Solid mensuration in CDS covers cuboids, right circular cylinders and cones, and spheres. The syllabus asks for surface areas and volumes, and the paper adds a few recurring twists. A solid is melted and recast into another shape, water flows through a pipe into a tank, or every dimension is scaled and you must say what happens to the volume.
Each twist rests on one principle. When a solid is melted or water is poured, the volume stays the same. Write "volume before = volume after", cancel π where you can, and the arithmetic usually comes out in whole numbers.
Formula table
| Solid | Volume | Curved (lateral) surface | Total surface |
|---|---|---|---|
| Cube, edge a | a³ | 4a² | 6a² |
| Cuboid, l, b, h | lbh | 2h(l + b) | 2(lb + bh + hl) |
| Cylinder, r, h | πr²h | 2πrh | 2πr(r + h) |
| Cone, r, h, slant l | ⅓πr²h | πrl | πr(l + r) |
| Sphere, r | ⁴⁄₃πr³ | 4πr² | 4πr² |
| Hemisphere, r | ⅔πr³ | 2πr² | 3πr² |
| Hollow cylinder, radii R and r | πh(R² − r²) | 2πh(R + r) | 2πh(R + r) + 2π(R² − r²) |
For a cone, the slant height is l = √(r² + h²). The radius, the height and the slant form a right triangle.
Why some formulas look the way they do
- A cone holds exactly one-third of a cylinder with the same base and height. So their volumes are in the ratio 1 : 3.
- A hemisphere's total surface is its curved part, 2πr², plus the flat circular face, πr², which gives 3πr².
- The cylinder's curved surface unrolls into a rectangle of length 2πr (the circumference) and height h.
Diagonals
- Cube: a√3.
- Cuboid: √(l² + b² + h²).
Scaling
If every length is multiplied by k, surface areas are multiplied by k² and volumes by k³. So doubling the radius of a sphere makes its surface 4 times and its volume 8 times larger. A 10% rise in every length raises the volume by 1.1³ − 1 = 33.1%.
Melting, recasting and flow
- Number of small solids = volume of the big solid ÷ volume of one small solid.
- Wire drawn from a solid: the wire is a long cylinder, so length = volume ÷ (π × wire radius²).
- Flow through a pipe: volume per unit time = cross-section area × speed of flow. Match units, since speeds are often in km/h and pipe radii in cm.
- When cubes of edges a, b and c are melted into one cube, its edge is the cube root of a³ + b³ + c³. Edges 3, 4 and 5 give 27 + 64 + 125 = 216, so the new edge is 6.
Worked questions
Question 1: A cuboid measures 12 cm by 4 cm by 3 cm. Find its diagonal, total surface area and volume.
- Diagonal = √(144 + 16 + 9) = √169 = 13 cm.
- Surface = 2(48 + 12 + 36) = 192 cm². Volume = 144 cm³.
Question 2: A cone has a base radius of 7 cm and a height of 24 cm. Find its curved surface area and volume.
- l = √(49 + 576) = √625 = 25 cm.
- Curved surface = 22/7 × 7 × 25 = 550 cm².
- Volume = ⅓ × 22/7 × 49 × 24 = 1,232 cm³.
Question 3: A metal sphere of radius 6 cm is melted and drawn into a wire of radius 2 mm. Find the length of the wire.
- Sphere volume = ⁴⁄₃π × 216 = 288π cm³.
- Wire radius = 0.2 cm, so the cross-section is π × 0.04.
- Length = 288π ÷ 0.04π = 7,200 cm = 72 m.
Question 4: A hemispherical bowl has an internal radius of 21 cm. How many litres does it hold?
- ⅔ × 22/7 × 21³ = ⅔ × 22/7 × 9,261 = 19,404 cm³.
- 1 litre = 1,000 cm³, so the bowl holds 19.404 litres.
Question 5: A cylinder of radius 6 cm and height 15 cm is melted and recast into cones of radius 3 cm and height 5 cm. How many cones are made?
- Cylinder = π × 36 × 15 = 540π. One cone = ⅓ × π × 9 × 5 = 15π.
- Number = 540 ÷ 15 = 36.
Question 6: Water flows at 5 km/h through a pipe of internal diameter 14 cm into a tank 50 m long and 44 m wide. How long will the water level take to rise by 7 cm?
- Pipe radius = 0.07 m. Water per hour = 22/7 × 0.0049 × 5,000 = 77 m³.
- Water needed = 50 × 44 × 0.07 = 154 m³.
- Time = 154 ÷ 77 = 2 hours.
Practice set
- A cuboid measures 10 cm by 8 cm by 5 cm. Find its volume.
- A cylinder has a radius of 14 cm and a height of 5 cm. Find its volume.
- A cube has a total surface area of 216 cm². Find its volume.
- A cube has a diagonal of 5√3 cm. Find its volume.
- A cone has a radius of 6 cm and a height of 8 cm. Find its curved surface area.
- A sphere has a volume of 36π cm³. Find its surface area.
- Three cubes of edges 3 cm, 4 cm and 5 cm are melted into a single cube. Find its edge.
- The radius and height of a cylinder are in the ratio 2 : 3, and its volume is 1,617 cm³. Find the radius and height.
Answers
- 400 cm³.
- 3,080 cm³. 22/7 × 196 × 5.
- 216 cm³. 6a² = 216 gives a = 6.
- 125 cm³. The edge is 5 cm.
- 60π cm², about 188.6 cm². The slant is 10 cm, and π × 6 × 10.
- 36π cm². ⁴⁄₃πr³ = 36π gives r³ = 27, so r = 3, and 4π × 9.
- 6 cm. 27 + 64 + 125 = 216.
- 7 cm and 10.5 cm. With r = 2k and h = 3k, 22/7 × 12k³ = 1,617, so k³ = 42.875 and k = 3.5.
What to do next
- Write the formula table from memory, including the hemisphere's two surface areas
- For every recasting question, write "volume before = volume after" as your first line
- Revise area and perimeter, since every solid has circles and rectangles in it
- Use the maths revision sheet for a one-page summary before the exam
- Solve the volume and surface area questions from the last five CDS papers under time
A note on dates and numbers. Exam patterns, vacancies and schedules change from year to year. Always confirm the current details in the latest notification on the Union Public Service Commission website .
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