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Cone and Sphere
1. A vessel in the form of an inverted cone is filled with water to the
brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid
cones are dropped in it so that they are fully submerged. As a result,
one third of the water in the original cone overflows. What is the
volume of each of the solid cones submerged?
Solution. Height of the cone, h = 20 cm
16.8
cm = 8.4 cm
Radius of the cone, r =
2
Volume of water in the vessel
1
1 22
× 8.4 × 8.4 × 20 cm3
= πr2h = ×
3
3
7
3
= 1478.4 cm
1478.4
Volume of water overflown =
cm3 = 492.8 cm3
3
3
Volume of two equal cones = 492.8 cm
492.8
Volume of each cone =
cm3 = 246.4 cm3.
2
2. A girl fills a cylindrical bucket 32 cm in height and 18 cm in radius
with sand. She empties the bucket on the ground and makes a
conical heap of the sand. If the height of the conical heap is 24 cm,
find (i) its radius (ii) its slant height (Leave your answer in square
root form).
Solution. Height of the cylinder, h = 32 cm
Radius of the cylinder, r = 18 cm
∴ Volume of the sand contained in the cylinder
= πr2h = π × 182 × 32 cm3
... (i)
(i) Height of the conical heap, H = 24 cm
Let radius of the conical heap be R cm.
1
Then, volume of the conical heap = πR 2 H
3
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1
× π × R2 × 24 cm3
... (ii)
3
From (i) and (ii), we have
1
π × 182 × 32 = × π × R2 × 24
3
18 × 18 × 32 × 3
⇒
R2 =
= 36 × 36
24
⇒
R
= 36 cm.
(ii) We know that slant height, L of the cone
= R 2 + H 2 = 362 + 242 = 1872 cm.
3. An exhibition tent is in the form of the cylinder surmounted by a
cone. The height of the tent above the ground is 85 m and the height
of the cylindrical part is 50 m. If the diameter of the base is 168 m,
find the quantity of canvas required to make the tent. Allow 20%
extra for folds and for stitching. Give your answer to the nearest m2.
Solution. Base radius of the tent = 84 m
Height of the cylindrical part = 50 m
Height of the conical part
= 35 m
∴ Slant height of the conical part
= r 2 + h2
= 842 + 352 m
= 8281 m
= 91 m
Curved surface area of the tent
= Curved surface area of the cylinder
+ Curved surface area of the cone
= 2πrh + πrl
= πr(2h + l)
22
=
× 84 (2 × 50 + 91)m2
7
= 22 × 12 × 191 m2 = 50424 m2
Area of canvas required for folds and stitching
=
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22 2
m = 10084.80 m2
100
∴ Quantity of canvas required to make the tent
= (50424 + 10084.80) m2 = 60508.80 m2
= 60509 m2.
4. From a solid cylinder whose height is 8 cm and radius 6 cm, a
conical cavity of height 8 cm and of base radius 6 cm is hollowed
out. Find the volume of the remaining solid correct to four
significant figures. Also find the total surface area of the remaining
solid.
Solution :
Radius of the cylinder = Radius of the cone
= 6 cm
Height of the cylinder = Height of the cone
= 8 cm
∴ Slant height of he cone = h 2 + r 2
= 82 + 62 cm = 100 cm = 10 cm = 10 cm
Volume of the remaining solid
= Volume of the cylinder – Volume of the cone
1
= πr2h – πr2h
3
2
2
= πr2h = × 3.1416 × 62 × 8 cm3
3
3
= 603.18 cm3 = 603.2 cm3.
Total surface area of the remaining solid
= Curved surface are of the cylinder
+ Area of the base of the cylinder
+ Curved surface are of the cone
= 2πrh + πr2 + πrl = πr(2h + r + l)
= 3.1416 × 6(2 × 8 + 6 + 10) cm2
= 3.1416 × 6 × 32 cm2
= 603.18 cm2 = 603.2 cm2.
= 20% of 50424 m2 = 50424 ×
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5.
