4.14H Cylinders: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 65 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the volume of a cylinder with radius 7 cm and height 10 cm. Give your answer in terms of π.

(Total for Question 1 is 2 marks)

Calculate the total surface area of a closed cylinder with radius 5 cm and height 12 cm. Give your answer in terms of π.

(Total for Question 2 is 3 marks)

A metal rod is made by drilling a cylindrical hole through the centre of a solid cylinder. The original cylinder has radius 8 cm and height 15 cm. The hole is a cylinder of radius 3 cm drilled along the full height. Work out the volume of metal remaining, giving your answer in terms of π.

(Total for Question 3 is 3 marks)

A cylinder has height 20 cm and volume 8000 π cm³. Work out the radius of the cylinder.

(Total for Question 4 is 3 marks)

Show that if every linear dimension of a cylinder is multiplied by a factor k then its volume is multiplied by k³. Write down any algebraic steps you use.

(Total for Question 5 is 4 marks)

A cylindrical column has radius 0.4 m and height 2.5 m. Only the curved surface is to be painted. One litre of paint covers 6 m². Calculate how many litres of paint are needed. Give your answer to the nearest whole litre.

(Total for Question 6 is 3 marks)

A cylinder has the property that its curved surface area is equal to the combined area of its two circular bases. If its volume is 4500 π cm³, find the radius and height of the cylinder. Give answers to 2 decimal places.

(Total for Question 7 is 4 marks)

Mark scheme · 4.14H Cylinders: Higher Tier Practice

Question 1

M1 use V = π * r² * h with r = 7, h = 10
A1 490 π cm³ cao
Answer: 490 π cm³

Question 2

M1 calculate curved surface area 2 * π * r * h = 2 * π * 5 * 12 = 120 π
M1 calculate area of two bases 2 * π * r² = 2 * π * 25 = 50 π
A1 170 π cm² cao
Answer: 170 π cm²

Question 3

M1 volume of original cylinder π * 8² * 15 = 960 π
M1 volume of drilled cylinder π * 3² * 15 = 135 π
A1 remaining volume = 960 π - 135 π = 825 π cm³ cao
Answer: 825 π cm³

Question 4

M1 use V = π * r² * h and rearrange to r² = V / (π * h)
M1 substitute V = 8000 π and h = 20 to get r² = 8000 / 20 = 400
A1 r = 20 cm cao
Answer: 20 cm

Question 5

M1 start with V = π * r² * h and substitute r' = k r, h' = k h giving V' = π * (k r)² * (k h)
M1 simplify to V' = π * k² * r² * k * h
A1 simplify to V' = k³ * (π * r² * h)
B1 conclude V' = k³ V, stating that volume is multiplied by k³ oe
Answer: V' = k³ V, so the volume is multiplied by k³.

Question 6

M1 calculate curved surface area 2 * π * r * h = 2 * π * 0.4 * 2.5 = 2 π m²
M1 divide by coverage: litres = (2 π) / 6 = π / 3
A1 π/3 approx 1.05 so 1 litre to the nearest whole litre awrt
Answer: 1 litre

Question 7

M1 state equation curved area = area of two bases so 2 * π * r * h = 2 * π * r² and deduce h = r
M1 use volume V = π * r² * h and substitute h = r to get V = π * r³
M1 rearrange r³ = 4500 and take cube root
A1 r = 16.51 cm, h = 16.51 cm (both to 2 d.p.) cao
Answer: r = 16.51 cm, h = 16.51 cm

Cylinders: Higher Tier Practice - Worksheets, Questions and Revision