Question Bank In Mathematics Class X (Term–II)
13
SURFACE AREAS AND VOLUMES
A. SUMMATIVE ASSESSMENT
(b) Total surface area = 2πr2 + 2πrh
= 2πr(r + h)
(c) Curved surface area of hollow cylinder
= 2πh(R – r),
where R and r are outer and inner radii
HA
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13.1 SURFACE AREA OF A
COMBINATION OF SOLIDS
1. Cuboid : For a cuboid of dimensions l,
b and h, we have :
AK
AS
(d) Total surface area of hollow cylinder
= 2πh(R + r) + 2π(R2 – r2)
4. Cone : For a cone of
height h, radius r and slant
height l, we have :
(a) Curved surface area
l 2 b2 h2
TH
ER
2. Cube : For a cube of edge l, we have :
= πrl = r h 2 r 2
(b) Total surface area
= πr2 + πrl
= πr (r + l)
S
(c) Length of diagonal =
PR
(a) Lateral surface area = 2h(l + b)
(b) Total surface area = 2(lb + bh + lh)
BR
O
5. Sphere : For a
sphere of radius r, we
have :
Surface area = 4πr2
(a) Lateral surface area = 4l2
(b) Total surface area = 6l2
(c) Length of diagonal = 3 l
3. Cylinder : For a cylinder
of radius r and height h, we
have :
(a) Area of curved surface
= 2πrh
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6. Hemisphere (solid) : For a hemisphere of
radius r we have :
(a) Curved surface area
= 2πr2
(b) Total surface area
= 3πr2
TEXTBOOK’S EXERCISE 13.1
Unless stated otherwise, take =
Sol. Let the side of cube = y cm
22
.
7
Volume of cube = 64 cm3
Then, volume of cube = side3 = y3
As per condition
y3 = 64
y3 = 4 3
Q.1. 2 cubes each of volume 64 cm3 are
joined end to end. Find the surface area of the
resulting cuboid.
[2011 (T-II)]
1
Q.2. A vessel is in the form of a hollow
hemisphere mounted by a hollow cylinder. The
diameter of the hemisphere is 14 cm and the
total height of the vessel is 13 cm. Find the inner
surface area of the vessel.
[2011 (T-II)]
Sol. Diameter of the hollow hemisphere
AS
(3.5)2 (12) 2 cm
PR
AK
= 156.25 cm = 12.5 cm
Total surface area of the toy
= Curved surface area of the hemisphere
+ Curved surface area of the cone
= 2 (3.5)2 cm2 + (3.5) (12.5) cm2
= 24.5 cm2 + 43.65cm2
S
= 68.25 cm2 =
68.25 22
cm2 = 214.5 cm2.
7
Q.4. A cubical block of side 7 cm is
surmounted by a hemisphere. What is the
greatest diameter the hemisphere can have? Find
the surface area of the solid.
[2011 (T-II)]
Sol. Side of cubical block = 7 cm
O
TH
ER
= Diameter of the hollow cylinder = 14 cm
14
Radius of the hollow hemisphere =
cm = 7 cm
2
Radius of the base of the
hollow cylinder = 7 cm
Total height of the vessel
= 13 cm
Height of the hollow
cylinder = (13 – 7) cm
= 6 cm
Inner surface area of the vessel
= Inner surface area of the
hemisphere + Inner surface area of the hollow cylinder
= 2 (7)2 cm2 + 2 (7) (6) cm2
= 98cm2 + 84cm2 = (98 + 84) cm2
22
= 182 cm2 = 182 ×
cm2 = 26 × 22 cm2
7
= 572 cm2.
=
HA
N
y = 4 cm
Hence, side of cube is 4 cm.
For the resulting cuboid
length (l ) = 4 + 4 = 8 cm
breadth (b) = 4 cm
height (h) = 4 cm
Surface area of the resulting cuboid
= 2(lb + bh + hl )
= 2(8 × 4 + 4 × 4 + 4 × 8) cm2
= 2(32 + 16 + 32) cm2 = 2(80) cm2
= 160 cm2.
