Cylinder Cone and Sphere Class 10 ICSE — Bansal Chapter 20
Cylinder cone and sphere Class 10 ICSE is the topic of Chapter 20 in the Bansal Mathematics book. The chapter teaches formulas for volume and surface area of a cylinder, a cone, a sphere, and related solids. It covers hollow versions, spherical shells, conversions, and combined shapes. Each section below follows the book’s order with a worked example. Worked examples use fresh numbers so you can practise the method.
Download Cylinder Cone and Sphere Class 10 ICSE Mathematics Chapter 20 PDF
Download the Cylinder Cone and Sphere Class 10 ICSE Mathematics chapter PDF from the Bansal textbook to revise offline or print for practice. It is the full chapter as printed in the book — 25 pages, 8 sections, 83 questions.
What is inside Chapter 20
25 pages · 8 sections · 29 worked examples · 83 exercise questions
| Section | Title |
|---|---|
| 20.1 | Cylinder |
| 20.2 | Hollow Cylinder |
| 20.3 | Cone |
| 20.4 | Sphere |
| 20.5 | Spherical Shell |
| 20.7 | Conversion of solids |
| 20.8 | Combination of Solids |
| 20.9 | Miscellaneous Problems |
- EXERCISE 20(A): 28 questions
- EXERCISE 20(C): 12 questions
- EXERCISE 20(E): 10 questions
- EXERCISE 20(F): 15 questions
- EXERCISE 20(G): 18 questions
20.1 Cylinder
A cylinder is a solid with a uniform circular cross-section. Stack identical circles vertically and you get one. Let r be the radius of the circular cross-section and h the height. This chapter builds on mensuration skills you met earlier in the Class 10 ICSE Maths book.
The formulas follow from two simple ideas. Curved surface area equals the perimeter of the cross-section times the height. Total surface area adds the two circular ends to the curved part.
- Area of cross-section: \pi r^2
- Perimeter of cross-section: 2\pi r
- Curved surface area: 2\pi rh
- Total surface area: 2\pi r(h + r)
- Volume: \pi r^2 h
Worked example: a cylinder with radius 7 cm and height 15 cm.
V = \pi r^2 h = \frac{22}{7} \times 7^2 \times 15 = 2310\ \text{cm}^3
TSA = 2\pi r(h + r) = 2 \times \frac{22}{7} \times 7 \times (15 + 7) = 968\ \text{cm}^2
Curved surface area covers only the wall. Total surface area includes both flat ends. If a question says “open at one end,” subtract one \pi r^2.
20.2 Hollow Cylinder
A hollow cylinder has two radii. Let R be the external radius, r the internal radius, and h the height. The wall thickness is R - r.
The key idea is subtraction. The ring-shaped cross-section has area \pi R^2 - \pi r^2 = \pi(R^2 - r^2). Volume of material is external volume minus internal volume:
V = \pi R^2 h - \pi r^2 h = \pi(R^2 - r^2) h
Total surface area adds the external curved surface 2\pi Rh, the internal curved surface 2\pi rh, and two ring faces: 2\pi Rh + 2\pi rh + 2\pi(R^2 - r^2).
Worked example: external radius 10 cm, internal radius 8 cm, height 20 cm.
V = \pi(10^2 - 8^2) \times 20 = \frac{22}{7} \times 36 \times 20 = 2262.9\ \text{cm}^3
Common mistake: using only the external radius in the volume formula. You need the difference of the squares, R^2 - r^2. If a pipe has internal diameter 11.2 cm and the metal is 0.4 cm thick, the external radius is 5.6 + 0.4 = 6.0 cm. Always add the thickness to the internal radius. Hollow cylinder problems appear often in the Class 10 ICSE syllabus.
20.3 Cone
A right circular cone is formed by revolving a right-angled triangle about one of its legs, not the hypotenuse. The height h, base radius r, and slant height l satisfy Pythagoras’ theorem:
l^2 = h^2 + r^2
The formulas are:
- Volume: \frac{1}{3}\pi r^2 h
- Curved surface area: \pi rl
- Total surface area: \pi r(l + r)
Worked example: cone with radius 6 cm and height 8 cm.
l = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\ \text{cm}
V = \frac{1}{3} \times \frac{22}{7} \times 36 \times 8 = 301.7\ \text{cm}^3
TSA = \frac{22}{7} \times 6 \times (10 + 6) = 301.7\ \text{cm}^2
For conical tent problems, the textbook uses a ratio method. If base area to curved surface area is 40:41, then r:l = 40:41. Take r = 40x and l = 41x, then use l^2 = h^2 + r^2 to find x.
20.4 Sphere
A sphere is the solid obtained by revolving a circle about any of its diameters. If the circle has radius r, the sphere also has radius r.
Only two formulas:
- Volume: \frac{4}{3}\pi r^3
- Surface area: 4\pi r^2
Worked example: sphere with radius 5 cm.
V = \frac{4}{3} \times \frac{22}{7} \times 125 = 523.8\ \text{cm}^3
A = 4 \times \frac{22}{7} \times 25 = 314.3\ \text{cm}^2
Watch the surface area formula. It is 4\pi r^2, not 2\pi r^2. A sphere has no flat faces, so surface area and curved surface area are the same quantity.
20.5 Spherical Shell
A spherical shell is the solid enclosed between two concentric spheres. With external radius R and internal radius r, its volume is
V = \frac{4}{3}\pi (R^3 - r^3)
When a solid sphere is cut through its centre, each half is a hemisphere. The hemisphere is a common shape in this ICSE chapter.
- Volume of hemisphere: \frac{2}{3}\pi r^3
- Total surface area of hemisphere: 3\pi r^2
The total surface area includes the curved part 2\pi r^2 plus the flat circular base \pi r^2.
