MathsCBSE Class 10 · Surface Areas and Volumes

Volume of a Combination of Solids for Class 10

Updated 14 October 2026 · Fact-checked

The volume of a solid made by joining two solids is the sum of the volumes of both parts, because joining them creates or destroys no space. If one part is cut out, you subtract. Board questions ask for the volume or capacity of toys, containers and tanks, often hiding a height, slant height or radius to find first.

Add the hemisphere and cylinder volumes; the total height is H + r, so find each part's own height first.

Illustration drawn for QuestionBuddy.

Understand Volume of a Combination of Solids

A combination of solids is one object built from two simple solids, such as a cylinder with a cone on top or a cuboid with a hemisphere on it. Its volume is the space it takes up, measured in cubic units.

Why do you simply add? Volume measures space. When you glue a hemisphere onto a cylinder, the hemisphere's space is not squeezed and the cylinder's space is not lost. They only touch at a flat circle, and a flat circle has no thickness, so it has no volume. So $V = V_{1} + V_{2}$ . Compare this with surface area, where the touching faces disappear and must be left out. Volume has no hidden face to worry about.

The reverse also holds. If one solid is cut out of another, the space left is $V = V_{1} - V_{2}$ . For example, a cone drilled out of a cylinder of the same radius and height leaves $\frac{2}{3}$ of the cylinder.

The real work is usually finding the dimensions. Each part needs its own height. In a cylinder with a hemisphere on top, the total height is the cylinder height plus one radius. In a capsule, the total length is the cylinder length plus two radii. A cone's volume uses its vertical height $h$ , never the slant height $l$ . If only $l$ is given, find $h = l^{2} - r^{2}$ first. Find every height first and write it on its own line.

When the parts share a radius, take $π r^{2}$ out as a common factor, for example $V = π r^{2} (H + \frac{2 r}{3})$ . This shortens the arithmetic with $π = \frac{22}{7}$ . Radii like 3.5, 7, 10.5 and 21 cm cancel neatly.

Finally, convert units if asked: $1000 cm^{3} = 1$ litre and $1 m^{3} = 1000$ litres. The syllabus limits you to combinations of two kinds of solid from cube, cuboid, sphere, hemisphere, cylinder and cone. A cylinder with a hemisphere at each end still counts as two kinds.

Volume of a Combination of Solids: formulas to remember

Volume of a cube: $V = a^{3}$; $a$ is the edge.
Volume of a cuboid: $V = l bh$; Length × breadth × height.
Volume of a cylinder: $V = π r^{2} h$; In a combination, $h$ is the height of the cylindrical part only.
Volume of a cone: $V = \frac{1}{3} π r^{2} h$; $h$ is the vertical height.
Volume of a sphere: $V = \frac{4}{3} π r^{3}$; Two hemispheres of the same radius make one sphere.
Volume of a hemisphere: $V = \frac{2}{3} π r^{3}$; Half the volume of a sphere.
Slant height of a cone: $l = r^{2} + h^{2}$; Volume uses $h$ , not $l$ . If $l$ is given, find $h = l^{2} - r^{2}$ first.
Volume of a combination (joined): $V = V_{1} + V_{2}$; No space is lost where solids touch.
Volume when a part is removed: $V = V_{1} - V_{2}$; Scooped, drilled or carved cavities.
Cylinder with cone on top (same r): $V = π r^{2} (H + \frac{h}{3})$; $H$ is the cylinder height, $h$ is the cone's vertical height.
Cylinder with hemisphere on one end (same r): $V = π r^{2} (H + \frac{2 r}{3})$; $H$ = total height $- r$ .
Cone on hemisphere (same r): $V = \frac{1}{3} π r^{2} (h + 2 r)$; Cone height $h$ = total height $- r$ .
Capsule: hemisphere at each end: $V = π r^{2} h + \frac{4}{3} π r^{3}$; $h$ = total length $- 2 r$ ; the two ends make one full sphere.
Cube or cuboid with a hemisphere on a face: $V = a^{3} + \frac{2}{3} π r^{3}$; Needs $2 r \leq a$ . Use $l bh$ for a cuboid.
Unit conversions: $1000 cm^{3} = 1 L, 1 m^{3} = 1000 L$; Capacity questions usually want litres.

How to solve Volume of a Combination of Solids questions

Use this method on any volume-of-combination question, from a 2-mark toy to a 5-mark case.