G
O
YA
L
BR
Side of cubical block = Diameter of hemisphere
= 7 cm
2R = 7
7
R = cm
2
Surface area of solid = Surface area of the cube
– Area of base of hemisphere
+ C.S.A. of hemisphere
2
= 6 × side – R2 + 2R2
= 6 (7)2 cm2 + R2
22 7 7 2
= 6 × 7 × 7 cm2 +
× × cm
7 2 2
7
= 6 49 11 cm2
2
77
588 77 2
= 294 cm2 =
cm .
2
2
Q.3. A toy is in the form of a cone of radius
3.5 cm mounted on a hemisphere of same radius.
The total height of the toy is 15.5 cm. Find the
total surface area of the toy.
[2011 (T-II)]
Sol. Radius of the cone = Radius of hemisphere
= 3.5 cm
Total height of the toy = 15.5 cm
Height of the cone = (15.5 – 3.5) cm = 12 cm
Slant height of the cone (l ) =
=
r 2 h2
2
665
cm2 = 332.50 cm2
2
= 45 mm2 + 25 mm2 = (45 + 25) mm2
= 70 mm2
22
= 70 ×
mm2 = 220 mm2.
7
Hence, surface area of capsule = 220 mm2
Q.5. A hemispherical depression is cut out
from one face of a cubical wooden block such
that the diameter l of the hemisphere is equal to
the edge of the cube. Determine the surface area
of the remaining solid.
Q.7. A tent is in the shape of a cylinder
surmounted by a conical top. If the height and
diameter of the cylindrical part are
2.1 m and 4 m respectively and the slant height
of the top is 2.8 m, find the area of the canvas
used for making the tent. Also, find the cost of
the canvas of the tent at the rate of Rs 500 per
m2 (note that the base of the tent will not be
covered with canvas.)
Sol. Radius of the cone = 2 m
Sol. Diameter of the hemisphere = l = Side of the
cube
l
2
Surface area of the remaining solid
= Surface area of hemisphere
+ Surface area of cube
– Area of base of hemisphere
HA
AS
2
AK
2
l
l
2
= 2 6l
2
2
N
Radius of the hemisphere =
Radius of the cylinder = 2 m
Total surface area of the tent
2
S
l2
24 .
4
ER
=
PR
l
l 2
2
6l 2
= 6l =
4
2
TH
Q.6. A medicine capsule is in the shape of a
cylinder with two hemispheres stuck to each of
its ends (see figure below). The length of the
entire capsule is 14 mm and the diameter of the
capsule is 5 mm. Find its surface area.
O
= Curved surface area of the cylinder
+ Curved surface area of the cone
YA
L
BR
= 2 (2) (2.1) m2 + (2) (2.8) m2 = (8.4 + 5.6)m2
22 2
= 14 m2 = 14 ×
m = 44 m2
7
Cost the canvas of the tent at the rate of Rs 500
per m2 = Rs 44 × 500 = Rs 22000
Hence, cost of the canvas is Rs 22000.
O
Q.8. From a solid cylinder whose height is
2.4 cm and diameter 1.4 cm, a conical cavity of
the same height and same diameter is hollowed
out. Find the total surface area of the remaining
solid to the nearest cm2.
Sol. Height of cylinder = 2.4 cm
G
Sol. Diameter of capsule
= Diameter of hemisphere
= Diameter of cylinder = 5 mm
5
Radius of the hemisphere = r = mm
2
Height of the cylinder
Height of cone = 2.4 cm
= [14 – (2.5 + 2.5)] mm = 9 mm
Radius of cylinder = r =
Surface area of the capsule = Surface area of
cylinder + 2 Surface area of hemisphere
1.4
cm = 0.7 cm
2
Radius of cone = 0.7 cm
Slant height, of the cone
5 2
5
2
2
= 2π (9) mm + 2 2π mm
2
2
l=
3
(0.7)2 (2.4) 2 cm = 2.5 cm
Q.9. A wooden article was made by scooping
out a hemisphere from each end of a solid
cylinder, as shown in figure. If the height of the
cylinder is 10 cm, and its base is of radius
3.5 cm, find the total surface area of the article.
Sol. Height of cylinder = 10 cm
S
PR
AK
AS
HA
N
Radius of cylinder = 3.5 cm
Total surface area of the article
= C.S.A of cylinder + 2 C.S.A. of hemisphere
= 2 (3.5 (10) cm2 + 2 [2 (3.5)2] cm2
= 70cm2 + 49 cm2 = (70 + 49) cm2
22 2
= 119 cm2 = 119 ×
cm
7
= 17 × 22 cm2 = 374 cm2.