Worked example: hemispherical bowl with internal radius 7 cm.
V = \frac{2}{3} \times \frac{22}{7} \times 343 = 718.7\ \text{cm}^3
TSA = 3 \times \frac{22}{7} \times 49 = 462\ \text{cm}^2
Here is a comparison of the formulas in this chapter so far:
| Solid | Volume | Curved surface area | Total surface area |
|---|---|---|---|
| Cylinder (r, h) | \pi r^2 h | 2\pi rh | 2\pi r(h+r) |
| Cone (r, h, l) | \frac{1}{3}\pi r^2 h | \pi rl | \pi r(l+r) |
| Sphere (r) | \frac{4}{3}\pi r^3 | 4\pi r^2 | 4\pi r^2 |
| Hemisphere (r) | \frac{2}{3}\pi r^3 | 2\pi r^2 | 3\pi r^2 |
20.7 Conversion of solids
When a solid is melted and recast into another shape, the volume does not change. Volume before = volume after.
Worked example: a sphere of radius 6 cm is melted into cones of radius 2 cm and height 4 cm. How many cones?
V_{\text{sphere}} = \frac{4}{3}\pi (6)^3 = 288\pi\ \text{cm}^3
V_{\text{cone}} = \frac{1}{3}\pi (2)^2 (4) = \frac{16}{3}\pi\ \text{cm}^3
n = \frac{288\pi}{\frac{16}{3}\pi} = \frac{288 \times 3}{16} = 54
So 54 cones are formed. Check units before substituting: if a radius is in mm, convert to cm first. The same method handles hollow spheres melted into cones; use the shell volume \frac{4}{3}\pi(R^3 - r^3) as the “before” volume. This ratio method connects to the earlier chapter on constructions of circles.
20.8 Combination of Solids
Composite solids combine two or more shapes. Examples are a cone mounted on a hemisphere, or a cylinder with a conical cavity. Add the volumes of the parts. For surface area, add only the exposed surfaces.
The base shared by two solids is not double-counted. A cone on a hemisphere shares the circular face, so the toy’s surface area is just \pi rl + 2\pi r^2.
Worked example: toy of cone on hemisphere, both with radius 6 cm, cone height 8 cm.
l = \sqrt{8^2 + 6^2} = 10\ \text{cm}
A = \pi rl + 2\pi r^2 = 3.14 \times 60 + 2 \times 3.14 \times 36 = 414.5\ \text{cm}^2
V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = 3.14 \times 96 + 3.14 \times 144 = 753.6\ \text{cm}^3
When a conical cavity is drilled into a cylinder, subtract the cone’s volume instead of adding it. The cavity’s slant height comes from l^2 = h^2 + r^2.
20.9 Miscellaneous Problems
This section mixes the solids with two extra tools.
Uniform cross-section: for a body with a constant cross-section, volume equals area of cross-section times length. Surface area excluding the ends equals perimeter of cross-section times length.
Water flow: volume discharged per second equals area of cross-section times speed of flow.
Worked example: a pipe with internal diameter 3.5 cm carries water at 2 m/s. How much water discharges in one minute?
r = 1.75\ \text{cm},\quad A = \frac{22}{7} \times (1.75)^2 = 9.625\ \text{cm}^2
\text{per second} = 9.625 \times 200 = 1925\ \text{cm}^3
\text{per minute} = 1925 \times 60 = 115{,}500\ \text{cm}^3 = 115.5\ \text{litres}
Remember 1\ \text{litre} = 1000\ \text{cm}^3 and 1\ \text{m}^3 = 1000\ \text{litres}. For tunnels and tanks, find the cross-section perimeter first, then multiply by the length.
What to practise from Cylinder Cone and Sphere Class 10 ICSE
Work through every exercise in this ICSE chapter. The question blocks are:
- EXERCISE 20(A) — 28 questions on cylinders and hollow cylinders
- EXERCISE 20(C) — 12 questions on spheres and hemispheres
- EXERCISE 20(E) — 10 questions on combinations of solids
- EXERCISE 20(F) — 15 questions on mixed solids and tunnels
- EXERCISE 20(G) — 18 questions on conversions and mixed problems
Many questions in EXERCISE 20(E) and EXERCISE 20(G) are drawn from previous year papers spanning 2000 to 2015. The syllabus pattern follows the official CISCE website at cisce.org.
Cylinder Cone and Sphere Class 10 ICSE frequently asked questions
What is the formula for the volume of a hollow cylinder?
For a hollow cylinder with external radius R, internal radius r, and height h, the volume of material is \pi(R^2 - r^2)h. This is the external volume minus the internal volume.
How do I find the slant height of a cone when I know its height and radius?
Use Pythagoras: l^2 = h^2 + r^2, so l = \sqrt{h^2 + r^2}. For a cone of height 8 cm and radius 6 cm, the slant height is 10 cm.
What is the difference between a sphere and a spherical shell?
A sphere is solid throughout, with volume \frac{4}{3}\pi r^3. A spherical shell is the region between two concentric spheres, with volume \frac{4}{3}\pi(R^3 - r^3).
How do I approach conversion of solids problems in cylinder cone and sphere class 10 ICSE?
Set the volume of the original solid equal to the total volume of the new solid. Then solve for the unknown dimension. Keep every unit the same before substituting.
When should I use total surface area instead of curved surface area?
Use total surface area when a question asks for the whole outer covering, including flat ends. Use curved surface area for the wall alone, like the inner surface of a well or the canvas of a conical tent.
Reference: ICSE Class 10 Mathematics textbook material, chapter 20: Cylinder Cone and Sphere — CISCE-aligned.
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