1Name the two solids and draw a rough sketch marking every radius and height.
2Find any hidden dimension first: cylinder height = total height minus r (one hemisphere) or minus 2r (two hemispheres); cone height = total height minus r on a hemisphere; cone height $h = l^{2} - r^{2}$ if only the slant height is given.
3Decide whether the parts are joined (add) or removed (subtract).
4Write the volume as a sum of named parts, for example V = volume of cylinder + volume of hemisphere.
5Take out a common factor such as $π r^{2}$ and cancel with $π = \frac{22}{7}$ before multiplying.
6Convert to the unit asked (litres, m³) and answer any follow-up such as the number of cups or bags.
7Write the final answer with units on its own line.

Volume of a Combination of Solids in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Volume of a combination: add the parts

Imagine your school installs a water cooler tank shaped like a cylinder with a hemispherical bottom. Its inner radius is 21 cm and the total inner depth, from the lowest point to the top rim, is 81 cm. Students fill 500 mL bottles from it.

Find the capacity of the tank in litres and the number of 500 mL bottles that can be filled completely from one full tank.

Try it first, then show the solutionSolution

Height of cylindrical part $H = 81 - 21 = 60$ cm.
$V = π r^{2} (H + \frac{2 r}{3})$ .
$π r^{2} = \frac{22}{7} \times 441 = 1386$ cm².
$H + \frac{2 r}{3} = 60 + 14 = 74$ cm.
$V = 1386 \times 74 = 102564$ cm³ $= 102.564$ L.
Bottles $= 102564 \div 500 = 205.128$ , so 205 full bottles.

Answer: Capacity 102.564 litres; 205 full 500 mL bottles.

Memory hook: Cooler tank: drum plus bowl, so add the cylinder and the hemisphere.

Tricks to remember Volume of a Combination of Solids

Short memory aids. Say each one aloud once, then use it on a question the same day.

Glue adds no space
Glue changes the paint, not the fill. Joining hides surface but never hides volume.
Use it when: Stops you subtracting a joined face from a volume.
One end, one r; two ends, two r
Cylinder height = total minus r for one hemisphere, minus 2r for a capsule.
Use it when: Finding the hidden cylinder height.
Cone, hemisphere, cylinder: 1 : 2 : 3
With equal radius and height equal to r, volumes are $\frac{1}{3} π r^{3}$ , $\frac{2}{3} π r^{3}$ , $π r^{3}$ .
Use it when: Quick MCQ and assertion-reason checks.
Slant is for skin
Slant height $l$ goes in surface area; vertical height $h$ goes in volume.
Use it when: Avoids using $l$ in a cone's volume.

Volume of a Combination of Solids: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Cylinder with a hemispherical top · 3 marks · Short answerBoard level

A solid is a right circular cylinder of diameter 21 cm with a hemisphere of the same diameter fixed on top. The total height of the solid is 23.5 cm. Find its volume in cm³ and in litres. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 10.5$ cm; cylinder height $H = 23.5 - 10.5 = 13$ cm. (1)
$V$ = volume of cylinder + volume of hemisphere $= π r^{2} (H + \frac{2 r}{3})$ . (½)
$π r^{2} = \frac{22}{7} \times 110.25 = 346.5$ cm²; $H + \frac{2 r}{3} = 13 + 7 = 20$ cm. (½)
$V = 346.5 \times 20 = 6930$ cm³ $= 6.93$ L. (1)

Answer: 6930 cm³, which is 6.93 litres.

Example 2: A 2-mark volume · 2 marks · Very short answerBoard level

A solid is a cylinder of radius 3.5 cm and height 10 cm with a hemisphere of the same radius on one end. Find its volume. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$V = π r^{2} H + \frac{2}{3} π r^{3} = π r^{2} (H + \frac{2 r}{3})$ , with $π r^{2} = \frac{22}{7} \times 12.25 = 38.5$ cm². (1)
$V = 38.5 \times (10 + \frac{7}{3}) = 38.5 \times \frac{37}{3} \approx 474.83$ cm³. (1)

Answer: About 474.83 cm³.

Example 3: Find the total height from the volume · 3 marks · Short answerBoard level

A toy is a right circular cone mounted on a hemisphere of the same radius 3.5 cm. The volume of the toy is 231 cm³. Find the total height of the toy. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$V = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r)$ . (1)
$π r^{2} = \frac{22}{7} \times 12.25 = 38.5$ , so $\frac{1}{3} \times 38.5 \times (h + 7) = 231$ . (½)
$h + 7 = \frac{693}{38.5} = 18$ , so $h = 11$ cm. (½)
Total height $= h + r = 11 + 3.5 = 14.5$ cm. (½)
Check: cone $= \frac{423.5}{3}$ , hemisphere $= \frac{269.5}{3}$ , sum $= \frac{693}{3} = 231$ cm³. (½)

Answer: Total height of the toy is 14.5 cm.