ER
Total surface area of the remaining solid
= C.S.A. of cylinder
+ C.S.A. of cone + Area of base
= 2rh + rl + r2
= r (2 h + l + r)
22
=
× 0.7 × (2 × 2.4 + 2.5 + 0.7) cm2
7
22 7
=
×
(4.8 + 3.2) cm2
7 10
22 7
=
×
× 8.0 cm2
7 10
176
=
cm2 = 17.6 cm2
10
Hence, total remaining surface area = 17.6 cm2
O
TH
= 18 cm2.
BR
OTHER IMPORTANT QUESTIONS
Q.1. A cylindrical pencil sharpened at one
edge is the combination of :
(a) a cone and a cylinder
(b) frustum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) two cylinders
Sol. (a) The given shape is a combination of a
YA
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Its surface area = 6 ×
9a 2
27a 2
=
4
2
O
Increase in surface area =
G
Per cent increase =
cone and a cylinder.
27a 2
15a 2
– 6a2 =
2
2
15a 2 100
× 2 = 125%
6a
2
Q.3. The total surface area of a hemisphere
of radius 7 cm is :
[2011(T-II)]
(a) 447 cm2
(b) 239 cm2
(c) 147 cm2
(c) 174 cm2
Sol. (c) Total surface area of the hemisphere
Q.2. If each edge of a cube is increased by
50%, the percentage increase in the surface area
is :
(a) 25% (b) 50% (c) 75% (d) 125%
Sol. (d) Let the edge of the cube be a.
= 3r2
= 3 × × 49 cm2
= 147 cm2
Then, its surface area = 6a2
150a
3a
New edge =
=
.
100
2
Q.4. If two solid hemispheres of same base
radius r are joined together along their bases,
4
then curved surface area of this new solid is :
(a) 4r2
(b) 6r2
2
(c) 3r
(d) 8r2
Sol. (a) The resulting solid will be a sphere of
Q.9. The total surface area of a
top (lattu) as shown in the figure is the
sum of total surface area of
hemisphere and the total surface area
of cone. Is it true?
Sol. No, the statement is false.
radius r.
Its curved surface area = 4r2.
Total surface area of the top (lattu) is the sum of
the curved surface area of the hemisphere and the
curved surface area of the cone.
Q.5. Volumes of two cubes are in the ratio
64 : 27. The ratio of their surface areas is :
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) 16 : 9
a2
32
N
Sol. Slant height of each cone =
16
=
= 16 : 9
9
Q.6. The diameter of a solid hemispherical
toy is 7 cm. Find its curved surface area and
total surface area.
Sol. Diameter of the hemispherical toy = 7 cm.
= 64 225 cm
= 17 cm.
Surface area of the resulting shape
22
× 8 × 17 cm2
7
= 854.85 cm2 = 855 cm2 (approx.)
= 2 (rl) = 2 ×
ER
.
. . Radius of the hemispherical toy, r = 3.5 cm
Curved surface area of the toy = 2πr2
22
=2×
× (3.5)2 cm2 = 77 cm2
7
Total surface area of the toy = 3πr2
22
=3×
× (3.5)2 cm2 = 115.50 cm2.
7
82 152 cm
AK
=
4
64
a1
=
3
a2
27
PR
6a22
42
=
S
6a12
3
HA
a13
AS
Sol. (d) We have
Q.10. Two cones with the same base radius
8 cm and height 15 cm are joined together along
their bases. Find the surface area of the shape so
formed.
BR
O
TH
Q.11. A tent of height 8.25 m is in the form
of a right circular cylinder with diameter of base
30 m and height 5.5 m, surmounted by a right
circular cone of the same base. Find the cost of
the canvas of the tent at the rate of Rs 45 per m2.
Sol.
YA
L
Q.7. If a solid cone of base radius r and
height h is placed over a solid cylinder having
same base radius and height as that of the cone,
then the curved surface area of the shape is
O
r r 2 h2 2rh . Is it true?
G
Sol. True. Since the curved surface area taken
together is same as the sum of curved surface areas
measured separately.
Height of the tent = 8.25 m.
Height of the cylindrical part = 5.5 m
.
. . Height of the conical part = (8.25 – 5.5) m
= 2.75 m.