A hollow cone is cut by a plane parallel to the base and the upper
portion is removed. If the curved surface area of the remainder is
8
9
of the curved surface area of the whole cone, find the ratio of the
line segments into which the altitude of the cone is divided by the
plane.
Solution.
OAB is a cone which is cut through
CD to get a smaller cone OCD.
Let the radius, height and slant height
of the original cone be r1, h1 and l1 respectively.
Also, let the radius, height and slant height
of smaller cone be r2, h2 and l2 respectively.
ΔOFB ~ ΔOED
OF
OB
FB
=
=
⇒
OE OD ED
h1
l
r
= 1 = 1
⇒
... (i)
h2
l2
r2
Curved surface area of the original cone = πr1l1
... (ii)
Curved surface area of the smaller cone = πr2l2
... (iii)
8
Now, curved surface area of the remainder = of the curved
9
surface area of the whole cone.
⇒ Curved surface area of the whole cone – curved surface area of
the smaller cone
8
= of the curved surface area of the whole cone.
9
⇒ Curved surface area of the smaller cone
8⎞
⎛
= ⎜1 – ⎟ curved surface area of the whole cone.
9⎠
⎝
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1
curved surface area of the whole cone.
9
1
⇒ πr2l2 = πr1l1
[From (ii) and (iii)]
9
9r2
h
l
⇒
[From (i)]
= 1 = 1
r1
h2
l2
h
h
⇒ 9 2 = 1
h1
h2
h12 9
h
3
⇒ 2 = ⇒ 1 =
h2 1
h2
1
h – h2
3–1
⇒ 1
[Applying dividendo]
=
h2
1
h2
1
⇒
=
h1 – h2
2
Hence, the required ratio is 1 : 2.
6. An open cylindrical vessel of internal diameter 7 cm and height 8
cm stands on a horizontal table. Inside this is placed a solid metallic
right circular cone, the diameter of whole base is 3.5 cm and height
8 cm. Find the volume of water required to fill the vessel. If the cone
3
is replaced by another cone, whose height is 1 cm and the radius
4
of whose base is 2 cm, find the drop in the water level.
Solution.
Radius of the cylinder, R = 3.5 cm
Height of the cylinder, H = 8 cm
Volume of the cylinder = πR2H
22
=
× (3.5)2 × 8 cm3
7
= 308 cm3 ... (i)
=
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Case I
3.5
cm
2
= 1.75 cm
Height of the cone, h = 8 cm
1
Volume of the cone = πr2h
3
1 22
= ×
× (1.75)2 × 8 cm3
3
7
77
cm3
... (ii)
=
3
∴ Volume of water required to fill the vessel
= Volume of the cylinder – Volume of the cone
77 ⎞
847
⎛
3
cm3.
= ⎜ 308 –
⎟ cm =
3 ⎠
3
⎝
Case II :
3
7
Height of the cone, h = 1 cm = cm
4
4
Radius of the cone, r = 2 cm
Let the height of the water level from the base be x cm.
1
Then, π × R2 × x = π × r2 × h + volume of water in the cylinder
3
1
7 847
1
7 539 553
2
×
= +
=
⇒ ( 3.5 ) × x = × 22 × +
3
4
3
3
6
6
π
553
1
553 2 2 158
⇒ x=
×
=
× × =
6
(3.5) 2
6
7 7
21
∴ Drop in the water level = H – x
158 ⎞
10
⎛
= ⎜8 –
cm
=
cm.
⎟
21 ⎠
21
⎝
Radius of the cone, r =
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7. A spherical metallic ball of radius 3 cm is melted and recast into
three spherical balls. The radii of two of these balls are 2.5 cm and 2
cm respectively. Find the radius of the third ball.
Solution.
Let the radius of the third ball be r cm.
4
∴ Volume of the bigger ball = π× 33 cm3 and volume of the three
3
4
4
4
smaller balls are π(2.5)3 cm3, π × 23cm3 and πr3 cm3
3
3
3
respectively.