Example 4: Capsule with equal parts · 5 marks · Long answerHarder board level

A solid is a cylinder with a hemisphere of the same radius at each end. The volume of the cylindrical part equals the combined volume of the two hemispheres. The total length of the solid is 35 cm. Find (i) the radius, (ii) the length of the cylindrical part and (iii) the volume of the solid in litres. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Two hemispheres make a sphere, so $π r^{2} h = \frac{4}{3} π r^{3}$ . (1)
Dividing by $π r^{2}$ : $h = \frac{4 r}{3}$ . (½)
Total length $h + 2 r = \frac{4 r}{3} + 2 r = \frac{10 r}{3} = 35$ , so $r = 10.5$ cm. (1)
$h = \frac{4}{3} \times 10.5 = 14$ cm. (½)
Cylinder volume $= 346.5 \times 14 = 4851$ cm³. (1)
Total $V = 2 \times 4851 = 9702$ cm³ $= 9.702$ L. (1)

Answer: (i) 10.5 cm (ii) 14 cm (iii) 9702 cm³ = 9.702 litres.

Example 5: Conical cavity drilled from a cylinder · 5 marks · Long answerHarder board level

A solid cylinder has radius 7 cm and height 24 cm. Take $π = \frac{22}{7}$ .
(i) A conical cavity of the same radius and height 24 cm is drilled out of it. Find the volume left.
(ii) An identical cylinder instead has a conical cavity of radius 7 cm drilled into one end, and the volume left is 3080 cm³. Find the depth of this cavity.
(iii) Find the ratio of the volume left in (i) to the volume of the whole cylinder.

Try it first, then show the solutionSolution

(i) Cylinder $= \frac{22}{7} \times 49 \times 24 = 154 \times 24 = 3696$ cm³. (1)
Cone $= \frac{1}{3} \times 3696 = 1232$ cm³, since it has the same base and height. (½)
Volume left $= 3696 - 1232 = 2464$ cm³. (½)
(ii) Cavity volume $= 3696 - 3080 = 616$ cm³. (½)
Let the depth be $d$ : $\frac{1}{3} \times 154 \times d = 616$ . (½)
$d = \frac{616 \times 3}{154} = 12$ cm. (1)
(iii) $2464 : 3696 = 2 : 3$ , dividing both by 1232. (1)

Answer: (i) 2464 cm³ (ii) 12 cm (iii) 2 : 3

Example 6: Case study: a grain silo with a conical roof · 4 marks · Case-basedHOTS / competency

Suppose a village cooperative builds a grain silo shaped like a cylinder with a conical roof of the same radius. The radius is 2.1 m, the height of the cylindrical part is 8.4 m, and the slant height of the roof is 3.5 m. Grain fills the whole silo, roof included. Take $π = \frac{22}{7}$ .
(i) Find the vertical height of the roof. (1 mark)
(ii) Find the volume of the roof. (1 mark)
(iii) Find the capacity of the silo. Would a plain cylindrical silo of the same radius and the same total height hold more or less, and by how much? (2 marks)

Try it first, then show the solutionSolution

(i) $h = 3. 5^{2} - 2. 1^{2} = 12.25 - 4.41 = 7.84 = 2.8$ m. (1)
(ii) $π r^{2} = \frac{22}{7} \times 4.41 = 13.86$ m²; roof $= \frac{1}{3} \times 13.86 \times 2.8 = 12.936$ m³. (1)
(iii) Cylinder $= 13.86 \times 8.4 = 116.424$ m³, so capacity $= 116.424 + 12.936 = 129.36$ m³. (1)
Total height $= 8.4 + 2.8 = 11.2$ m. A plain cylinder holds $13.86 \times 11.2 = 155.232$ m³, which is $155.232 - 129.36 = 25.872$ m³ more: the roof holds only one-third of a cylinder of the same height. (1)

Answer: (i) 2.8 m (ii) 12.936 m³ (iii) 129.36 m³; a plain cylinder of the same height would hold 25.872 m³ more.