30
Base radius of the tent =
m = 15 m.
2
.
. . Slant height of the conical part
Q.8. Two identical solid cubes of side a are
joined end to end. Then find the total surface
area of the resulting cuboid.
Sol. The resulting solid is a cuboid of dimensions
2a × a × a.
Total surface area of the cuboid
= 2 (lb + bh + hl)
= 2 (2a × a + a × a + a × 2a)
= 10a2.
(15)2 + (2.75)2 m =
= 15.25 m.
=
5
225 + 7.5625 m
Curved surface area of the tent = curved surface
area of the cylindrical part + curved surface area of
the conical part
= 2πrh + πrl = πr (2h + l)
22
=
× 15 (2 × 5.5 + 15.25) m2
7
22
=
× 15 × 26.25 m 2
7
Sol.
Height of the toy = 31 cm
Base radius of the cone
= radius of the hemisphere
= 7 cm
Height of the cone = (31 – 7) cm = 24 cm
N
= 1237.50 m2.
AS
= Rs 55687.50.
Q.12. A cone of maximum size is carved out
from a cube of edge 14 cm. Find the surface area
of the cone and of the remaining solid left out
after the cone carved out.
Sol. Diameter of the cone = 14 cm
22
× 7 ( 7 5 + 7) cm2
7
49 576 cm
AK
PR
S
BR
=
O
TH
2
=
=
22
× 7 (14 + 25) cm = 858 cm2.
7
Q.14. A solid is in the form of a right
circular cylinder with hemispherical ends. The
total height of the solid is 58 cm and the
diameter of the cylinder is 28 cm. Find the total
surface area of the solid.
[2006]
= 7 14 cm = 245 cm = 7 5 cm.
Total surface area of the cone = rl + r2
= r (l + r)
2
72 242 cm
ER
r 2 h2
=
= 625 cm = 25 cm
Total surface area of the toy
= Curved surface area of the hemisphere
+ Curved surface area of the cone
= 2r2 + rl = r (2r + l)
and height of the cone = 14 cm
Slant height of the cone =
r 2 h2
HA
Slant height of the cone =
Rate of the canvas = Rs 45 per m2
.
. . Cost of the canvas = Rs (1237.50 × 45)
Sol.
YA
L
= 154 ( 5 + 1) cm2
Surface area of the cube = 6 × 142 cm2
= 1176 cm2
Surface area of the remaining solid left out
after the cone is carved out
= surface area of the cube
– area of base of the cone
+ curved surface area of the cone
22 2
= 1176 7 154 5 cm2
7
G
O
Radius of the each hemisphere = base
radius of the cylinder = 14 cm
Total height of the toy = 58 cm
Height of the cylinder
= [58 – (14 + 14)] cm = 30 cm
Total surface area of the solid
= 2r2 + 2rh + 2r2 = 2r (2r + h)
22
=2×
× 14 (2 × 14 + 30) cm2
7
= 88 × 58 cm2 = 5104 cm2.
= 1022 154 5 cm2.
Q.13. A toy is in the form of a cone mounted
on a hemisphere of common base radius
7 cm. The total height of the toy is 31 cm. Find
the total surface area of the toy.
[2007, 2011 (T-II)]
Q.15. A toy is in the shape of a right circular
cylinder with a hemisphere on one end and a
cone on the other. The radius and height of the
6
= 2r2 + 2rh + rl = r (2r + 2h + l )
cylindrical part are 5 cm and 13 cm respectively.
The radii of the hemisphercial and conical parts
are the same as that of the cylindrical part. Find
the surface area of the toy if the total height of
the toy is 30 cm.
[2002]
=
22
× 5 (2 × 5 + 2 × 13 + 13) cm2
7
=
22
× 5 × 49 cm2 = 770 cm2.
7
Q.16. The internal and external diameters of
a hollow hemispherical vessel are 24 cm and
25 cm respectively. If the cost of painting 1 cm2
of the surface area is Rs 5.25, find the total cost
of painting the vessel all over.
[2001]
Sol. External radius (R) of the vessel = 12.5 cm
HA
N
Sol.
Radius of the cone = Radius of the cylinder
= radius of the hemisphere
= 5 cm.
Total height of the toy = 30 cm
Height of the cylinder h = 13 cm
Height of the cone = [30 – (13 + 5)] cm
= 12 cm.
