Density of irregular solids

Pick up a stone, a bent nail, or a lump of wax and try to find its volume with a ruler — you cannot, because there is no simple length, breadth or height to multiply. Yet every one of these objects is made of matter packed at some definite tightness, its density, and density is one of the most useful numbers in physics: it tells a jeweller whether a "gold" ring is really gold, and tells an engineer whether a component will float or sink. The trick for an irregular solid is to stop trying to measure its shape directly and instead measure how much water it pushes out of the way. This article builds the density formula from mass and volume, shows exactly how the displacement method turns an odd shape into a clean number using a measuring cylinder or a eureka can, and works through the arithmetic so you can repeat it confidently in a practical exam.

The core insight

The density of an irregular solid is found as ρ=m/V\rho = m/V, where the mass is weighed directly and the volume is found indirectly as the volume of water the solid displaces when fully submerged.

The core idea

Density describes how much mass is squeezed into each unit of volume of a substance — a dense material like iron packs a lot of mass into a small space, while a light material like cork packs very little.

ρ=mV\rho = \frac{m}{V}

where: ρ\rho (rho) is density, mm is mass, and VV is volume. The SI unit of density is kilograms per cubic metre (kg/m³); the CGS unit, more common in a school lab, is grams per cubic centimetre (g/cm³), and the two are linked by 1 g/cm3=1000 kg/m31\ \text{g/cm}^3 = 1000\ \text{kg/m}^3. For a regular solid such as a cube or a cylinder, VV comes straight from a geometric formula — volume of a cube =l3= l^3, for instance. An irregular solid, such as a stone or a bunch of keys, has no such formula, so we borrow a fact from Archimedes: a solid lowered fully into a liquid pushes aside, or displaces, a volume of liquid exactly equal to its own volume. Measuring that displaced volume is the whole trick.

Using a measuring cylinder. Partly fill the cylinder with water and read the initial level, V1V_1. Tie a thread to the solid and lower it in gently until fully submerged, without letting it touch the sides or the bottom — touching can trap air bubbles or scrape displaced water out of the reading. Read the new, higher level, V2V_2. The solid's volume is V=V2−V1V = V_2 - V_1.

Using a eureka can. If the solid is too big for a narrow cylinder, a eureka (displacement) can is used instead. The can is filled with water exactly up to its side spout, so it is completely full before the solid enters. When the solid is lowered in, the water it displaces has nowhere to go but out through the spout, and this overflow is caught in an empty measuring cylinder placed underneath. The volume collected in that cylinder is the volume of the solid — no subtraction is needed here, because the can started already full.

Figure
A measuring cylinder before and after an irregular stone is lowered into it, water level rising from a lower mark to a higher mark, with the rise labelled as the volume of the stone
Displacement method: water level rises by the solid's volume before immersion after immersion V1 = 50 V2 = 65 thread V = V2 minus V1 = 15 cm3 The stone's volume equals the rise in water level, not the final reading alone.

Once VV is known, the mass mm is measured separately on a beam or spring balance, and density follows from ρ=m/V\rho = m/V. Notice that the two measurements — mass and volume — are always taken independently and only combined at the last step.

Derivation AFinding the density of an irregular solid from its displaced volume

The goal is to show, step by step and with the numbers of a real lab reading, exactly how a mass measurement and a displacement measurement combine into a single density value, so that in an exam you can justify every figure you write rather than just quoting a memorised formula. Take a stone whose mass on a balance reads m=45 gm = 45\ \text{g}, placed in a measuring cylinder that initially reads V1=50 cm3V_1 = 50\ \text{cm}^3 and rises to V2=65 cm3V_2 = 65\ \text{cm}^3 once the stone is fully submerged on its thread.

Step 1. Isolate the volume reading. The measuring cylinder does not tell us the stone's volume directly — it only tells us the water level before and after. The stone's volume is the difference between the two readings, because that extra 15 cm315\ \text{cm}^3 of apparent water is really the space the stone now occupies: V=V2−V1=65−50=15 cm3V = V_2 - V_1 = 65 - 50 = 15\ \text{cm}^3.

Step 2. Confirm the mass is independent. The mass m=45 gm = 45\ \text{g} came from a balance and has nothing to do with the water reading — mixing up which quantity comes from which instrument is the single most common slip in this experiment, so it is worth stating explicitly before substituting.

Step 3. Substitute into the density formula. \rho = \frac{m}{V} = \frac{45\ \text{g}}{15\ \text{cm}^3} = 3\ \text{g/cm}^3 Both quantities were measured in the lab's natural units (grams and cubic centimetres), so no conversion was needed at this stage.

