Equating Volumes: Melting and Recasting

When a solid is melted or water is poured, the volume stays the same: set the two volumes equal and solve for the missing length.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024, Edexcel Further Pure Maths 4PM1 · 10 real past-paper questions below

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Same Stuff, New Shape

Melt a candle and pour the wax into a new mould. The shape changes, but there is exactly as much wax as before.

Rule

Melted, recast or poured: the volume stays the same. So volume before equals volume after.

You already know each volume from its own lesson. Questions usually print the sphere and cone formulas for you.

Solid	Volume
Prism or cuboid	end area × length
Cylinder	$π r^{2} h$
Cone	$\frac{1}{3} π r^{2} h$
Sphere	$\frac{4}{3} π r^{3}$

Common mistake

Setting the surface areas equal. The outside changes completely when the shape changes. Only the volume is kept.

Melted and Made Into a New Solid

The 3 or 4-mark question: one solid is melted and made into another. Find the old volume, set it equal to the new one, and solve.

A cylinder of radius 2 cm and height 9 cm is melted and made into a sphere. Find the radius of the sphere.

Worked example

1
$π \times 2^{2} \times 9 = 36 π$ — the metal in the cylinder. Keep π.
2
$\frac{4}{3} π r^{3} = 36 π$ — the sphere has the same volume
3
$\frac{4}{3} r^{3} = 36$ — π is on both sides, so it cancels
4
$r^{3} = 36 \times \frac{3}{4} = 27$ — times by 3, divide by 4
5
$r = 327 = 3 cm$ — r is cubed, so cube root

Extended papers do the same with letters. A cone of radius $3 x$ and height $12 x$ is made into a sphere.

Worked example

1
$\frac{1}{3} π (3 x)^{2} (12 x) = 36 π x^{3}$ — the cone: square 3x first
2
$\frac{4}{3} π r^{3} = 36 π x^{3}$ — set the volumes equal
3
$r^{3} = 27 x^{3}$ — cancel π, times 3, divide by 4
4
$r = 3 x$ — the cube root of 27 is 3

Common mistake

Stopping at $r^{3} = 27$ and writing 27. The radius was cubed, so you must cube root it.

Poured Into Another Container

The 3-mark question: water is poured from one container into another. The water does not change, so its two volumes are equal.

A tank 30 cm by 20 cm holds water 12 cm deep. It is poured into a cylinder, where it is 18 cm deep. Find the radius of the cylinder.

Worked example

1
$30 \times 20 \times 12 = 7200$ — the water in the tank, in cm³
2
$π r^{2} \times 18 = 7200$ — the same water in the cylinder
3
$r^{2} = \frac{7200}{18 π} = 127.32 \dots$ — divide by 18π
4
$r = 127.32 \dots = 11.3 cm$ — square root, 3 significant figures

Common mistake

Using the height of the tank instead of the depth of the water. Only the water is poured, so use 12, not the full height.

Drop It In or Take It Out

The 3-mark question: a solid sinks into water, or is lifted out. The water level moves by the solid's volume, spread over the base of the tank.

Formula

rise = \frac{volume of the solid}{base area of the tank}

The sphere from the last card (radius 3 cm) sinks into a cylinder of water of radius 10 cm. The water was 15 cm deep.

Worked example

1
$\frac{4}{3} π \times 3^{3} = 36 π$ — the sphere pushes up its own volume
2
$π \times 1 0^{2} \times d = 36 π$ — a thin layer of water, depth d
3
$d = \frac{36 π}{100 π} = 0.36 cm$ — π cancels
4
$15 + 0.36 = 15.36 cm$ — the new depth

Common mistake

Only half the ball is under water? Then it pushes up only half its volume. Take the sphere out and the level falls by the same amount.

Key idea

The story goes that Archimedes saw his bath water rise as he got in, and shouted "Eureka!" ("I have found it!"). He had found a way to measure the volume of an odd shape, like the king's gold crown.

How Many Can You Make?

The how-many question: one big solid is melted into many small ones. Divide the big volume by one small volume, then round down.

A metal block 10 cm by 8 cm by 6 cm is melted into balls of radius 1.5 cm.

Worked example

1
$10 \times 8 \times 6 = 480$ — the block, in cm³
2
$\frac{4}{3} π \times 1. 5^{3} = 4.5 π$ — one ball, about 14.14
3
$480 \div 4.5 π = 33.95 \dots$ — how many balls of metal
4
$33 balls$ — round down: no part balls

Common mistake

Rounding 33.95 up to 34. There is not enough metal for the 34th ball, however close it looks.

Key idea

Asked how much metal is left? Take away what was used: $480 - 33 \times 4.5 π$ is about 13.5 cm³.

Units First

Before you set two volumes equal, check that both use the same unit. Litres and metres hide in these questions.

Rule

$1 litre = 1000 cm^{3}$ and $1 m^{3} = 1000000 cm^{3}$ .

2 litres of water are poured into a cylinder of radius 8 cm. How deep is the water?

Worked example

1
$2 litres = 2000 cm^{3}$ — change units first
2
$π \times 8^{2} \times h = 2000$ — same water in the cylinder
3
$h = \frac{2000}{64 π} = 9.95 cm$ — divide by 64π

Common mistake

Thinking $1 m^{3}$ is 100 cm³. A metre is 100 cm in each of three directions, so it is $10 0^{3}$ .

Which Method? Read the Question

Find the two things that hold the same amount. Then write that volume before equals volume after.

You see	You do
Melted, recast, made into	set the two volumes equal
Poured into another container	the water's volume stays the same
Sinks in, or is taken out	solid's volume ÷ base area gives the rise
How many can be made	divide the volumes, round down
Litres or metres	change to cm³ first
One solid, find a length	Volume of a Prism
Then find the surface area	Surface Area of a Cuboid and Cylinder

Key idea

Keep π as a symbol. It is on both sides, so it usually cancels and you round only once, at the end.

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