AS
AK
=
Required cost of painting
= Rs 5.25 × 1925.78 = Rs 1010.38.
ER
= 144 25 cm = 13 cm.
Total surface area of the toy
= curved surface area of the hemisphere
+ curved surface area of the cylinder
+ curved surface area of the cone
22
× 612.75 cm2 = 1925.78 cm2.
7
PR
122 52 cm
S
Slant height of the cone =
Internal radius (r) of the vessel = 12 cm
Total surface area of the vessel
= 2R2 + 2r2 + (R2 – r2)
= [2 × (12.5)2 + 2 × 122 + (12.52 – 122)] cm 2
= [312.5 + 288 + 12.25] cm 2
TH
PRACTICE EXERCISE 13.1 A
canvas required to built the tent is :
O
Choose the correct option (Q 1 – 7) :
1. A funnel is the combination of :
(a) a cone and a cylinder
(b) frustrum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) a hemisphere and a cone.
2. A plumbline (shahul) is the combination of :
(a) a cone and a cylinder
(b) a hemisphere and a cone
(c) frustrum of a cone and a cylinder
(d) a sphere and a cylinder
BR
(a) 490 m (b) 525 m (c) 665 m (d) 860 m
G
O
YA
L
5. The ratio of the total surface area to the
lateral surface area of a cylinder with base diameter
160 cm and height 20 cm is :
(a) 1 : 2 (b) 2 : 1 (c) 3 : 1 (d) 5 : 1
6. The radius of the base of a cone is 5 cm and
its height is 12 cm. Its curved surface area is :
(a) 30π cm2
(b) 65π cm2
2
(c) 80π cm
(d) none of these
7. A right circular cylinder of radius r cm and
height h cm (h > 2r) just encloses a sphere of
diameter :
(a) r cm (b) 2r cm (c) h cm (d) 2h cm
3. A shuttle cock used for playing badminton
has the shape of the combination of : [2011 (T-II)]
(a) a cylinder and a cone
(b) a cylinder and a hemisphere
(c) a sphere and a cone
(d) frustrum of a cone and a hemisphere
8. Two identical solid hemispheres of equal
base radius r cm are stuck together along their
bases. The total surface area of the combination is
6r2. Is it true?
4. The height of a conical tent is 14 m and its
floor area is 346.5 m2. The length of 1.1 m wide
7
9. A solid cylinder of radius r and height h is
placed over other cylinder of same height and
radius. The total surface area of the shape formed
is 4rh + 4r2. Is it true?
10. A solid ball is exactly fitted inside the
cubical box of side a. Surface area of the ball is
4a2. Is it true?
11. From a solid cylinder whose height is 2.4
cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out.
Find the total surface area of the remaining solid
to the nearest cm2.
12. A decorative block shown below, is made
of two solids – a cube and a hemisphere. The
base of the block is a cube with edge 5 cm, and
the hemisphere fixed on the top has a diameter
4.2 cm. Find the total surface area of the block.
22
π = .
7
[2011 (T-II)]
N
13. A tent of height 3.3 m is in the form of
a right circular cylinder of diameter 12 m and
height 2.2 m, surmounted by a right circular cone
of the same diameter. Find the cost of canvas of
the tent at the rate of Rs 500 per m2.
AS
HA
14. Three cubes each of side 5 cm are joined
end to end. Find the surface area of the resulting
cuboid.
PR
AK
15. A solid is composed of a cylinder with
hemispherical ends. If the whole length of the
solid is 108 cm and the diameter of
hemispherical ends is 36 cm, find the cost of
polishing the surface at the rate of
7 paise per cm2.
ER
S
16. A rocket is in the form of a cone of height
28 cm, surmounted over a right circular cylinder
of height 112 cm. The radius of the bases of cone
and cylinder are equal, each being 21 cm. Find
22
BR
O
TH
the total surface area of the rocket. π =
7
4. Volume of a cone of base radius r and height
1
h = r2h.
3
4 3
5. Volume of a sphere of radius r =
r .
3
2
6. Volume of a hemisphere of radius r = r3.
3
TEXTBOOK’S EXERCISE 13.2
G
O
YA
L
13.2 VOLUME OF A COMBINATION OF
SOLIDS
1. Volume of a cuboid of dimensions l, b and
h = l × b × h.
