Volume of a Combination of Solids Class 10: Method & Examples

Quick answer: To find the volume of a combination of solids, simply add the volumes of the individual solids. Unlike surface area, no volume is lost when solids are joined together — the space filled by the combined object is exactly the space of one part plus the space of the other. If part of the object is hollow or scooped out, you subtract that empty volume instead.

A combination of solids is an object built from two or more basic solids (cone, cylinder, sphere, hemisphere, cuboid or cube). A solid toy, a gulab jamun, an industrial shed, a juice glass — each is just basic solids put together.

This guide gives you the method and worked examples for the exam. It assumes you know the basic-solid volume formulas; if you need them, see the surface area and volume formulas guide. For the tougher problems (floating solids that displace water, lead shots, iron poles and mass) and full practice, you can go further on examfront.

The key idea: volumes add, they don’t disappear

This is the one line that makes the whole section easy:

Volumes add. When two solids are joined, the volume of the new solid = volume of solid 1 + volume of solid 2. Nothing is lost.

Compare this with surface area, where the faces at the join do disappear (see the surface area of a combination guide). For volume, joining changes nothing about the space inside — so you never subtract at a simple join. You only subtract when material is genuinely removed (a cavity, depression, or hollow).

A step-by-step method you can trust

For any volume-of-combination problem:

1. Identify the basic solids in the object (cone, cylinder, hemisphere, cuboid …).
2. Find the radius/height of each part — watch for a shared radius, and split a “total height” into each solid’s own height.
3. Add the volumes of the parts that are solid.
4. Subtract the volume of any hollow or scooped-out part (a cavity, a hemispherical depression).
5. Keep units cubic (cm³, m³) and compute; if capacity in litres is asked, use 1000 cm³ = 1 litre.

Worked example 1: a solid toy (cone on a hemisphere)

Problem. A solid toy is a cone standing on a hemisphere, both of radius 7 cm. The height of the cone is 12 cm. Find the volume of the toy. (Use π = 22/7.)

Solution. The toy is solid, so add the two volumes.

• Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 7² × 12 = 616 cm³
• Volume of hemisphere = (2/3)πr³ = (2/3) × (22/7) × 7³ = 2156/3 ≈ 718.67 cm³
• Volume of toy = 616 + 718.67 = 1334.67 cm³ (approx.)

No subtraction is needed — the toy is completely solid, so the volumes just add.

Want the same shape with new numbers until it’s automatic? examfront’s Topic Practice gives you graded cone-and-hemisphere volume questions with instant feedback.

🎬 Quick challenge: Can you find the total volume when two solids are joined? 🧊

Worked example 2: a glass with a hemispherical depression (a cavity)

Problem. A cylindrical glass has inner radius 3.5 cm and height 10 cm, but its base has a hemispherical depression (radius 3.5 cm) that reduces how much it holds. Find its actual capacity. (Use π = 22/7.)

Solution. Start with the full cylinder, then subtract the scooped-out hemisphere.

• Apparent capacity (full cylinder) = πr²h = (22/7) × 3.5² × 10 = 385 cm³
• Volume of the hemispherical depression = (2/3)πr³ = (2/3) × (22/7) × 3.5³ = 539/6 ≈ 89.83 cm³
• Actual capacity = 385 − 89.83 = 295.17 cm³ (approx.)

Here apparent capacity is what the glass looks like it holds (a plain cylinder), and actual capacity is what it really holds after removing the bump. The difference is exactly the hemisphere’s volume.

Add or subtract? Ask one question: is material there or missing? Two solid pieces joined → add. A hollow, cavity or depression → subtract that empty volume.

What’s reserved for deeper practice

The exam pushes further with: a shed (cuboid + half-cylinder) and then removing machinery/people space; a gulab jamun (cylinder + two hemispheres) where you find the syrup as a percentage of volume; lead shots dropped into a cone of water (using displaced volume); and a solid iron pole where you convert volume to mass. All of them rest on the same rule — add solid volumes, subtract hollow ones.

Practise the full set of these mixed problems, step by step, inside examfront’s Chapter Quizzes, and use Mistake Identification to catch where you add when you should subtract.

🎬 Quick challenge: How much juice can the glass actually hold? 🥤

Key Takeaways

• Volume of a combined solid = sum of the volumes of its parts — nothing is lost at a join.
• Volumes add; surface areas don’t — that is the core difference from section 12.2.
• For a cavity, depression or hollow, subtract the empty volume.
• Actual capacity = apparent capacity − removed volume (e.g. πr²h − (2/3)πr³).
• Volume is always in cubic units (cm³, m³); 1000 cm³ = 1 litre.

Quick Facts

• Cone volume = (1/3)πr²h; hemisphere volume = (2/3)πr³; cylinder volume = πr²h.
• Toy (cone on hemisphere): volume = (1/3)πr²h + (2/3)πr³.
• Glass with hemispherical depression: capacity = πr²h − (2/3)πr³.
• Joining solids never changes total volume; only removing material does.
• Sphere volume = (4/3)πr³ — twice a hemisphere’s volume.

Common Mistakes

1. Subtracting at a simple join — for a solid combination you add; you only subtract for a cavity.
2. Forgetting to split the total height — a cone’s own height is not always the object’s full height.
3. Using slant height in a volume — cone volume uses vertical height h, not slant height l.
4. Confusing apparent and actual capacity — actual = apparent − removed volume.
5. Writing volume in cm² instead of cm³ — volume is a cubic unit.

FAQ

How do you find the volume of a combination of solids in Class 10? Add the volumes of the individual solids. No volume is lost at a join, so the combined volume is exactly the sum. For a cone on a hemisphere, volume = (1/3)πr²h + (2/3)πr³.

Why do volumes add up but surface areas do not, when solids are joined? Surface area is hidden at the join, so surface areas don’t simply add. But no space disappears, so the total volume is just one solid’s space plus the other’s — volumes always add exactly.

How do you find the volume when a solid has a cavity or depression? Find the volume of the full outer solid and subtract the empty part. A glass with a hemispherical depression holds πr²h − (2/3)πr³.

What is the difference between apparent capacity and actual capacity of a glass? Apparent capacity treats the glass as a plain cylinder (πr²h). Actual capacity subtracts any raised or hollowed part, so actual = πr²h − (2/3)πr³ for a hemispherical bump at the base.

What is the volume of a toy shaped like a cone on a hemisphere? For a cone of height h on a hemisphere of radius r, volume = (1/3)πr²h + (2/3)πr³. The volumes simply add because joining removes no space.

Related Concepts

• Surface Area and Volume Formulas (12.1) — the six basic-solid volume formulas this method combines; see the formula sheet.
• Surface Area of a Combination of Solids (12.2) — the contrast where faces disappear at the join; see the surface area of a combination guide.
• Area of Sector and Segment of a Circle (Chapter 11) — the circle-area basis for circular cross-sections; see the area of sector and segment guide.
• Class 10 Maths All Formulas — the complete cross-chapter formula reference.

Ready to test yourself? Start with examfront’s Surface Areas and Volumes quiz, track which combined shapes trip you up with Progress Tracking, and get personalised help exactly where you slip.