Volume and capacity

Pour water from a tall glass into a wide bowl and something strange seems to happen — the water looks like less. But the amount of water has not changed at all; only its container has. This is the everyday confusion that the topic of volume and capacity clears up with one precise idea: volume is the space a solid occupies, capacity is how much liquid a hollow container can hold, and the two are tied together by a single conversion factor, $1$ litre $= 1000\ \text{cm}^3$ . Once you know that one number, every tank, can, and bottle problem in this chapter becomes an exercise in careful unit-conversion and careful bookkeeping of a quantity that never disappears — it only changes shape.

The core insight

Capacity is just the volume of a container's hollow interior measured in litres or millilitres instead of cubic centimetres, connected to volume by the fixed conversion $1\ \text{L} = 1000\ \text{cm}^3$ , and that fixed quantity of liquid is conserved even as it is poured between containers of completely different shapes.

The core idea

Volume tells you how much space a solid fills; capacity tells you how much a hollow container can hold. They are measured in different-looking units — volume usually in $\text{cm}^3$ or $\text{m}^3$ , capacity usually in litres (L) or millilitres (mL) — but underneath they measure exactly the same thing: three-dimensional space. The link between the two systems of units is one number you must know cold:

1\ \text{L} = 1000\ \text{cm}^3, \qquad 1\ \text{mL} = 1\ \text{cm}^3

where: L stands for litre, mL for millilitre, and $\text{cm}^3$ for cubic centimetre. Because a millilitre equals a cubic centimetre exactly, it also follows that $1\ \text{L} = 1000\ \text{mL}$ — the same factor of $1000$ shows up whether you are comparing $\text{cm}^3$ to L or mL to L. This chapter restricts capacity problems to the three solids you already know how to measure — the cube, the cuboid, and the right circular cylinder — so once a container's volume is found with the usual formulas, converting it to a capacity is just a division by $1000$ :

V_{cube} = a^3, \qquad V_{cuboid} = lbh, \qquad V_{cylinder} = \pi r^2 h

where: $a$ is the edge of a cube, $l$ , $b$ , $h$ are the length, breadth and height of a cuboid, and $r$ , $h$ are the radius and height of a cylinder, all measured in the same length unit so the volume comes out consistently in that unit cubed. For large real-world tanks — water tanks on a rooftop, swimming pools, storage drums — it is often more natural to work in metres, and a second, larger conversion applies:

1\ \text{m}^3 = 1{,}000{,}000\ \text{cm}^3 = 1000\ \text{L} = 1\ \text{kilolitre (kL)}

So a cubic metre is worth a thousand litres, or one kilolitre. Picture a cube exactly $10\ \text{cm}$ on each side sitting inside a larger vessel — its volume is $10^3 = 1000\ \text{cm}^3$ , and that cube of water is precisely one litre. Scaling this cube up by a factor of $10$ in every direction turns it into a $1\ \text{m}$ cube holding $1000$ litres, one kilolitre.

Figure

A small cube of side ten centimetres shown equal to a filled measuring jar marked one litre, illustrating that a thousand cubic centimetres of volume is exactly one litre of capacity

Derivation AWhere the factor of 1000 in the litre conversion actually comes from

The goal is to show that $1\ \text{L} = 1000\ \text{cm}^3$ is not an arbitrary number to memorise but falls directly out of the definition of the litre combined with the ordinary cube-volume formula $V = a^3$ , and that the same reasoning, scaled up, gives the metre-based conversion $1\ \text{m}^3 = 1000\ \text{L}$ used for large tanks.

Step 1. Start from the definition. A litre is defined as one cubic decimetre, and a decimetre equals $10\ \text{cm}$ . So a litre is, by definition, the capacity of a cube whose edge is exactly $10\ \text{cm}$ .

Step 2. Apply the cube-volume formula to that defining cube. Using $V = a^3$ with $a = 10\ \text{cm}$ , the volume works out to $V = 10^3 = 1000\ \text{cm}^3$ . This single computation is the entire justification for the conversion factor — it is not a rounded or approximate number, it is exact by construction.

Step 3. Extend the same reasoning to the millilitre. Since a millilitre is one-thousandth of a litre, it must correspond to a volume one-thousandth the size of $1000\ \text{cm}^3$ , which is exactly $1\ \text{cm}^3$ — the volume of a cube with edge $1\ \text{cm}$ . This is why the millilitre and the cubic centimetre are numerically identical, and it follows automatically once Step 2 is accepted.

Step 4. Scale the whole argument up to metres. A cube of side $1\ \text{m} = 100\ \text{cm}$ has volume $100^3 = 1{,}000{,}000\ \text{cm}^3$ by the same formula, and dividing by the $1000\ \text{cm}^3$ -per-litre factor from Step 2 gives $1{,}000{,}000 \div 1000 = 1000$ litres, which is defined as one kilolitre. Every conversion used in this topic — $\text{cm}^3$ to litres, litres to kilolitres, $\text{m}^3$ to litres — is this same idea of expressing both units as cubes and cubing the ratio of their edge lengths.

