Volume by water displacement

A stone has no straight edges, no length-times-width-times-height. Neither does a key, a bolt, or a fistful of nails. So how do you measure the volume of something that refuses to look like a box? The answer is one of the oldest tricks in science: drop it in water and watch the level rise. Any solid lowered into water pushes water out of the way — it displaces a volume of water exactly equal to its own volume, because two objects cannot occupy the same space at once. Read the water level before and after, subtract, and the irregular solid's volume falls out with no ruler needed at all. This article builds the water-displacement method from that one idea: how to read a measuring cylinder correctly, why the curved meniscus matters, how the method extends naturally to finding density once mass is known, and where the technique quietly breaks down. Two worked examples carry the same numbers used to introduce the idea, taken further.

The core insight

An irregular solid submerged in water raises the water level by an amount exactly equal to its own volume, so Volume=Final level−Initial level\text{Volume} = \text{Final level} - \text{Initial level}, and pairing that volume with a measured mass gives density.

The core idea

Regular vs irregular solids. A cuboid, cube, or cylinder is a regular solid — its volume follows directly from a geometric formula, such as length times width times height for a cuboid. A stone, a key, a screw, or a lump of clay is an irregular solid — it has no consistent edges or faces to measure, so no such formula applies. The water-displacement method sidesteps the need for a formula entirely.

Why displacement works. Water and a solid object cannot occupy the same region of space simultaneously. When a solid is lowered into water, the water already there has nowhere to go except up, so the water level rises by precisely the volume of the submerged solid — not an approximation, but an exact geometric fact as long as the solid is fully underwater and none of the water escapes.

Volume of object=Final water level−Initial water level\text{Volume of object} = \text{Final water level} - \text{Initial water level}

where: the initial water level is the reading taken before the solid enters the water, and the final water level is the reading taken once the solid is fully submerged; both are read on the same measuring cylinder in the same units (commonly millilitres, mL).

Reading the cylinder correctly. A measuring cylinder is a narrow, graduated glass or plastic vessel. Between two numbered marks (say, 10 mL and 20 mL) there are usually several unlabelled divisions; the smallest quantity you can read off depends on how many such divisions are marked — ten divisions between 10 mL and 20 mL means each division is worth 1 mL. Reading finer than the smallest marked division is guesswork, not measurement.

The meniscus. Water climbs slightly up the inside wall of a narrow container due to surface effects, so its surface is not flat but curved — this curve is called the meniscus. For water and other colourless liquids, the surface curves so the middle dips lower than the edges, forming a "smiley" shape. The correct reading is always taken level with the eye, at the bottom of this curve — reading at the top edge, or reading with your eye above or below the water line, introduces a small but avoidable error called parallax error.

Figure
A measuring cylinder shown before and after a stone is lowered in, with the water level rising from fifty millilitres to fifty five millilitres and the meniscus bottom marked at eye level
Water displacement: level rises by the volume of the stone before (initial level) 50 mL lower stone in after (final level) 55 mL eye level, bottom of meniscus Rise = 55 mL − 50 mL = 5 mL = 5 cm³ = volume of the stone

From volume to density. Once the volume of an irregular solid is found by displacement, and its mass is measured on a balance, its density follows exactly as it would for a regular solid:

Density=MassVolume\text{Density} = \frac{\text{Mass}}{\text{Volume}}

where: mass is measured in grams (g) and volume in cubic centimetres (cm³), with 1 mL=1 cm31\ \text{mL} = 1\ \text{cm}^3, so density comes out directly in g/cm³.

Derivation AWhy the rise in water level equals the object's volume

Show, from the simple fact that two solids (here, water and the submerged object) cannot share the same space at the same time, that the increase in water level in a measuring cylinder after fully submerging an object exactly equals the object's own volume, with no approximation involved.

Step 1. Before the object is added, the water in the cylinder occupies a certain volume, read off the graduated scale as the initial level — call it ViV_i. This volume of water fills the cylinder up to a certain height, and nothing else occupies space below that height.

Step 2. When the solid object is lowered in and fully submerged, it physically occupies a region of space that was previously filled only by water (or empty, above the original water surface). Since water is a fluid, it cannot be compressed into a smaller volume — it must move to make room for the solid.

Step 3. The water that used to occupy the space now taken up by the submerged object has nowhere to go but upward, since the cylinder's sides are rigid. The water surface therefore rises until the total volume enclosed below the new surface — water plus submerged solid — equals the new, higher reading, VfV_f.