4
4
4
4
∴
π × 33 = π (2.5)3 + π × 23 + π r 3
3
3
3
3
4
4
3
⇒
π × 33 = π ⎡( 2.5 ) + 23 + r 3 ⎤
⎦
3
3 ⎣
3
⇒ 27 = 15.625 + 8 + r
⇒ r3 = 27 – 23.625 = 3.378
⇒ r = 1.5
Hence, radius of the third circle = 1.5 cm.
8. With reference to the figure given alongside,
a metal container in the form of a cylinder
is surmounted by a hemisphere of the same
radius. The internal height of the cylinder
is 7 m and the internal radius is 3.5 m. Calculate :
(i) The total area of the internal surface, excluding the base.
(ii) The internal volume of the container in m3.
22
(Take π =
)
7
Solution. Height of the cylindrical part = 7 m
Radius of the cylinder = Radius of the hemisphere = 3.5 m
(i) Total area of the internal surface excluding base
= Internal curved surface area of the cylinder
+ Internal curved surface area of the hemisphere
= 2πrh + 2πr2 = 2πr (h + r)
Math Class X
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Question Bank
22
× 3.5 (7 + 3.5) m2
7
= 22 × 10.5 m2 = 231 m2.
(ii) The internal volume of the container
= Volume of the cylinder + Volume of the hemisphere
2
2 ⎞
⎛
= πr2h + πr3 = πr2 ⎜ h + r ⎟
3 ⎠
3
⎝
22
2
⎛
⎞
=
× (3.5)2 ⎜ 7 + × 3.5 ⎟ m3
7
3
⎝
⎠
22
28 3
× 3.5 × 3.5 ×
m
=
7
3
22 × 3.5 × 0.5 × 28 3 1078 3
=
m
m =
3
3
1
= 359 m3
3
9. A cylindrical can whose base is horizontal and of radius 3.5 cm
contains sufficient water so that when a sphere is placed in the can,
the water just covers the sphere. Given that the sphere just fits into
the can, calculate :
(i) The total surface area of the can in contact with water when the
sphere is in it.
(ii) The depth of water in the can before the sphere was put into the
22
and give your answer as proper fractions.
can. Take π to be
7
Solution.
Radius of the can = Radius of the sphere
= 3.5 cm
[∴ Sphere just fits into the can]
Height of water = Diameter of the sphere
= 7 cm
[∴ Water just covers the sphere]
(i) Required surface area of the can
=2×
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= 2πrh + πr2 = 2πr(2h + r)
22
=
× 3.5(2 × 7 + 3.5) cm2
7
= 22 × 0.5 × 17.5 cm2 = 11 ×
35
cm2
2
1
cm2.
2
(ii) Let the depth of water in the can before the sphere was put into
it be H cm.
Volume of water in the can before the sphere was put into it
= Volume of water in the can with sphere
– Volume of sphere.
4
⇒
πr2H = πr2h – πr3
3
4 ⎞
⎛
π r2 ⎜ h – r ⎟
4
3 ⎠
⎝
h
r
⇒
H=
=
–
2
πr
3
4
4 7⎞
⎛
⎞
⎛
= ⎜ 7 – × 3.5 ⎟ cm = ⎜ 7 – × ⎟ cm
3
3 2⎠
⎝
⎠
⎝
42 – 28
7
1
cm = cm = 2 cm
=
6
3
3
Hence, depth of water in the can before the sphere was put into it =
1
2 cm.
3
10. The internal and external radii of a hollow sphere are 3 cm and 5 cm
respectively. The sphere is melted to form a solid cylinder of height
2
2 cm. Find the diameter and curved surface area of the cylinder.
3
Solution.
External radius of the sphere, R = 5 cm
Internal radius of the sphere, r = 3 cm
= 192
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4
π(R3 – r3)
3
4
= π(53 – 33) cm3
3
4
= π(125 – 27) cm3
3
4
= π × 98 cm3 ..... (i)
3
2
8
Height of the cylinder = 2 cm = cm
3
3
Let r1 be the radius of the cylinder.