2. Volume of a cube of edge l = l3.
3. Volume of a cylinder of base radius r and
height h = r2h.
Unless stated otherwise, take =
Sol.
22
.
7
Q.1. A solid is in the shape of a cone
standing on a hemisphere with both their radii
being equal to 1 cm and the height of the cone
is equal to its radius. Find the volume of the
solid in terms of .
8
Radius of the hemisphere
= Radius of cone = 1 cm
Height of cone = h = 1 cm
2
2
Volume of hemisphere = r3 = (1)3 cm3
3
3
2
= cm3
..(i)
3
1
1
Volume of cone = r2h = (1)2 (1) cm3
3
3
1
= cm3
..(ii)
3
Volume of the solid
= Volume of the hemisphere + Volume of cone
1 2
r h1
3
1
(1.5)2 (2) cm3 = 1.5 cm3
...(i)
3
1
Volume of cone ABO = r2h1
3
1
= (1.5)2 × (2) cm3 = 1.5 cm3 ...(ii)
3
For cylindrical portion :
Radius of the base (r) = 1.5 cm
Height of cylinder h2= 12 cm – (2 + 2) cm = 8 cm
Volume of cylinder = r2h2
= (1.5)2 (8) cm3 = 18 cm3 ..(iii)
Adding equations (i), (ii) and (iii), we have
Total volume of the model = volume of the two
cones + volume of the cylinder.
= 1.5 cm3 + 1.5 cm3 + 18 cm3 = 21 cm3
HA
N
=
AS
1 3
2
π cm = cm3.
3
3
Q.2. Rachel, an engineering student, was
asked to make a model shaped like a cylinder
with two cones attached at its two ends by using
a thin aluminium sheet. The diameter of the
model is 3 cm and its length is 12 cm. If each
cone has a height of 2 cm, find the volume of air
contained in the model that Rachel made.
(Assume the outer and inner dimensions of the
model to be nearly the same.)
Sol.
Volume of cone OAB =
PR
AK
= π+
22 3
cm = 66 cm3
7
Hence, the volume of the air contained in the model
that Rachel made is 66 cm3.
ER
S
= 21 ×
O
YA
L
BR
O
TH
Q.3. A gulab jamun,
contained sugar syrup up to
about 30% of its volume. Find
approximately how much syrup
would be found in 45 gulab
jamuns, each shaped like a
cylinder with two hemispherical
ends with length 5 cm and
diameter 2.8 cm (see figure).
[2011 (T-II)]
Sol. Gulab jamun is in the shape of cylinder with
G
two hemispherical ends.
Diameter of cylinder = 2.8 cm
Radius of cylinder = 1.4 cm
Height of cylindrical part
= (5 – 1.4 – 1.4) cm = (5 – 2.8) cm = 2.2 cm
For conical portion :
Radius of the base (r) =
3
cm = 1.5 cm
2
Height of cone (h1) = 2 cm
We know that, volume of cone =
1 2
r h
3
9
Volume of a gulab jamun
=
2
2
= (1.4)3 cm3 + (1.4)2 (2.2) cm3 + (1.4)3cm3
3
3
11
cm3
30
Volume of four conical depressions
=
11 3 22 3
cm =
cm = 1.47 cm3
30
15
Volume of the wood in the pen stand
= (525 – 1.47) cm3 = 523.53 cm3.
HA
N
=4×
PR
AK
AS
Q.5. A vessel is in the form of an inverted
cone. Its height is 8 cm and the radius of its top,
which is open, is 5 cm. It is filled with water up
to the brim. When lead shots, each of which is a
sphere of radius 0.5 cm are dropped into the
vessel, one-fourth of the water flows out. Find
the number of lead shots dropped in the vessel.
S
Sol. Radius of cone = 5 cm
ER
4
= (1.4)3 cm3 + (1.4)2 (2.2)cm3
3
4 1.4
2.2 cm3
= (1.4)2
3
5.6 6.6
= (1.96)
cm3
3
(1.96) (12.2)
=
cm3
3
Volume of 45 gulab jamuns
(1.96) (12.2)
= 45 ×
cm3
3
= 15 (1.96) (12.2) cm3
22
× 1.96 × 12.2 cm3
= 15 ×
7
= 15 × 22 × 0.28 × 12.2 = 1127.28 cm3
30
Volume of syrup = 1127.28 ×
cm3
100
= 338.184 = 338 cm3 (approximately)
1
22
14 3
×
× 0.25 ×
cm
3
7
10
Height of cone = 8 cm
Volume of cone =
TH
Q.4. A pen stand made of wood is in the
shape of a cuboid with four conical depressions
to hold pens. The dimensions of the cuboid are
15 cm by 10 cm by 3.5 cm. The radius of each of
the depressions is 0.5 cm and the depth is 1.4
cm. Find the volume in the entire stands. (See
figure).