Step 4. Convert to SI if asked. Using 1 g/cm3=1000 kg/m31\ \text{g/cm}^3 = 1000\ \text{kg/m}^3, the same density is 3×1000=3000 kg/m33 \times 1000 = 3000\ \text{kg/m}^3. Reporting both forms shows you understand the unit relationship, not just the arithmetic.

Worked example

A metal key ring is too bulky to fit in a narrow measuring cylinder, so a eureka can is used instead, and its mass is m=60 gm = 60\ \text{g}.

Step 1. Set up the eureka can. The can is filled with water exactly to the level of its side spout, so it is brimming full before the key ring goes in — this is essential, because only a completely full can guarantees that every drop the key ring pushes aside actually overflows and gets collected.

Step 2. Read the collected overflow. As the key ring is lowered in on a thread, water spills out of the spout into an empty measuring cylinder placed to catch it. The cylinder collects 24 cm324\ \text{cm}^3 of water. Since the can was already full, this collected volume is the volume displaced, so V=24 cm3V = 24\ \text{cm}^3 directly — there is no earlier reading to subtract, unlike the plain-cylinder method.

Step 3. Apply the density formula. \rho = \frac{m}{V} = \frac{60\ \text{g}}{24\ \text{cm}^3} = 2.5\ \text{g/cm}^3

Step 4. Convert and state the answer. In SI units, 2.5 g/cm3=2500 kg/m32.5\ \text{g/cm}^3 = 2500\ \text{kg/m}^3. The density of the key ring's metal is 2.5 g/cm³, equivalent to 2500 kg/m³.

Competitive stretch

Suppose the same key ring were instead lowered into a partly filled measuring cylinder reading V1=40 cm3V_1 = 40\ \text{cm}^3, and the level rose to V2=64 cm3V_2 = 64\ \text{cm}^3. Check that this gives the same volume: V=64−40=24 cm3V = 64 - 40 = 24\ \text{cm}^3, matching the eureka-can result exactly. This confirms an important idea — the two methods are just two different ways of catching the same displaced water, and either one, done carefully, must give the identical volume for the identical object.

Common traps

  • Forgetting to subtract the initial reading. With a measuring cylinder, the solid's volume is V2−V1V_2 - V_1, never the final reading V2V_2 alone.
  • Letting the solid touch the container. Touching the sides or bottom can trap air bubbles or prevent full displacement, giving a volume that is too small.
  • Treating the eureka can's overflow as a "start from zero" reading. The can must already be completely full to the spout before immersion, or some displaced water stays inside instead of overflowing, and the collected volume undercounts the true volume.
  • Mixing units when converting. Mass is usually in grams and volume in cm³ in the lab; remember 1 g/cm3=1000 kg/m31\ \text{g/cm}^3 = 1000\ \text{kg/m}^3 when an SI answer is demanded, and do the conversion at the end, not partway through.

Ask the tutor: Want to try a density problem with a different irregular object, or see how this connects to whether an object floats or sinks in water?

Why do we need the thread when lowering the solid into the water

The thread lets you control exactly how far and how gently the solid goes in, so it can be fully submerged without touching the sides or the bottom of the cylinder or can. If the solid slides down a wall or scrapes the base, it can trap air bubbles against its surface or disturb the water level in ways that make the reading unreliable. The thread also lets you lift the solid straight back out afterward without spilling water or knocking the container, keeping the whole measurement clean and repeatable.

Does it matter whether I use a measuring cylinder or a eureka can for the same stone

No — both methods measure exactly the same physical quantity, the volume of water the solid displaces, so a correctly performed experiment gives the same volume either way, as the competitive-stretch check above shows. The choice is purely practical: a measuring cylinder is simpler and more precise for small solids that fit comfortably inside it, while a eureka can is used when the solid is too large or oddly shaped to fit in a narrow cylinder. Neither method is more "correct" than the other; pick whichever apparatus actually fits your solid.

If two objects have the same volume but different masses, are their densities the same

No. Density depends on both mass and volume together, through ρ=m/V\rho = m/V, so two objects can share a volume yet have completely different densities if their masses differ — a wooden block and a steel block of exactly the same size have very different densities because the steel packs far more mass into that same space. This is exactly why density, not just volume, tells you what a material actually is: a jeweller comparing a suspect "gold" ring to a genuine one would find they can have the same volume but different masses, revealing different densities and different metals.

Before this

Where this leads