Exercise 10.3 — Volume of Combined Solids
Volumes of combination of solids.
Exercise 10.3 — Volume of Combination of Solids
Exercise 10.3 from Chapter 10, Mensuration, of Class 10 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) is about finding the volume of solids that are built by joining two or more basic 3D shapes together, or by removing one shape from inside another. The seven problems in this exercise cover an iron pillar, a wooden toy, a cone carved from a cube, a solid lowered into a tub of water, a cylinder with conical holes drilled into it, marbles dropped into a beaker, and a pen stand with conical depressions.
Every one of these situations looks different on the surface, but they are all solved the same way: break the combined solid into the simple shapes you already know, find the volume of each piece using the standard formula, and then add or subtract those volumes depending on whether the pieces are joined together or cut away.
What Is a "Combination of Solids"?
A combination solid is simply an object made up of two or more basic solids — usually a cylinder, cone, hemisphere, sphere, or cuboid — joined where their flat circular or rectangular faces match up exactly. An ice-cream cone with a scoop on top, a test tube, a pen stand with pen-holes drilled into it, and a medicine capsule are all everyday combination solids.
Two Patterns You'll See Again and Again
Once you look past the wording, almost every question in this exercise follows one of three patterns:
| Pattern | What You Do | Used In |
|---|---|---|
| Joined solids | Add the volumes of the pieces that are stuck together. | Q1, Q4 (first step) |
| Carved-out solids | Subtract the volume that has been removed from the bigger solid. | Q4 (second step), Q5, Q7 |
| Equal volumes | Set two volumes equal to each other and solve for an unknown length or count. | Q2, Q6 |
Volume & Surface Area Formulas You Need for This Exercise
Keep this table close by while solving the exercise — almost every question is just these formulas applied one or two at a time.
| Solid | Volume | Curved / Lateral Surface Area | Total Surface Area |
|---|---|---|---|
| Cuboid (l, b, h) | l × b × h |
— | 2(lb + bh + hl) |
| Cube (edge a) | a³ |
— | 6a² |
| Cylinder (r, h) | πr²h |
2πrh |
2πr(h + r) |
| Cone (r, h, slant l) | ⅓πr²h |
πrl |
πr(l + r) |
| Sphere (r) | 4/3 πr³ |
— | 4πr² |
| Hemisphere (r) | 2/3 πr³ |
2πr² |
3πr² |
l = √(r² + h²) — this comes from the right triangle formed by the cone's radius, height, and slanted edge. You'll use this in Question 2.
Question 1 — Weight of an Iron Pillar (Cylinder + Cone)
An iron pillar is made of a cylindrical shaft 2.8 m tall and 20 cm in diameter, topped with a cone-shaped spike 42 cm tall. Since 1 cm³ of iron weighs 7.5 g, we need the pillar's total volume before we can find its weight.
Question 2 — Height and Surface Area of a Toy (Hemisphere + Cone)
A toy is shaped like a hemisphere with a cone of the same base radius mounted on top — picture an ice-cream cone. The radius of both the cone and the hemisphere is 7 cm, and the cone's volume is given as 3/2 times the hemisphere's volume. We need the cone's height, and then the toy's total surface area correct to two decimal places.
Question 3 — Largest Cone That Can Be Cut From a Cube
A cube has an edge of 7 cm. We need the volume of the largest right circular cone that can be carved out of it. The key insight: the biggest cone that fits inside a cube has its circular base inscribed exactly in one face of the cube, and its height equal to the cube's edge length — its apex just touches the centre of the opposite face.
Question 4 — Water Left in a Tub After a Solid Is Immersed
A cylindrical tub of radius 5 cm and height 9.8 cm is completely full of water. A solid made of a cone mounted on a hemisphere (hemisphere radius 3.5 cm, cone height 5 cm) is lowered into the tub. Since the solid is fully submerged, it pushes out a volume of water exactly equal to its own volume — so we need the volume of the tub, the volume of the solid, and then the difference.
Question 5 — Remaining Volume After Two Conical Holes Are Cut
A solid cylinder is 10 cm tall with a diameter of 7 cm. Two identical cone-shaped holes, each of radius 3 cm and height 4 cm, are drilled into the cylinder from its two flat ends, pointing inward. We need the volume of the solid that's left over.
Question 6 — How Many Marbles Raise the Water Level?
Spherical marbles of diameter 1.4 cm are dropped, one by one, into a cylindrical beaker of diameter 7 cm that already contains some water. We need to find how many marbles must be dropped in for the water level to rise by 5.6 cm.
Question 7 — Volume of Wood in a Pen Stand
A wooden pen stand is shaped like a cuboid measuring 15 cm by 10 cm by 3.5 cm, with three identical cone-shaped depressions cut into its top to hold pens. Each depression has a radius of 0.5 cm and a depth of 1.4 cm. We need the volume of wood that remains after the three depressions are removed.
Common Mistakes to Avoid
- Mixing units: Always convert every length in a problem to the same unit (usually cm) before adding or subtracting volumes — Question 1 is a classic trap because the height is in metres while the diameter is in centimetres.
- Confusing "added" with "removed": Read carefully whether a shape is being joined on (Q1, Q2, Q4's first step — add the volumes) or carved/drilled out (Q4's second step, Q5, Q7 — subtract the volumes).
- Writing the wrong type of unit: A volume answer must always be in cubic units (cm³, m³), never square units (cm², m²) — it's an easy slip to make when you're focused on the numbers, as in Questions 3 and 5.
- Adding flat faces that are actually hidden: When two solids are joined (like the cone and hemisphere in Question 2), the flat circular face where they meet is sealed inside the object — only the curved surfaces are part of the toy's visible surface area.
- Misreading "largest solid that fits inside": In Question 3, the cone's height equals one edge of the cube — not the cube's diagonal — and its base circle is inscribed in a square face, so the diameter equals the side length.
- Rounding too early: Keep a couple of extra decimal places in intermediate steps (like the slant height in Question 2) and round only the final answer, so small rounding errors don't add up in a "correct to two decimal places" question.
Quick Reference — All Answers at a Glance
| Question | What's Combined | Answer |
|---|---|---|
| Q1 | Cylinder + Cone (iron pillar) | 693 kg |
| Q2 | Hemisphere + Cone (toy) | h = 21 cm, S.A. ≈ 794.99 cm² |
| Q3 | Cone cut from a cube | ≈ 89.83 cm³ |
| Q4 | Hemisphere + Cone immersed in a tub | 616 cm³ water left |
| Q5 | Cylinder − 2 conical holes | ≈ 309.57 cm³ |
| Q6 | Spherical marbles raising water level | 150 marbles |
| Q7 | Cuboid − 3 conical depressions | 523.9 cm³ wood |
What This Exercise Prepares You For
Exercise 10.3 builds directly on the surface-area ideas from Exercise 10.2 — Surface Area of Combination of Solids, where the same cylinder–cone–hemisphere combinations appear but the question asks for outer surface area instead of volume. Once volume of combinations feels comfortable, the next step is Exercise 10.4 — Conversion of Solids, where a solid is melted down and recast into a completely different shape, using the same volume formulas but setting two volumes equal instead of adding or subtracting them.
A solid grip on the basic volume formulas from this exercise also pays off well beyond Chapter 10 — it's the same toolkit used in Introduction to Mensuration and in real-world estimation problems that show up across CBSE, Telangana, and Andhra Pradesh board exams.