200
cm3
3
O
YA
L
BR
O
=
1 2
1
r h = (5)2 8 cm3
3
3
G
Radius of spherical lead shot, r1 = 0.5 cm
Volume of a spherical lead shot
4 3
4
3
r = (0.5)3 cm3 =
cm
3 1
3
6
Volume of water that flows out
Sol. Length of cuboid, l = 15 cm
=
Width of cuboid, b = 10 cm
Height of cuboid, h = 3.5 cm
Volume of the cuboid
= 15 × 10 × 3.5 cm3 = 525 cm3
Volume of a conical depression
=
1
(0.5)2 (1.4) cm3
3
10
=
1
× volume of the cone
4
=
1 200
50
cm3
=
4 3
3
50 6
n =
3
n = 100
Hence, the number of lead shots dropped in the
vessel is 100.
BR
O
TH
ER
S
Q.6. A solid iron pole consist of a cylinder of
height 220 cm and base diameter 24 cm, which
is surmounted by another cylinder of height 60
cm and radius 8 cm. Find the mass of the pole,
given that 1 cm3 of iron has approximately 8 g
mass.
(Use = 3.14)
Q.7. A solid consisting of a right circular cone
of height 120 cm and radius 60 cm
standing on a hemisphere of radius 60 cm
is placed upright in a right circular
cylinder full of water such that it touches
the bottom. Find the volume of water left in
the cylinder, if the radius of the cylinder is
60 cm and its height is 180 cm.
Sol.
N
n
50
=
6
3
HA
As per condition,
AS
n
cm3
6
AK
Volume of n lead shots =
= 31680 cm3 + 3840 cm3
= 35520 cm3
= 35520 × 3.14 cm3 = 111532.8 cm3
Mass of the pole = 111532.8 × 8 g
= 892262.4 g = 892.26 kg
Hence, the mass of the pole is 892.26 kg
(approximately).
PR
Let the number of lead shots dropped in the
vessel be n.
YA
L
Radius of the cone OAB (r) = 60 cm
Height of cone OAB (h1) = 120 cm
Volume of cone OAB
Sol. Diameter of cylinder ABCD = 24 cm
24
cm3
2
= 12 cm
Height of cylinder ABCD (h) = 220 cm
Volume of cylinderABCD
= r2h = (12)2 (220)cm3 = 31680 cm3
Base radius of cylinder ABCD, R = 8 cm
Height of cylinder ABCD (H) = 60 cm
Volume of cylinder ABCD
= R2h = (8)2 (60) cm3 = 3840 cm3
Volume of solid iron pole
= Volume of the cylinder ABCD
+ Volume of the cylinder ABCD
1 2
1
r h1 = (60)2 (120) cm3
3
3
= 144000 cm3
Radius of the hemisphere (r) = 60 cm
Base radius of cylinder ABCD, r =
G
O
=
Volume of hemisphere =
=
=
Radius of the cylinder (r) =
Height of cylinder (h2) =
Volume of cylinder =
11
2 3
r
3
2
(60)3 cm3
3
144000 cm3
60 cm
180 cm
r2h2
= (60)2 (180) cm3
= 648000 cm3
Volume of water left in the cylinder
= Volume of the cylinder – [Volume of the cone
+ Volume of the hemisphere]
= 648000 cm3 – [144000 + 144000] cm3
= 648000 cm3 – 288000 cm3
= 360000 cm3
360000π
=
m3 = 0.36 m3
100 × 100 × 100
22 3
= 0.36 ×
m = 1.131 m3 (approx.)
