid pressure increases with depth.

4. A body floats when upthrust = weight; the fraction submerged is ρ_body/ρ_fluid.

5. Stokes' law F = 6πηrv gives the drag on a small sphere moving slowly in laminar flow. The viscosity η (Pa s) of a liquid falls steeply as it warms.

6. At terminal velocity in a liquid W = U + F, which gives v = 2r²g(ρₛ - ρ_f)/9η; Core Practical 2 finds η from the gradient of v against r².

Because 1 cm³ = 10⁻⁶ m³ and 1 g = 10⁻³ kg, 1 g cm⁻³ = 10⁻³ ÷ 10⁻⁶ = 1000 kg m⁻³. One litre is 1000 cm³ = 10⁻³ m³.

Measuring density

Regular solid: mass on a top-pan balance; dimensions with a ruler, vernier calipers or a micrometer; then calculate the volume (cuboid lwh, sphere (4/3)πr³, cylinder πr²h).

Irregular solid: find the volume by displacement. Lower it into water in a measuring cylinder and record the rise in level, or use a displacement (eureka) can and collect the overflow.

Liquid: weigh an empty measuring cylinder, add a measured volume of liquid, weigh again; the difference is the mass of the liquid.

Read a measuring cylinder at eye level, from the bottom of the meniscus, to avoid parallax.

WORKED EXAMPLE · QUESTION

A pebble of mass 64.8 g is lowered into a measuring cylinder containing water. The water level rises from 50.0 cm³ to 74.0 cm³. Calculate the density of the pebble in kg m⁻³.

SOLUTION

Volume of pebble = volume of water displaced: V = 74.0 - 50.0 = 24.0 cm³ = 24.0×10⁻⁶ m³. Mass: m = 64.8×10⁻³ kg.

ρ= m/V = 64.8×10⁻³ ÷24.0×10⁻⁶ → ρ = 2.70 × 10³ kg m⁻³.

Quick check: 64.8 ÷24.0 = 2.70 g cm⁻³, and × 1000 gives the same answer.

A fluid (liquid or gas) pushes on every surface in contact with it, at right angles to the surface. The pressure in a fluid increases with depth, because there is more fluid above pushing down. So the bottom of an immersed body is pushed up harder than its top is pushed down. The sideways pushes at each depth cancel. The resultant of all these pushes is the upthrust, acting vertically upwards.

The upthrust does not depend on what the body is made of, only on the volume it displaces and the density of the fluid.

That is why it equals the weight of the fluid displaced: the same pushes would exactly hold up a "body" made of the fluid itself, which simply sits in equilibrium.

For a fully submerged body V_displaced is the whole volume of the body. For a floating body it is only the volume below the surface.

WORKED EXAMPLE · QUESTION

An aluminium block measures 5.0 cm × 4.0 cm × 3.0 cm. It hangs from a newton-meter and is lowered until it is completely under water. Density of aluminium = 2700 kg m⁻³; density of water = 1000 kg m⁻³. Calculate the reading on the newton-meter when the block is fully submerged.

SOLUTION

Volume: V = 5.0×4.0×3.0 = 60 cm³ = 6.0×10⁻⁵ m³.

Weight: W = ρVg = 2700×6.0×10⁻⁵×9.81 = 1.589 N. Upthrust = weight of water displaced: U = 1000×6.0×10⁻⁵×9.81 = 0.589 N.

Equilibrium, so the newton-meter supplies the rest: T = W - U → T = 1.00 N.

The block's weight has not changed; the newton-meter reads less because the water now carries part of the weight.

When a body is released in a fluid, compare its weight W with the upthrust U_(max) it would feel if fully submerged:

U_(max) > W: it rises, and floats with part of it above the surface.

U_(max) = W: it stays wherever it is put, like a submarine trimmed for neutral buoyancy.

U_(max) < W: it sinks.

Both forces depend on the same volume, so this is the same as comparing densities: a body floats if its average density is less than the fluid's. A steel ship floats because its average density, steel plus the air inside the hull, is less than that of water.

A floating body is in equilibrium, U = W. It sinks just far enough to displace its own weight of fluid. Setting ρ_fluidV_subg = ρ_bodyVg and cancelling g gives the fraction submerged:

WORKED EXAMPLE · QUESTION

A cylindrical navigation buoy has a mass of 85 kg and a radius of 0.30 m. It floats upright in sea water of density 1030 kg m⁻³. (a) Calculate the depth of the buoy below the surface. (b) Calculate the extra mass that must be attached on top so that its base is 0.50 m below the surface.

SOLUTION

(a) Floating, so weight of sea water displaced = weight of buoy, and g cancels: ρ(πr²h) = m, so h = 85 ÷(1030×π×0.30²) = 85 ÷(1030×0.283) = 0.292 m → h = 0.29 m.

(b) Mass of sea water displaced at 0.50 m depth: 1030×0.283×0.50 = 145.6 kg. This must equal the total floating mass, so the extra mass = 145.6 - 85 = 60.6 kg → 61 kg.

The hydrometer

A hydrometer is a sealed glass float: a weighted bulb with a narrow, uniform stem marked with a density scale. Its weight is fixed, so it always floats with the same weight of liquid displaced. In a denser liquid a smaller volume is enough, so it floats higher. The density is read where the liquid surface crosses the stem, which means higher densities are marked lower down the stem.

The scale is not uniform. Each equal step in density needs a smaller change of submerged volume at higher densities, so the marks crowd together towards the bulb. A narrow stem makes it sensitive.

WORKED EXAMPLE · QUESTION

The hydrometer in the figure is placed in a sample of fruit juice. It floats with 6.2 cm of its stem below the surface. Calculate the density of the juice in kg m⁻³.

SOLUTION

Volume of juice displaced = bulb + submerged part of the stem: V = 16.0 + 0.50×6.2 = 19.1 cm³.

Floating, so mass of juice displaced = mass of hydrometer: ρ= 20.0 ÷19.1 = 1.047 g cm⁻³ → ρ = 1.05 × 10³ kg m⁻³.

The hot-air balloon

The envelope displaces its own volume of the cool outside air, so the upthrust is ρ_coolVg. Heating the air inside makes it expand; some leaves through the open bottom, so the air left inside is less dense and weighs ρₕₒₜVg. The balloon rises when the upthrust is greater than the total weight: hot air + envelope + basket + burner + people.

The useful lift per cubic metre is (ρ_cool - ρₕₒₜ)g, only a few newtons, which is why envelopes are so large. On a hot day, or at altitude, the possible load is smaller.

WORKED EXAMPLE · QUESTION

A balloon envelope holds 2800 m³ of air. The outside air has a density of 1.20 kg m⁻³ and the hot air inside has a density of 0.950 kg m⁻³. Calculate (a) the upthrust on the balloon and (b) the maximum total mass of envelope, basket, burner and passengers for the balloon to lift off. Ignore the volume of everything except the envelope.

SOLUTION

(a) Upthrust = weight of cool air displaced: U = 1.20×2800×9.81 = 32 962 N → U = 3.30 × 10⁴ N.

(b) Weight of the hot air: 0.950×2800×9.81 = 26 095 N. Lift-off when U ≥ weight of hot air + Mg: M = (32 962 - 26 095)÷9.81 → M = 700 kg.

Shortcut: M = (ρ_cool - ρₕₒₜ)V = (1.20 - 0.950)×2800 = 700 kg.

Viscosity is a measure of a fluid's resistance to flow, how "thick" it is. Treacle and glycerol have high viscosity; water and air have low viscosity. When a body moves through a fluid, the fluid next to its surface moves with it and drags on the layers beyond. The result is a viscous drag force opposing the motion.

The coefficient of viscosity η (eta) has the unit Pa s (equivalently N s m⁻² or kg m⁻¹ s⁻¹). Typical values at 20 °C: air 1.8 × 10⁻⁵ Pa s, water 1.0 × 10⁻³ Pa s, glycerol about 1.4 Pa s.

Laminar and turbulent flow

Viscosity depends on temperature

For liquids, viscosity decreases sharply as the temperature rises: glycerol is about 1.4 Pa s at 20 °C but only about 0.6 Pa s at 30 °C. (Gas viscosity rises slowly with temperature.) So any viscosity value must be quoted with its temperature, and in an experiment the temperature must be measured and kept constant.

Three forces act on a sphere falling through a liquid: weight W down, upthrust U up, and viscous drag F up, opposing the motion. W and U are constant, but F starts at zero and increases with speed. The resultant force W - U - F therefore shrinks as the sphere speeds up, so the acceleration falls. When F has grown until W = U + F, the resultant force is zero and the sphere moves at its terminal velocity.

With Stokes' law and V = (4/3)πr³:

Here ρₛ is the density of the sphere, ρ_f the density of the fluid and v the terminal velocity. This result is not in the formulae list: derive it from W = U + F, and learn the volume of a sphere. Terminal velocity is proportional to r²: double the radius and (weight − upthrust) goes up 8 times, while the drag at a given speed goes up only 2 times, so v goes up 4 times.

WORKED EXAMPLE · QUESTION

A steel ball bearing of radius 1.0 mm falls through glycerol. Density of steel = 7800 kg m⁻³; density of glycerol = 1260 kg m⁻³; viscosity of glycerol = 1.4 Pa s. Calculate the terminal velocity of the ball bearing.

SOLUTION

Volume: V = (4/3)π(1.0×10⁻³)³ = 4.19×10⁻⁹ m³.

Weight: W = 7800×4.19×10⁻⁹×9.81 = 3.21×10⁻⁴ N. Upthrust: U = 1260×4.19×10⁻⁹×9.81 = 5.18×10⁻⁵ N.

At terminal velocity F = W - U = 2.69×10⁻⁴ N. Stokes' law: v = F/(6πηr) = 2.69×10⁻⁴ ÷(6π×1.4×1.0×10⁻³) = 0.0102 m s⁻¹ → v = 1.0 × 10⁻² m s⁻¹.

Check with the formula: 2×(1.0×10⁻³)²×9.81×(7800 - 1260)÷(9×1.4) = 0.0102 m s⁻¹. The speed is low and the ball is small, so the flow really is laminar and Stokes' law is justified.

Idea: measure the terminal velocity of small spheres falling through the liquid, then use W = U + F with Stokes' law to find η.

Apparatus: a tall, wide tube of the liquid with marks (rubber bands) at equal spacings, small steel balls, a stopwatch, a metre rule, a micrometer, a thermometer, a balance and a measuring cylinder.

Diameter: micrometer, several readings at different orientations, averaged; halve for the radius.

Densities: the steel from the mass and volume of a batch of balls; the liquid from the mass of a measured volume.

Terminal velocity: set the top band far enough below the surface that the ball is already at terminal velocity. Check that the times for successive equal intervals are the same, then v = distance ÷ time.

Release each wetted ball at the centre of the surface, away from the walls; repeat each size several times, use five or six sizes, and keep the temperature constant.

Safety: tube clamped in a tray; wipe up slippery spills at once.

Analysing the results

Rearranged, v = [(2g(ρₛ - ρ_f))/9η] r², so a graph of v against r² should be a straight line through the origin, with gradient 2g(ρₛ - ρ_f)/9η. Hence η= 2g(ρₛ - ρ_f)/(9×gradient). The graph uses every ball and shows up any anomaly, such as a large ball slowed by the walls.

WORKED EXAMPLE · QUESTION

Use the graph above to determine the viscosity of the liquid.

SOLUTION

Gradient from the large triangle: 41.4 mm s⁻¹ ÷3.50 mm² = 11.83 mm s⁻¹ mm⁻².

Convert to SI: 1 mm s⁻¹ ÷ 1 mm² = 10⁻³ ÷ 10⁻⁶ = 10³ m⁻¹ s⁻¹, so gradient = 11.83×10³ m⁻¹ s⁻¹.

η= (2g(ρₛ - ρ_f))/(9×gradient) = (2×9.81×(7800 - 1260))/(9×11.83×10³) = 1.205 Pa s → η = 1.21 Pa s.

Uncertainties

Diameter: a micrometer reads to 0.01 mm, so the uncertainty is half of that, 0.005 mm. For a 2.00 mm ball: 0.005 ÷ 2.00 × 100 = 0.25%.

Time: repeat readings vary by more than the stopwatch resolution (reaction time), so use half the range. For times of 7.62 s, 7.50 s and 7.56 s: mean 7.56 s, uncertainty ± 0.06 s, which is 0.8%.

The timing is usually the largest source of uncertainty. Longer distances between the bands, or filming the fall with a timer in view, reduce it.

2 Forces and Newton's laws

Free-body diagrams, resultant force and the chain of reasoning for terminal velocity all carry over; here the drag has a formula and upthrust is added to the diagram.

8 Hooke's law and 3.3 The AS core practicals

Density and viscosity describe materials in bulk; the next pages describe how solids stretch. The falling-ball analysis (rearrange to y = mx, plot v against r², read a large gradient, find percentage uncertainties) is a pattern that recurs throughout practical work.

Thermal physics (Unit 5)

Hot-air balloons rely on air becoming less dense as it is heated at constant pressure, which the gas laws describe quantitatively.

1. 250 cm³ of olive oil has a mass of 229 g. Calculate its density in kg m⁻³.

2. Explain why a steel ship floats although steel is much denser than water.

3. A glass bead of radius 0.50 mm moves at 4.0 mm s⁻¹ through an oil of viscosity 0.80 Pa s. Calculate the viscous drag on it.

4. In Core Practical 2 the liquid warms up during the lesson. Explain the effect on the measured times.

5. A hydrometer floats with more of its stem below the surface in liquid X than in water. Is X more or less dense than water? Explain.

1. ρ= 229 ÷250 = 0.916 g cm⁻³ = 916 kg m⁻³.

2. The hull encloses a large volume of air, so the average density of the ship is less than that of water. It sinks until the water it displaces weighs as much as the ship; then upthrust = weight and it floats.

3. F = 6π×0.80×0.50×10⁻³×4.0×10⁻³ = 3.02×10⁻⁵ N ≈3.0×10⁻⁵ N.

4. The viscosity of the liquid decreases as it warms, so the drag at a given speed is smaller and the terminal velocity is larger. The times between the bands become shorter, and the result for η applies to a temperature that was not constant.

5. Less dense. The hydrometer must displace its own weight of liquid; it displaces a larger volume of X, so each cubic metre of X weighs less than a cubic metre of water.