8 min readConcept sheetPart I MCQ, Part II (2 marks), and often a Part III answer (5 marks)

Conversion of solids: melting and recasting

When a solid is melted and recast, its shape changes but its volume stays the same. Write volume before equals volume after and solve.

Key ruleVolume before melting = total volume after recasting

Number of new solidsVolume of the big solid ÷ volume of one small solid

Water through a pipeVolume per hour = πr² × distance the water flows in one hour

Earth spread outHeight = volume of earth ÷ area it covers

Method for any recasting sum

1
Write: volume of the original solid = total volume of the new solid or solids.
2
Write the volume formula for each shape.
3
For many small solids: number = volume of the big solid ÷ volume of one small solid.
4
Cancel π and common factors before you multiply.
5
Solve for the unknown and write the unit.

Example 1: sphere into a cone

Problem

A metal sphere of radius 6 cm is melted and recast into a cone of base radius 6 cm. Find the height of the cone.

Working

Volume of cone = volume of sphere.
(1/3)π × 6² × h = (4/3)π × 6³
Cancel (1/3)π on both sides: 36h = 4 × 216
36h = 864
h = 864 ÷ 36 = 24 cm

Answer

Height of the cone = 24 cm (4 times the radius)

Example 2: how many small spheres?

Problem

A solid copper cylinder of radius 6 cm and height 32 cm is melted and made into small spheres of radius 2 cm. How many spheres are made?

Working

Number of spheres = volume of cylinder ÷ volume of one sphere.
= (π × 6² × 32) ÷ ((4/3)π × 2³)
= (36 × 32) ÷ (32/3)
= 36 × 32 × 3/32
= 108

Answer

108 spheres

Example 3: earth from a well

Problem

A well of diameter 7 m is dug 20 m deep. The earth taken out is spread evenly to make a platform 22 m long and 14 m wide. Find the height of the platform. (Take π = 22/7.)

Working

Radius of the well = 7 ÷ 2 = 3.5 m.
Volume of earth = πr²h = 22/7 × 3.5 × 3.5 × 20
= 38.5 × 20 = 770 m³
Volume of platform = 22 × 14 × height = 308 × height
308 × height = 770
height = 770 ÷ 308 = 2.5 m

Answer

Height of the platform = 2.5 m

Example 4: water flowing into a tank

Problem

Water flows at 10 km per hour through a pipe of internal radius 7 cm into a rectangular tank 22 m long and 14 m wide. How long will the water level take to rise by 1 m? (Take π = 22/7.)

Working

In 1 hour the water in the pipe moves 10 km = 10000 m. Radius = 7 cm = 0.07 m.
Water per hour = πr² × length = 22/7 × 0.07 × 0.07 × 10000
= 22/7 × 49 = 154 m³
Water needed = 22 × 14 × 1 = 308 m³
Time = 308 ÷ 154 = 2 hours

Answer

2 hours

In real life

A goldsmith melting old gold bangles into new jewellery, or earth dug from a village well spread as a platform: the shape changes but the volume stays the same.

Exam angle

1 marks: “A sphere of radius r is melted and recast into a cone of the same radius. What is the height of the cone?” (1/3)πr²h = (4/3)πr³ gives h = 4r.

2 marks: “How many spheres of radius 1 cm can be made by melting a sphere of radius 4 cm?” Number = (4/3)π × 4³ ÷ ((4/3)π × 1³) = 64 ÷ 1. Answer: 64 spheres.

5 marks: “A solid metal hemisphere of radius 9 cm is melted and recast into small cones, each of radius 3 cm and height 3 cm. How many cones are made?” Volume of hemisphere = (2/3)π × 729 = 486π cm³. Volume of one cone = (1/3)π × 9 × 3 = 9π cm³. Number = 486π ÷ 9π = 54 cones.

Remember it

Melt it, mould it: the volume never changes. Make 'volume before = volume after' your first line.