Spheres, cones and cylinders Quiz (Grade 8+)

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Topic:

Score: 0 / 10

Spheres, cones and cylinders — Notes & Worked Examples Show / Hide

Explanation:

Key 3D curved solid formulas: Cylinder volume = πr²h, curved surface area = 2πrh; Sphere volume = 43πr³, surface area = 4πr²; Cone volume = 13πr²h, curved surface area = πrl (where slant height l = √(r² + h²)).

For a solid hemisphere or solid cone, remember to include the flat circular base (+πr²), so the total surface area of a solid hemisphere is 2πr² + πr² = 3πr². When melting and recasting shapes, equate their exact volumes in terms of π so π cancels cleanly.

Example 1: Find the exact total surface area of a solid cone with base radius 5 cm and vertical height 12 cm.

Solution:

l = √(5² + 12²) = 13 cm

CSA + Base = π(5)(13) + π(5)² = 65π + 25π = 90π cm²

Example 2: A metal sphere of radius 6 cm is melted down to form a cylinder of radius 4 cm. Find the height h of the cylinder.

Solution:

V_sphere = 43π(6)³ = 288π; V_cyl = π(4)²h = 16πh

16πh = 288π ⇒ h = 288 ÷ 16 = 18 cm

One Page Guide (PDF) Spheres, cones and cylinders Worksheets

Question 1

[1 mark]

Work out
A solid toy consists of a hemisphere of radius 3 cm joined to the base of a cone of radius 3 cm and vertical height 6 cm.
Calculate the exact total volume of the toy in terms of π.

Worked Solution & Answer

V = 23π(3)³ + 13π(3)²(6) = 18π + 18π = 36π cm³ B1

Answer: 36π cm³

Question 2

[1 mark]

Work out
A solid toy consists of a hemisphere of radius 6 cm joined to the base of a cone of radius 6 cm and vertical height 12 cm.
Calculate the exact total volume of the toy in terms of π.

Worked Solution & Answer

V = 23π(6)³ + 13π(6)²(12) = 144π + 144π = 288π cm³ B1

Answer: 288π cm³

Question 3

[4 marks]

The diagram shows a solid shape.
The shape is a cone on top of a hemisphere.

The height of the cone is 10 cm.
The base of the cone has a diameter of 6 cm.
The hemisphere has a diameter of 6 cm.
The total volume of the shape is kπ cm³, where k is an integer.
Work out the value of k.

k =

Worked Solution & Answer

r = 6 ÷ 2 = 3 cm M1
Volume of cone = 1/3 × π × 3² × 10 = 30π M1
Volume of hemisphere = 2/3 × π × 3³ = 18π M1
Total volume = 30π + 18π = 48π ⇒ k = 48 A1

Answer: k = 48

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Question 4

[4 marks]

The diagram shows a solid shape made from a cylinder and a cone.

Work out the vertical height of the cone.
You must show all your working.

The cone has a base radius of 8 cm.
The cylinder has a radius of 8 cm.
The vertical height of the cone is three times the height of the cylinder.
The volume of the solid shape is 640π cm³

Worked Solution & Answer

(The volume of a cylinder is πr²h)
Let cylinder height = H M1
Cone height h = 3H
Radius r = 8 cm M1
Volume of cylinder = π r² H = π × 8² × H = 64π H
Volume of cone = 1/3 π r² h = 1/3 π × 8² × 3H = 64π H M1
Total volume = 64π H + 64π H = 128π H = 640π ⇒ H = 640 ÷ 128 = 5 cm
Height of cone h = 3 × 5 = 15 cm. A1

Answer: 15 cm

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Question 5 AQA Jun 2023 2H Q11

[4 marks]

@@TABULAR₀@@
A bowl is a hemisphere with radius 12 cm

Water is poured into the bowl
at a rate of 325 cm³ per second
for 8 seconds.
Does the water fill **more than** 70% of the bowl?
You **must** show your working.

Worked Solution & Answer

Volume of hemisphere = 1/2 × 4/3 × π × 12³ = 1152π ≈ 3619.11 cm³ M1
Volume of water = 325 × 8 = 2600 cm³. M1
70% of bowl = 0.7 × 3619.11 = 2533.38 cm³ M1
Since 2600 > 2533.38 (or 71.8% > 70%), Yes. C1

Answer: Yes — volume of bowl = 3619.1 cm³, water = 2600 cm³, 2600 / 3619.1 = 71.8% > 70%

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Question 6 IGCSE Specimen 1H Q18

[4 marks]

A container is made from a hemisphere on top of a cylinder, as shown in the diagram.

The hemisphere and the cylinder both have radius 6 cm.
The height of the cylinder is 10 cm.
There is water to a depth of t cm in the cylinder.
The volume of water in the container is half the total volume of the container.
Work out the value of t.