Example 7: Cone height hidden behind the slant height · 3 marks · Short answerBoard level

A solid is a cylinder of radius 7 cm and height 10 cm with a cone of the same radius on top. The slant height of the cone is 25 cm. Find the volume of the solid. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Cone height $h = 2 5^{2} - 7^{2} = 625 - 49 = 576 = 24$ cm. (1)
$V$ = volume of cylinder + volume of cone $= π r^{2} (H + \frac{h}{3})$ . (½)
$π r^{2} = \frac{22}{7} \times 49 = 154$ cm²; $H + \frac{h}{3} = 10 + 8 = 18$ cm. (½)
$V = 154 \times 18 = 2772$ cm³. (1)

Answer: 2772 cm³.

Example 8: Cuboid with a hemisphere on top · 3 marks · Short answerHarder board level

A decorative block is a cuboid with a square base of side 21 cm, with a hemisphere of diameter 21 cm fixed flat-side down on its top face. The total height of the block is 18.5 cm. Find its volume. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 10.5$ cm. (½)
Height of cuboid $= 18.5 - 10.5 = 8$ cm. (½)
Cuboid $= 21 \times 21 \times 8 = 3528$ cm³. (½)
Hemisphere $= \frac{2}{3} \times \frac{22}{7} \times 1157.625 = 2425.5$ cm³. (1)
Total $V = 3528 + 2425.5 = 5953.5$ cm³. (½)

Answer: 5953.5 cm³.

Where students lose marks in Volume of a Combination of Solids

The slips examiners see most often, and what to write instead.

Mistake: Using total height as the cylinder height
The hemisphere adds one radius to the height.
Wrong working
Total 23.5 cm, r = 10.5: $V = 346.5 \times (23.5 + 7) = 10568.25$ cm³
Right working
$H = 23.5 - 10.5 = 13$ ; $V = 346.5 \times 20 = 6930$ cm³
Do this: Write H = total minus r as its own line before any volume.
Usually costs: Can cost the method mark and the final answer mark.
Mistake: Using slant height in a cone's volume
Students mix up the formulas $π r l$ and $\frac{1}{3} π r^{2} h$ .
Wrong working
r = 7, l = 25: $V = \frac{1}{3} \times 154 \times 25$
Right working
$h = 625 - 49 = 24$ ; $V = \frac{1}{3} \times 154 \times 24 = 1232$ cm³
Do this: Volume always uses vertical height; find h by Pythagoras if only l is given.
Usually costs: Can cost the method mark and the final answer mark.
Mistake: Subtracting a joined face from the volume
Carrying over the surface area rule.
Wrong working
$V = V_{cyl} + V_{hemi} - π r^{2}$
Right working
$V = V_{cyl} + V_{hemi}$ ; an area cannot be subtracted from a volume.
Do this: Check units: every term in a volume must be in cm³.
Usually costs: Method marks for the step.
Mistake: Wrong litre conversion
Dividing by 100 instead of 1000.
Wrong working
$6930$ cm³ $= 69.3$ L
Right working
$6930 \div 1000 = 6.93$ L
Do this: Remember 1 L = 1000 cm³ and 1 m³ = 1000 L.
Usually costs: The final ½ to 1 mark.

Exam tips

Write the volume as a sum of named parts before substituting; it earns the method mark.
State every found dimension (cylinder height, cone height, radius) on its own line with units.
Use $π = \frac{22}{7}$ and cancel early; numbers like 3.5, 7, 10.5 and 21 are chosen to cancel.
In case studies, reuse earlier sub-part answers, but give each part its own final line with units.
If a question asks for both surface area and volume, remember that only the surface area loses hidden faces.

