Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.
Step-by-step solution
To find: Volume of liquid (nearest whole number)
Idea: The liquid fills the space in the box not taken up by the spheres: liquid = box volume − total volume of the spheres.
- Volume of box = 16 × 8 × 8 = 1024 cm3. (The 16 spheres fit as 4 × 2 × 2 of diameter 4 cm.)1 mark
- Volume of one sphere = 43π × 23 = 32π3 cm3.1 mark
- 16 spheres: 16 × 32π3 = 512π3 ≈ 512 × 3.143 ≈ 535.89 cm3.1 mark
- Liquid = 1024 − 535.89 = 488.11 ≈ 488 cm3.1 mark
Check: With π = 227: 1024 − 536.38 = 487.62 ≈ 488 ✓. The spheres fill about 52% of the box, as expected for this square packing.
Answer to write in the exam
Volume of box = 16 × 8 × 8 = 1024 cm3
Volume of 16 spheres = 16 × 43π × 23 = 512π3 ≈ 535.89 cm3
Liquid = 1024 − 535.89 = 488.11
∴ Volume of liquid ≈ 488 cm3
Common mistakes that cost marks
- Using surface area 4πr2 instead of volume.
- Forgetting to multiply by 16.
- Rounding the sphere volume early (e.g. 33 cm3 each), which gives 496 cm3.
How this can come in the exam
A shop packs 8 glass marbles of radius 1.5 cm in a cubical box of side 6 cm and fills the gaps with foam beads.
(i) Check that 8 marbles fit. (ii) Find the volume of the box. (iii) Find the total volume of the marbles. (iv) Find the volume left for foam. (π = 3.14)
Show answer
(i) Diameter 3 cm: 2 × 2 × 2 = 8 fit (1 mark). (ii) 63 = 216 cm3 (1 mark). (iii) 8 × 43 × 3.14 × 3.375 = 113.04 cm3 (1 mark). (iv) 216 − 113.04 = 102.96 cm3 (1 mark).Try one yourself
Four balls of radius 3 cm are packed in a box 12 cm × 12 cm × 6 cm and the gaps are filled with sand. Find the volume of sand. (π = 3.14)
Show answer
Box 864; balls 4 × 113.04 = 452.16; sand ≈ 411.84 cm3.
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