Worked Solution & Answer

(The volume of a cylinder is πr²h)
Volume of hemisphere = (2/3)π(6³) = 144π cm³ M1
Volume of cylinder = π(6²)(10) = 360π cm³ M1
Total volume = 504π cm³ M1
Half total volume = 252π cm³
Volume of water = π(6²)t = 36πt = 252π ⇒ t = 252 ÷ 36 = 7. A1

Answer: 7

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Question 7 IGCSE Jun 2024 1H Q23

[5 marks]

A solid shape is made by removing a hemisphere, shown shaded, from a cone as shown in the diagram.

The radius of the hemisphere is 2x cm
The radius of the base of the cone is 5x cm
The vertical height of the cone is 6x cm
The volume of the solid shape is 6948π cm³
Work out the **total** surface area of the solid hemisphere that has been removed from the cone.
Give your answer correct to the nearest integer.

cm²

Worked Solution & Answer

Volume of cone = ⅓π(5x)²(6x) = 50πx³ M2
Volume of hemisphere = ½ × (4/3)π(2x)³ = (16/3)πx³ M1
Remaining volume = (50 − 16/3)πx³ = (134/3)πx³ = 6948π ⇒ x³ = 6948 × 3 ÷ 134 = 10422/67 ≈ 155.552 ⇒ x = 5.378 cm M1
Total surface area of solid hemisphere = 3πR² = 3π(2x)² = 12πx² = 12π(5.378)² ≈ 1090.35... ≈ 1090 cm² A1

Answer: 1090 cm²

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Question 8 Specimen Set 2 1H Q11

[5 marks]

A solid is made by putting a hemisphere on top of a cone.
The total height of the solid is 5x
The radius of the base of the cone is x
The radius of the hemisphere is x

A cylinder has the same volume as the solid.
The cylinder has radius 2x and height h
All measurements are in centimetres.
Find a formula for h in terms of x
Give your answer in its simplest form.

h =

Worked Solution & Answer

(The volume of a cylinder is πr²h)
Height of cone = 5x – x = 4x M1
Volume of cone = 1/3 πx²(4x) = 4/3 πx³ M1
Volume of hemisphere = 1/2 × 4/3 πx³ = 2/3 πx³ M1
Total solid volume = 2πx³ M1
Cylinder volume = π(2x)²h = 4πx²h
Equating: 4πx²h = 2πx³ → h = x/2. A1

Answer: h = x2

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Question 9

[5 marks]

A solid cone is joined to a solid hemisphere to make the solid **T** shown below.

The diameter of the base of the cone is 7 cm.
The diameter of the hemisphere is 7 cm.
The total volume of **T** is 120π cm³
The total height of **T** is y cm.

(a) Calculate the value of y.
Give your answer correct to 3 significant figures. [4 marks]

y =

(b) explain the effect this would have on your answer to part (a). [1 mark]

Worked Solution & Answer

(a)

r = 3.5 cm M1
Hemisphere volume = (2/3)π(3.5³) = (343/12)π M1
Cone volume = (1/3)π(3.5²)(y – 3.5) = (49/12)π(y – 3.5) M1
Total: 343/12 + 49/12(y – 3.5) = 120 ⇒ y – 3.5 = 1097/49 ≈ 22.388 ⇒ y ≈ 25.888 → 25.9 A1

Answer (a): y = 25.9

(b)

The value of y (the total height) would decrease

Answer (b): The value of y (the total height) would decrease

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Question 10 OCR Nov 2022 4H Q12

[6 marks]

The diagram shows a sphere and a cone.

The sphere has radius 4 cm.
The cone has radius 15 cm and height 30 cm.
The sphere is completely filled with water.
The same amount of water is poured into the cone.
Work out the depth, d cm, of the water in the cone.
You must show your working.
[The volume V of a sphere with radius r is V = 43π r³.
The volume V of a cone with radius r and height h is V = 13π r² h.]

d =

Worked Solution & Answer

Volume of sphere = (4/3)π(4³) = 256π/3 ≈ 268.08 cm³ M2
In the cone, radius is half the height (15 / 30 = 1/2), so at depth d the water radius is r = 0.5d M2
Volume of water in cone = (1/3)π(0.5d)²d = πd³ / 12 M1
Equating volumes: πd³ / 12 = 256π / 3 ⇒ d³ = 1024 ⇒ d = ∛1024 ≈ 10.08 ≈ 10.1 cm. A1

Answer: d = 10.1

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Score: 0 / 10 (0 / 39 marks)

Spheres, cones and cylinders Worksheets Next: Volume of a frustum →

Grade 8+ Quizzes

14 Topics

8A - Quad. sim. equations	8B - Transforming graphs y=f(x)
8C - Algebraic proof	8D - Completing the square
8E - Quadratic inequalities	8F - Quadratic sequences
8G - Velocity time graphs	8H - Circles and tangents
8J - Circle theorem proof	8K - Vector proof
8L - Probability equation questions	8M - Invariant points
8N - Spheres, cones and cylinders	8P - Volume of a frustum