Practice questions on Volume of a Combination of Solids

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerBoard level
An open vessel is a cylinder with a hemispherical bottom. Its inner radius is 10.5 cm and its total inner depth is 30.5 cm. Find its capacity in litres and how many full 200 mL glasses it can fill. Take $π = \frac{22}{7}$ .
Hint
How much of the 30.5 cm depth belongs to the hemispherical bottom?
Answer
$H = 20$ cm; $V = 346.5 \times (20 + 7) = 9355.5$ cm³ $= 9.3555$ L; $9355.5 \div 200 = 46.78$ , so 46 full glasses.
Q2 · 3 marks · Short answerHOTS / competency
A toy is a cone on a hemisphere, both of radius 3.5 cm, with total height 11.5 cm. A student finds the volume as $\frac{1}{3} π r^{2} (11.5) + \frac{2}{3} π r^{3}$ . Find the error and the correct volume. By what percentage did the student overestimate the volume? Take $π = \frac{22}{7}$ .
Hint
Is 11.5 cm really the cone's height?
Answer
The student used the total height as the cone height. Cone height $= 11.5 - 3.5 = 8$ cm. $V = \frac{1}{3} \times 38.5 \times (8 + 7) = 192.5$ cm³. The student's value is $\frac{1}{3} \times 38.5 \times 11.5 + \frac{2}{3} \times 38.5 \times 3.5 \approx 147.58 + 89.83 = 237.42$ cm³, too much by $44.92$ cm³, which is $\frac{44.92}{192.5} \times 100 \approx 23.3%$ .
Q3 · 1 mark · MCQBoard level
A cone with the same radius and height as a solid cylinder is drilled out of the cylinder. The volume left is what fraction of the volume of the cylinder?
(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{3}{4}$
Hint
Compare a cone with a cylinder on the same base and height.
Answer
(c) $\frac{2}{3}$ , since $π r^{2} h - \frac{1}{3} π r^{2} h = \frac{2}{3} π r^{2} h$ .
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): The volume of a solid formed by joining a cylinder and a hemisphere of the same radius is the sum of their volumes.
Reason (R): When two solids are joined, the total surface area of the new solid is the sum of their total surface areas.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What happens to the faces that touch when the solids are joined?
Answer
(c) A is true but R is false. No space is lost on joining, so volumes add, but the joined faces are hidden, so they are not part of the new surface area.
Q5 · 5 marks · Long answerHarder board level
A solid wooden piece is a cylinder with a hemisphere of the same radius at each end. Its total length is 42 cm, and the volume of the cylindrical part is to the combined volume of the two hemispheres as 3 : 2. Find (i) the radius, (ii) the length of the cylindrical part and (iii) the volume of the piece. Take $π = \frac{22}{7}$ .
Hint
Turn the ratio into an equation linking the cylinder length and the radius, then use the total length.
Answer
$\frac{π r ^{2} h}{\frac{4}{3} π r ^{3}} = \frac{3}{2}$ gives $h = 2 r$ ; $h + 2 r = 4 r = 42$ , so (i) $r = 10.5$ cm and (ii) $h = 21$ cm. (iii) Cylinder $= 346.5 \times 21 = 7276.5$ cm³; sphere $= \frac{4}{3} \times \frac{22}{7} \times 1157.625 = 4851$ cm³; total $= 12127.5$ cm³.
Q6 · 3 marks · Short answerBoard level
A solid cube of edge 12 cm has a hemispherical hollow scooped out of its top face. The volume left is $(1728 - 144 π)$ cm³. Find the radius of the hollow and say whether it is the largest hollow possible on that face.
Hint
The removed part is the difference between the cube and what is left.
Answer
Cube $= 1728$ cm³, so the hollow is $144 π$ cm³. $\frac{2}{3} π r^{3} = 144 π$ gives $r^{3} = 216$ , $r = 6$ cm. The diameter 12 cm equals the edge, so it is the largest possible hollow.
Q7 · 3 marks · Short answerHarder board level
A cylinder and a cone of the same radius 7 cm are joined to form a solid with the cone on top. The volume of the cylindrical part is 4 times the volume of the conical part, and the total height is 21 cm. Find the height of each part and the volume of the solid. Take $π = \frac{22}{7}$ .
Hint
Use the volume condition to link the two heights first.
Answer
$π r^{2} H = 4 \times \frac{1}{3} π r^{2} h$ gives $H = \frac{4 h}{3}$ ; $\frac{4 h}{3} + h = 21$ , so $h = 9$ cm and $H = 12$ cm. $V = 154 \times (12 + 3) = 2310$ cm³.
Q8 · 3 marks · Short answerHOTS / competency
Suppose a juice stall serves juice in glasses that look like cylinders of inner radius 3 cm and inner depth 12 cm (rim to lowest inside point). Inside, the bottom is actually shaped as a hemisphere of radius 3 cm, with thick glass below it. Leave answers in terms of $π$ .