∴
Volume of the cylinder = π r12 h
8
= π r12 × cm3 ... (ii)
3
4
8
From (i) and (ii), we have π × 98 = π r12 ×
3
3
98
= 49
⇒
= r12 =
2
⇒
r1
=7
⇒
Diameter of the cylinder = 14 cm.
Now, curved surface area of the cylinder
22
8
× 7 × cm2
= 2πr1h = 2 ×
7
3
352
1
cm2 = 117 cm2.
=
3
3
11. A conical vessel of radius 6 cm and height 8 cm is completely filled
with water. A sphere is lowered into the water and its size is such
that when it touches the sides, it is just emersed. What fraction of
water overflows?
Solution.
Height of the vessel CD = H = 8 cm
Radius of the conical vessel BD = R = 6 cm
∴
Math Class X
Volume of the material =
10
Question Bank
Let the radius of the sphere be r cm.
Then, OE = OD = r cm.
BE = BD = 6 cm
[Tangents from an external
point to a circle are equal]
Slant height of the conical vessel
BC = BD 2 + DC2
= 62 + 82 cm = 100
= 10 cm.
∴
CE = BC – BE
= (10 – 6) cm = 4 cm.
OC = CD – OD = (8 – r) cm.
In right triangle OCE, we have, OC2 = OE2 + CE2
⇒ (8 – r)2 = r2 + 42
⇒ 64 + r2 – 16r = r2 + 16
⇒ 16r = 48 ⇒ r = 3 cm.
Volume of water in the conical vessel before immersing the sphere
= volume of the vessel
1
= π × 36 × 8 cm3 = 96π cm3
... (i)
3
Volume of water overflown = Volume of the sphere
4
... (ii)
= π × 33 cm3 = 36π cm3
3
∴ Fraction of water overflown
Volume of water overflown
=
original volume of water
36π
3
= .
=
96π
8
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12. A solid sphere and a solid hemisphere have the same total surface
area. Find the ratio of their volumes.
Solution. Let the radius of the sphere be R and that of the
hemisphere be r.
Then, total surface area of the sphere = 4πR2.
And, total surface area of the hemisphere = 2πr2 + πr2 = 3πr2
But
4πR2 = 3πr2
R
3
⇒
…. (i)
=
r
8
4
Volume of the sphere = π R3
3
2
Volume of the hemisphere = π r 3
3
4
3
π R 3 2R 3
⎛
⎞
Volume of the sphere
3
3 3
= 3
= 3 =2 ⎜
=
⎟
2 3
Volume of the hemisphere
r
2
4
⎝
⎠
πr
3
Hence, the required ratio = 3 3 : 4.
13. A hemispherical cup of radius 4 cm is filled to the brim with a
liquid. The liquid is then poured into a vertical cone of radius 8 cm
and height 16 cm. Find the radius and height of the top surface of
the liquid in the cone. What percentage of the cone remains empty?
Solution. Radius of the
hemispherical cup = 4 cm.
Volume of liquid in the cup
2
128
π cm3
... (i)
= π × 43 cm3 =
3
3
Let the height of the liquid in
the cone be h cm and radius of
the top surface of the liquid be r cm.
ΔOAB ~ ΔOCD
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Question Bank
OA
OC
=
AB
CD
16 h
h
⇒
⇒
=
=2
8
r
r
⇒ h = 2r
⇒
... (ii)
1
Volume of liquid in the cone = π r 3h
3
128
1
⇒
π × π r 2h
[From (i)]
3
3
⇒ r2h = 128 ⇒ 2r3 = 128 ⇒ r = 4 cm
From (ii), h = 8 cm.
Hence, radius and height of the top surface of the liquid in the cone
are 4 cm and 8 cm respectively.
1
1024
π cm3
Volume of the whole cone = π × 83 × 16 cm3 =
3
3
Volume of the empty part of the cone = volume of the whole cone –
volume of liquid in the cone
128 ⎞ 896
⎛ 1024
=⎜
π –
π⎟ =
π cm3
3 ⎠
3
⎝ 3
∴ Percentage of the cone which remains empty
896
π
89600
3
=
× 100% =
% = 87.5%.
1024
1024
π
3
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