Grade 11 Volume of Solids: Short Notes

11 ශ්‍රේණිය · 5 පරිච්ඡේදය · ඝන වස්තුවල පරිමාව

90%

of Grade 11 term tests include it (55 of 61)

534

marks in past papers (57 questions)

11 min

full lesson with quick checks

🎯 What you'll learn

Find the volume of a right pyramid and a cone
Find the volume of a sphere and a hemisphere
Find the volume of a solid made of two simple solids
Solve melting and recasting problems by equating volumes
Find the rise in water level when a solid is immersed

Melting old pots

A small foundry in Moratuwa melts a solid metal cylinder and casts it into little balls for bearings. How many balls will one cylinder give? Metal is not lost when it melts, so the answer comes from comparing volumes. This chapter gives you the formulas you need.

Recall: prisms and cylinders

For any solid with the same cross-section all the way up:Volume = base area × height• Cuboid: l × b × h• Triangular prism: (½ × base × height of triangle) × length• Cylinder: πr²hThe new solids in this chapter (pyramid, cone, sphere) narrow down, so they need new formulas.

🧮 Right pyramid

h is the perpendicular height, from the apex straight down to the base, not the slant height. A pyramid holds exactly one third of a prism with the same base and height.

V = ⅓ × base area × h

✏️ A square pyramid

A paperweight is a right pyramid on a square base of side 12 cm, with perpendicular height 10 cm. Find its volume.

1Base area = 12 × 12 = 144 cm².
2V = ⅓ × 144 × 10 = 480 cm³.

✅ 480 cm³

A cone inside a cylinder

A cone and a cylinder with the same radius r and height h. Three cones of water exactly fill the cylinder, so V(cone) = ⅓πr²h.

🧮 Sphere and hemisphere

Sphere: 4/3 πr³. Hemisphere: 2/3 πr³. Note the cube, r³. A hemisphere is half a sphere, so 4/3 halves to 2/3. If you are given the diameter, halve it first.

V = 4/3 πr³ , V = 2/3 πr³

✏️ A hemispherical bowl

A large hemispherical brass bowl used at a dansala has an inner radius of 21 cm. How much kola kenda can it hold, in litres? (π = 22/7)

1V = 2/3 × 22/7 × 21³.
221³ = 9261, so V = 2 × 22 × 9261/21 = 19 404 cm³.
31000 cm³ = 1 l, so about 19.4 litres.

✅ 19 404 cm³ ≈ 19.4 l

Melting, recasting and water

Two ideas behind most Paper II questions:• Recasting: volume of the original solid = total volume of the new solids. Number of pieces = big volume ÷ one small volume.• Displacement: when a solid sinks fully in water, the water rises by a volume equal to the solid's volume. In a cylinder, rise × πR² = volume of the solid.

⚠️ Putting in numbers too early

Examiners note that students substitute π = 22/7 at the start, then drown in fractions. When two volumes are compared or set equal, keep π as a letter: it cancels. Write the equation first, e.g. πR² × rise = 4/3πr³, cancel π and common factors, then calculate.

⚠️ r² or r³? Radius or diameter?

Common slips:• Using r² for a sphere. Spheres and hemispheres use r³; cones and cylinders use r².• Using the diameter as the radius.• For a hemisphere, forgetting to halve the sphere's 4/3.• If a vessel is not full, the solid first fills the empty space; only the rest overflows.

📌 Remember

• Prism, cylinder: V = base area × h.• Pyramid, cone: V = ⅓ × base area × h (perpendicular height); cone V = ⅓πr²h.• Sphere 4/3πr³, hemisphere 2/3πr³.• Recasting: volumes are equal. Displacement: rise × base area = volume sunk.• Keep π as a letter until it cancels. When the numbers are awkward, exams ask you to finish with log tables.

🧾 Quick revision sheet

Volume of a right pyramid	⅓ × base area × perpendicular height
Volume of a cone	⅓πr²h
Volume of a sphere	4/3 πr³
Volume of a hemisphere	2/3 πr³
Volume of a cylinder	πr²h
Melting and recasting: what stays the same?	The total volume.
Sphere sinks in a cylinder of radius R: rise?	rise = volume of sphere ÷ πR²
1 litre = ? cm³	1000 cm³