(i) Find how much juice a full glass really holds.
(ii) By what fraction is this less than a flat-bottomed glass of the same radius and depth? Explain why the difference arises.
Hint
Split the inside into the part above the curved bottom and the curved bottom itself.
Answer
(i) Cylindrical part height $= 12 - 3 = 9$ cm; juice $= π \times 9 \times 9 + \frac{2}{3} π \times 27 = 81 π + 18 π = 99 π$ cm³. (ii) Flat glass $= π \times 9 \times 12 = 108 π$ cm³; shortfall $= 9 π$ , which is $\frac{9}{108} = \frac{1}{12}$ less. The bottom 3 cm is a hemisphere ( $18 π$ ) instead of a cylinder of height 3 cm ( $27 π$ ).
Q9 · 4 marks · Case-basedHOTS / competency
Imagine your school's art club makes a large wooden top for a display: a cone mounted on a hemisphere of the same radius 10.5 cm. The slant height of the cone is 17.5 cm. The wood has a mass of 0.8 g per cm³. Take $π = \frac{22}{7}$ .
(i) Find the vertical height of the cone. (1 mark)
(ii) Find the total height of the top. (1 mark)
(iii) Find the volume of the top and its mass in kg. How much heavier would a solid cylinder of the same radius and the same total height be? (2 marks)
Hint
The slant height and radius form a right triangle with the cone's height.
Answer
(i) $h = 306.25 - 110.25 = 196 = 14$ cm. (ii) $14 + 10.5 = 24.5$ cm. (iii) $V = \frac{1}{3} \times 346.5 \times (14 + 21) = 4042.5$ cm³; mass $= 4042.5 \times 0.8 = 3234$ g $= 3.234$ kg. A cylinder of height 24.5 cm has volume $346.5 \times 24.5 = 8489.25$ cm³ and mass $6791.4$ g, so it would be about $3.557$ kg heavier.
Q10 · 3 marks · Short answerHarder board level
A solid toy is a cylinder of radius 7 cm with a cone of the same radius on top. The slant height of the cone is 25 cm, and the volume of the cylindrical part is twice the volume of the cone. Find the total height and the volume of the toy. Take $π = \frac{22}{7}$ .
Hint
Find the cone height first, then use the volume ratio to get the cylinder height.
Answer
Cone height $= 625 - 49 = 24$ cm. $π r^{2} H = 2 \times \frac{1}{3} π r^{2} h$ gives $H = \frac{2 h}{3} = 16$ cm. Total height $= 16 + 24 = 40$ cm. Volume $= 154 \times (16 + 8) = 3696$ cm³.
Q11 · 3 marks · Short answerHarder board level
A cylindrical hole is drilled straight through a solid cuboid 20 cm by 15 cm by 14 cm, parallel to the 14 cm edges. The volume left is 3661 cm³. Find the radius of the hole and the percentage of wood removed. Take $π = \frac{22}{7}$ .
Hint
The hole is a cylinder of height 14 cm.
Answer
Cuboid $= 20 \times 15 \times 14 = 4200$ cm³, so the hole $= 4200 - 3661 = 539$ cm³. $\frac{22}{7} r^{2} \times 14 = 539$ gives $44 r^{2} = 539$ , $r^{2} = 12.25$ and $r = 3.5$ cm (the 7 cm diameter fits in the 20 cm by 15 cm face). Wood removed $= \frac{539}{4200} \times 100 \approx 12.83%$ .
Q12 · 2 marks · Very short answerBoard level
A solid toy is a cone of radius 3.5 cm and height 12 cm mounted on a hemisphere of the same radius. Find its volume. Take $π = \frac{22}{7}$ .
Hint
Use $\frac{1}{3} π r^{2} (h + 2 r)$ .
Answer
$V = \frac{1}{3} \times 38.5 \times (12 + 7) = \frac{731.5}{3} \approx 243.83$ cm³.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Volume of two joined solids?Show answer
$V_{1} + V_{2}$ .
2.Cylinder height in a capsule of total length L?Show answer
$L - 2 r$ .
3.Volume of cone on hemisphere, same r?Show answer
$\frac{1}{3} π r^{2} (h + 2 r)$ .
4.1 m³ in litres?Show answer
1000 litres.
5.Ratio of volumes of cone, hemisphere, cylinder with h = r?Show answer
1 : 2 : 3.

Volume of a Combination of Solids: frequently asked questions

Why do volumes add but surface areas do not?

When two solids are joined, the touching faces are hidden, so surface area loses them. But no space is lost or created, so the volumes simply add.

Which NCERT exercise has the volume questions?

Volume of combinations is in Exercise 12.2 of the NCERT book, after section 12.3. Exercise 12.1 covers surface area of combinations.

How do I find the cylinder height in a capsule?

Subtract the two hemisphere radii from the total length. So the cylinder length is the total length minus twice the radius.

Are frustum and melting-recasting questions in the 2026-27 syllabus?

No. The frustum of a cone, conversion of solids by melting and recasting as a separate section, and combinations of more than two kinds of solid are not in the 2026-27 syllabus.