The most important habit in this topic is remembering that liquid volume is conserved. Water poured from a full jug into three small glasses does not change in total quantity — only its shape and spread change. Whether water is poured from a tap into a tank, transferred from a wide tub into narrow bottles, or drained from a cylindrical drum into a rectangular container, the total volume stays fixed. So the working strategy for almost every capacity word problem is the same: find the volume (or capacity) of whichever container's dimensions are fully given, convert it into the unit the question asks for, and only then use that fixed number to answer questions about a second container — a missing dimension, a count of how many smaller containers can be filled, or how much liquid is left over.

Worked example

The same $2\ \text{m} \times 1.5\ \text{m} \times 1.2\ \text{m}$ cuboidal tank you may already have met holds $3600$ litres when full — this is the base fact this example builds on. Suppose that empty tank is now being filled by a pipe that delivers water at a steady rate of $150$ litres per minute. Find how long the pipe takes to fill the tank completely, and how many litres remain to be filled after $15$ minutes.

Step 1. Confirm the tank's capacity using the cuboid volume formula and the $\text{m}^3$ -to-litre conversion. $V = l \times b \times h = 2 \times 1.5 \times 1.2 = 3.6\ \text{m}^3$ , and since $1\ \text{m}^3 = 1000\ \text{L}$ , the capacity is $3.6 \times 1000 = 3600$ litres. This is the fixed target quantity the pipe must deliver.

Step 2. Find the total filling time by dividing the tank's capacity by the pipe's rate, since rate is litres delivered per minute. $\text{Time} = \dfrac{\text{Capacity}}{\text{Rate}} = \dfrac{3600}{150} = 24$ minutes.

Step 3. Find how much water has entered after $15$ minutes, using the same rate. $\text{Volume delivered} = 150 \times 15 = 2250$ litres.

Step 4. Subtract to find what remains. $\text{Remaining} = 3600 - 2250 = 1350$ litres.

The tank fills completely in 24 minutes, and 1350 litres remain to be filled after 15 minutes.

Competitive stretch

If a second, identical tank must also be filled from the same pipe, but only after the first tank is full, find the total time for both tanks together. Since the pipe fills one tank at a time at the same steady rate, the second tank also takes $24$ minutes, so the combined time is $24 + 24 = 48$ minutes — the rate does not change, so filling time simply adds when tanks are filled one after another rather than simultaneously.

Common traps

Reporting the volume instead of the capacity: calculating $V$ correctly in $\text{cm}^3$ or $\text{m}^3$ but forgetting to divide by $1000$ (or by $1{,}000{,}000$ then multiply by $1000$ ) when the question specifically asks for litres or kilolitres.
Inverting the conversion factor: writing $1\ \text{L} = 100\ \text{cm}^3$ or treating $1000\ \text{L} = 1\ \text{cm}^3$ instead of the correct $1\ \text{L} = 1000\ \text{cm}^3$ — always sanity-check with the definition, a litre is a small cube of side $10\ \text{cm}$ , not $1\ \text{cm}$ .
Rounding the wrong way in how-many-containers problems: if capacity divided by bottle size does not come out even, only whole bottles can be completely filled, so the answer must be rounded down, never up or to the nearest whole number.
Ignoring internal versus external dimensions: capacity is always about the internal hollow space of a container, so if a problem states wall thickness along with external measurements, the internal dimensions must be found first before any volume formula is applied.

Ask the tutor: Stuck on when to divide by 1000 versus 1000000, or want to try a filling-and-emptying pipe problem step by step? Ask the AI tutor to walk through one with you.

Why is a litre defined using a 10 cm cube instead of some other size

The litre is defined as one cubic decimetre, and a decimetre is exactly $10\ \text{cm}$ , so a cube of side $10\ \text{cm}$ has volume $10^3 = 1000\ \text{cm}^3$ by definition, and that is called one litre. It is not a coincidence you need to memorise separately — it falls directly out of the metric system's design, where each larger capacity unit is built from a cube of the matching length unit. Once you see the $10\ \text{cm}$ cube picture, the factor of $1000$ stops feeling arbitrary.

If I pour water from a tall thin glass into a short wide bowl, does the capacity change

No. The amount of water — its volume — stays exactly the same; only the shape of the space it occupies changes, since the bowl is wider but shallower than the glass. This is the same conservation principle used throughout this chapter: the total volume of a fixed quantity of liquid never changes just because you move it into a differently shaped container. What looks like "less water" in a wide bowl is really the same water spread thinner.

Do cone, sphere, and hemisphere capacities belong in this chapter

Not at this level. In Class 8 ICSE, volume-and-capacity problems are restricted to the cube, cuboid, and right circular cylinder — the three solids covered earlier in this same chapter. Capacity problems built on the cone, sphere, and hemisphere use volume formulas that are introduced only in Class 10, so you will not be expected to combine those shapes with litre conversions yet, even though the underlying conversion factor of $1000$ would work exactly the same way for them too.

Volume and capacity

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads