At some jewellers' shops, a gold chain is weighed twice on a sensitive balance. First it hangs in air. Then it hangs in a small cup of water, without touching the cup. The jeweller divides one reading by the difference of the two, and knows at once whether the gold has been mixed with a cheaper metal.
Density from two readings
Weigh a body in air and call the reading \(W\). Weigh it again fully under water and call the reading \(W'\). The loss of weight, \(W - W'\), is the buoyant force. By Archimedes' principle it is the weight of water of the same volume as the body. So the ratio \(W/(W - W')\) compares the weight of the body with the weight of an equal volume of water.
That ratio is the relative density of the body, its density divided by the density of water. It has no unit. Multiply it by \(1000\,\text{kg/m}^3\) to get the density. The method needs no ruler at all. So it works for a body of any shape, a crown, a chain or a lump of rock.
A crown has a mass of \(1\,\text{kg}\) in air and an apparent mass of \(0.9\,\text{kg}\) in water. Pure gold has a density of \(19\,300\,\text{kg/m}^3\). Find the density of the crown, and decide whether it is pure gold.
The loss of mass in water is \(0.1\,\text{kg}\). This is the mass of the water displaced, so the crown's volume is \(0.1/1000 = 10^{-4}\,\text{m}^3\).
The density is \(1/10^{-4} = 10\,000\,\text{kg/m}^3\), a relative density of \(1/0.1 = 10\).
Pure gold would be \(19\,300\,\text{kg/m}^3\). A crown of pure gold of \(1\,\text{kg}\) would have lost only about \(0.052\,\text{kg}\). This crown is gold mixed with a lighter metal, such as silver.
The jeweller's test is the same. A chain of pure gold that reads \(19.3\,\text{g}\) in air must read \(18.3\,\text{g}\) in water, a relative density of \(19.3/1.0 = 19.3\). A chain mixed with copper has the same mass but a larger volume. It loses more in water, and its relative density comes out smaller.
A metal block reads \(60\,\text{N}\) on a spring balance in air and \(45\,\text{N}\) when fully under water. What is its relative density?
An old story says that Archimedes found the fraud in King Hiero's crown by the water it spilled from a full vessel. Galileo, then twenty-two, doubted it. The overflow from a crown is so small that no one could measure it well. In a short essay called The Little Balance, he proposed a better way. Hang the crown and an equal mass of pure gold from the two arms of a balance, and lower both into water. The crown, being larger, loses more weight, and the balance tips.
The density of a liquid
The same idea measures a liquid. Take a sinker, any solid that sinks in both liquids, and weigh it three times: in air, in water and in the liquid. It displaces the same volume \(V\) each time. In water it loses \(\rho_w V g\), and in the liquid it loses \(\rho_l V g\). The ratio of the two losses is \(\rho_l/\rho_w\), and \(V\) never has to be known.
A body weighs \(100\,\text{N}\) in air, \(80\,\text{N}\) in water and \(84\,\text{N}\) in oil. Take \(g = 10\,\text{m/s}^2\). Find the volume of the body and the density of the oil.
Loss in water: \(100 - 80 = 20\,\text{N}\). Loss in oil: \(100 - 84 = 16\,\text{N}\).
\(20 = \rho_w V g\) gives \(V\). Then \(16 = \rho_{oil} V g\) gives the oil's density.
\(V = 20/(1000 \times 10) = 2 \times 10^{-3}\,\text{m}^3\). \(\rho_{oil} = 16/(2 \times 10^{-3} \times 10) = 800\,\text{kg/m}^3\).
The body has a volume of \(2\) litres, and the oil a density of \(800\,\text{kg/m}^3\), like kerosene.
A glass sinker reads \(30\,\text{N}\) in air, \(20\,\text{N}\) in water and \(20.8\,\text{N}\) in mustard oil. Find the density of the mustard oil, in kg/m³.
A float that reads density
A hydrometer is a weighted glass float with a long, thin stem. It floats upright, and by the law of flotation it always displaces its own weight. In a less dense liquid it must displace a larger volume, so it sinks deeper. The surface then meets the stem higher up. So the density marks on the stem increase downward.
The thin stem makes the instrument sensitive. A small change in the displaced volume moves the surface a long way along a narrow stem. A lactometer is a hydrometer made for milk. Its stem covers the narrow range of densities near that of milk, about \(1030\,\text{kg/m}^3\).
In water the lactometer of Fig 3.1 displaces \(31.5\,\text{cm}^3\): the bulb's \(30\,\text{cm}^3\) and \(7.5\,\text{cm}\) of stem. In pure milk of \(1030\,\text{kg/m}^3\) it needs only \(31.5/1.03 = 30.58\,\text{cm}^3\). So only \(2.9\,\text{cm}\) of stem goes under. Milk with a quarter of its volume water has a density of \(1022.5\,\text{kg/m}^3\). The stem then sinks to \(4.0\,\text{cm}\), between the marks, and the added water shows.
A hydrometer floats in water with its mark A at the surface. In petrol, of density \(700\,\text{kg/m}^3\), its mark B is at the surface. Which mark is higher on the stem?
Pure milk has a density of \(1030\,\text{kg/m}^3\). A milkman mixes \(4\) litres of it with \(1\) litre of water. What is the density of the mixture, in kg/m³?
At a village milk collection centre, each farmer's milk is tested before it is paid for. A lactometer is floated in a sample, and the mark at the surface gives the density. Together with a test of the fat, it tells the centre whether water has been added. Water lowers the density, and the lactometer sinks deeper.
Where the lost weight goes
The water pushes up on a block with the buoyant force. By Newton's third law, the block pushes down on the water with an equal force. The water level rises as the block goes in. The deeper water presses harder on the floor of the beaker, and this extra push reaches the scale under it.
The block's volume is \(2/4000 = 5 \times 10^{-4}\,\text{m}^3\), so fully under water it feels \(F_B = 5\,\text{N}\). The balance falls from \(20\,\text{N}\) to \(15\,\text{N}\), and the scale rises from \(30\,\text{N}\) to \(35\,\text{N}\). The weight that leaves the balance appears on the scale. At every instant the two readings add to \(50\,\text{N}\), the weight of all that the two instruments hold up.
A glass of water stands on a kitchen scale. You dip a finger into the water without touching the glass. What happens to the reading?
- Dividing the weight in air by the reading in water to get the relative density.Divide by the loss of weight, \(W - W'\), which is the weight of an equal volume of water.
- Reading a hydrometer as if its marks rise upward like a thermometer's.It floats deeper in a lighter liquid, so the smaller densities are higher on the stem.
- Forgetting that the buoyant force acts on the liquid too.The body pushes down on the liquid with \(F_B\), so the vessel's support carries it.