1.
A sphere has radius r.
A cone has base radius r and perpendicular height x.
The volume of the sphere is double the volume of the cone.
Not drawn accurately
r
x
r
(a)
Show that x =2r
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(2)
(b)
Calculate the ratio of the surface area of the sphere to the curved surface area of the cone.
Give your answer in surd form.
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Answer ......................................................................
(4)
(Total 6 marks)
The Robert Smyth School
1
2.
A thin-walled glass paperweight consists of a hollow cylinder with a hollow cone on top as
shown.
The paperweight contains just enough sand to fill the cylinder.
2 cm
4 cm
6 cm
The paperweight is now turned upside down.
x
Calculate the depth of the sand, (marked x in the diagram).
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Answer ........................................................... cm
(Total 5 marks)
The Robert Smyth School
2
3.
The first diagram shows a cone of base radius 12 cm and perpendicular height 10 cm.
A small cone of base radius 6 cm and perpendicular height 5 cm is cut off the bottom
to leave a frustum.
The frustum has a lower radius of 6 cm, an upper radius of 12 cm and a perpendicular
height of 5 cm (see second diagram).
12 cm
12 cm
5 cm
10 cm
6 cm
6 cm
Frustum
5 cm
Not to scale
(a)
Find the volume of the frustum, giving your answer in terms of .
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Answer ....................................... cm3
(4)
(b)
The frustum has the same volume as another cone of perpendicular height 35 cm.
Calculate the radius of this cone.
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Answer ...................................... cm
(3)
(Total 7 marks)
The Robert Smyth School
3
1.
(a)
4
1
πr³=2×
πr²x
3
3
Must include the factor of 2
Allow use of h instead of x
M1
Simplified to give x = 2r
Alternatively
Allow substitution of 2r for height of cone and verification of
result
ie
2 × Vol cone
=
(b)
=2×
A1
1
× π × r ²× 2r M1
3
4
π r ³ (must be seen) A1
3
(l) ² = r ² + 4r ²
(l) ² = r ² + (2r)² is M1 (l) ² = r ² + 2r² is
M1
M0
(l) = √5 r
A1
Surface area cone = π × r × √5 r
Using their l if from an attempt at Pythagoras
M1
4 : √5
A1
Allow √5 : 4
SC2 for a complete numerical solution
[6]
2.
volume cylinder = 113.(...) cm
Accept 36
3
B1
volume cone = 18.8(...)cm3
Accept 6
B1
Volume (their cylinder – their cone) ÷ 9
Accept 30 ÷ 9
M1
3.3(3)
= 5.3(3...)
A1
A1ft
Accept fraction. (5 1/3)
f.t. iff M1 awarded.
Consistent use of diameter for radius gives 144 (= 452.39) for
cylinder and 24 (=75.40) for cone.
Volume = 120 (376.99). Volume ÷ 36 = 3.333.. + 2 = 5.333
Give B0, B1, M1, A1, A1 f.t.
Hence do not give full marks if answer seen on answer line.
Check working before awarding full marks.
Do not accept 5 as a answer.
[5]
3.
(a)
Vol large cone = 1/3 ×  × 122 × 10
Or
Vol small cone = 1/3 ×  × 62 × 5
Subtraction of cone volumes
480 or 60
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M1
M1
A1
4
(b)
Vol frustum = 420
Or Vol frustum = 7 8 of Vol large cone
1/31260 scores 3 marks altogether
allow 420
A1
1/3 ×  × r2 × 35 = their 420
r2 = 36
r=6
M1 for equating volumes
M1
A1
A1
[7]
The Robert Smyth School
5