Step 4. The volume of water itself has not changed; only the solid has been added into the same container. So the extra volume enclosed between the old surface and the new surface, Vf−ViV_f - V_i, must be occupied by something that was not there before — and that something is exactly the submerged solid. Hence Volume of object=Vf−Vi\text{Volume of object} = V_f - V_i, with equality holding exactly, not approximately, as long as the object is fully underwater and no water has spilled out.

Worked example

A small metal key has a mass of 16.400 g16.400\ \text{g}. A student pours water into a measuring cylinder and reads the initial level as 40 mL40\ \text{mL}. She then lowers the key in until it is fully submerged and reads the new water level as 45 mL45\ \text{mL}. Find (a) the volume of the key and (b) its density.

Given: mass of key =16.400 g= 16.400\ \text{g}; initial water level =40 mL= 40\ \text{mL}; final water level =45 mL= 45\ \text{mL}.

Step 1. Apply the displacement formula to find the volume: Volume=Final level−Initial level=45 mL−40 mL=5 mL\text{Volume} = \text{Final level} - \text{Initial level} = 45\ \text{mL} - 40\ \text{mL} = 5\ \text{mL}.

Step 2. Convert to cubic centimetres using 1 mL=1 cm31\ \text{mL} = 1\ \text{cm}^3, so the key's volume is 5 cm35\ \text{cm}^3.

Step 3. Apply the density formula with the given mass: Density=MassVolume=16.400 g5 cm3\text{Density} = \dfrac{\text{Mass}}{\text{Volume}} = \dfrac{16.400\ \text{g}}{5\ \text{cm}^3}.

Step 4. Carry out the division: Density=3.28 g/cm3\text{Density} = 3.28\ \text{g/cm}^3.

The key has a volume of 5 cm³ and a density of 3.28 g/cm³.

Competitive stretch

Suppose the student had misread the final level as 46 mL46\ \text{mL} instead of 45 mL45\ \text{mL} (a one-division parallax error). The computed volume would jump to 6 mL6\ \text{mL}, and the density would come out as 16.400/6=2.73 g/cm316.400/6 = 2.73\ \text{g/cm}^3 — a change of over 16% in the final answer from a single misread division. This shows why careful, eye-level meniscus reading matters far more here than it might for a length measurement: in a narrow cylinder, a small height error corresponds to a proportionally larger volume error.

Common traps

  • Letting the solid splash water out while lowering it in — any water lost makes the final reading, and hence the calculated volume, come out too low.
  • Not fully submerging the object — if any part stays above the surface, the level rise underestimates the true volume, since that part of the solid never displaced any water.
  • Reading the meniscus at its top edge, or with the eye above or below the water line, instead of level with the eye at the bottom of the curve — this parallax error can shift the reading by a division or more.
  • Applying the method to a solid that dissolves in water (like a sugar cube) or reacts with it (like reactive metals) — displacement only measures true volume when the solid stays completely unchanged while submerged.

Ask the tutor: Want to try a displacement problem where the object floats, or see exactly how a misread meniscus changes the final density — ask the AI tutor to work through it with you.

Why does the method not simply work for salt or sugar

Salt and sugar dissolve in water, so lowering a sugar cube into a cylinder does not just displace water — the solid itself disappears into the liquid, mixing with it rather than staying as a separate submerged object. The water level might barely change, or change in a way that has nothing to do with the sugar's actual volume. The water-displacement method only gives a correct answer for solids that remain completely solid and insoluble while underwater, such as stones, metal keys, or glass beads.

What if the object floats instead of sinking

The method as described assumes full submersion, so a floating object needs a small trick: it must be gently pushed under the surface, often using a thin needle or pin, until it is completely covered by water but the needle itself displaces a negligible amount. The level is then read exactly as before. Without pushing it under, a floating object only displaces a volume of water equal to the part below the surface, not its full volume, and the reading would be wrong.

Does the shape of the measuring cylinder affect the answer

No — the volume calculated by subtracting the initial reading from the final reading is always the true volume of the object, regardless of how narrow or wide the cylinder is, because both readings are taken on the very same graduated scale. What the cylinder's width does affect is precision: a narrower cylinder shows a larger rise in water level for the same object volume, making the reading easier to take accurately, which is why measuring cylinders are deliberately made narrow rather than wide.

Before this

Where this leads