7
HA
N
Radius of cylindrical neck = 1 cm
Height of cylindrical neck = 8 cm
AK
3
4 8.5
π cm3 + π 12 (8) cm3
3 2
PR
=
4
× 3.14 × 4.25 × 4.25 × 4.25 cm3 + 8 × 3.14 cm3
3
= 321.39 cm3 + 25.12 cm3 = 346.51 cm3
=
8.5
cm
2
ER
Hence, she is correct. The correct volume is
346.51 cm3.
TH
So, r =
Amount of water it holds
S
Sol. Diameter of sphere = 8.5 cm
AS
Q.8. A spherical glass vessel has a
cylindrical neck 8 cm long, 2 cm in diameter; the
diameter of the spherical part is 8.5 cm. By
measuring the amount of water it holds, a child
finds its volume to be 345 cm3. Check whether
she is correct, taking the above as the inside
measurements, and = 3.14.
O
OTHER IMPORTANT QUESTIONS
Q.1. Volume of the largest right circular cone
that can be cut out from a cube of edge
4.2 cm is :
(a) 9.7 cm3
(b) 77.6 cm3
3
(c) 58.2 cm
(d) 19.4 cm3
YA
L
BR
remains unfilled. Then the number of marbles
that the cube can accommodate is :
(a) 142296
(b) 142396
(c) 142496
(d) 142596
Sol. (a) Volume of the cube
O
Sol. (d) Radius of the cone =
4.2
cm = 2.1 cm.
2
= 223 cm3 = 10648 cm3
Space which remains unfilled
G
Height of the cone = 4.2 cm.
10648
cm3 = 1331 cm3
8
Remaining space = (10648 – 1331) cm3
1
Volume of the cone= r2h
3
=
=
1 22
×
× 2.1 × 2.1 × 4.2 cm3 = 19.404 cm3.
3
7
= 9317 cm3
4
(0.25)3 cm3
3
Let n marbles can be accommodated.
Q.2. A hollow cube of internal edge 22 cm is
filled with spherical marbles of diameter 0.5 cm
and it is assumed that
Volume of 1 marble =
1
space of the cube
8
Then, n ×
12
4
22
×
× (0.25)3 = 9317
3
7
9317 3 7
n=
4 22 (0.25) 3
= 142296.
(b) 0.35 cm3
(c) 0.34 cm3
(d) 0.33 cm3
Q.6. The ratio of the volumes of two spheres
is 8 : 27. The ratio between their surface areas
is :
[2011 (T-II)]
(a) 2 : 3
(b) 4 : 27
(c) 8 : 9
(d) 4 : 9
Sol. (d )
HA
(a) 0.36 cm3
4 3
r = 4 r 2
3
r = 3 d = 2r = 2 × 3 = 6 cm
Sol. (b)
N
Q.3. A medicine capsule is in the shape of a
cylinder of diameter 0.5 cm with two
hemispheres stuck to each of its ends. The length
of entire capsule is 2 cm. The capacity of the
capsule is :
Q.5. The volume of a sphere (in cu.cm) is
equal to its surface area (in sq. cm). The
diameter of the sphere (in cm) is : [2011 (T-II)]
(a) 3
(b) 6
(c) 2
(d) 4
AS
Sol. (a)
S
Q.7. The ratio between the radius of the base
and the height of the cylinder is 2 : 3. If its
volume is 1617 cm3, the total surface area of the
cylinder is :
[2011 (T-II)]
(a) 208 cm2
(b) 77 cm2
(c) 707 cm2
(d) 770 cm2
Sol. (d) Let the radius and height of the cylinder
ER
O
TH
Height of the cylindrical part
= (2 – 0.5) cm = 1.5 cm
Radius of each hemispherical part
= Radius of the cylindrical part
= 0.25 cm.
Capacity of the capsule
4
4
= r3 + r2h = r2 r h
3
3
PR
AK
4 3
πr1
r
8
2
3
=
⇒ 1 =
4 3 27
r2 3
πr
3 2
Ratio between surface areas = 4 : 9
22
4
× (0.25)2 0.25 1.5 cm3
3
7
=
22
5.5
× (0.25)2 cm3 = 0.36 cm3
7
3
YA
L
BR
=
be 2x and 3x respectively.
Then, volume of the cylinder = r2h
22
1617 =
× (2x)2 × 3x
7
1617 7
343
=
22 4 3
8
x = 3.5 cm.
Total surface area of the cylinder
= 2r (h + r)
Q.4. A solid piece of iron in the form of a
cuboid of dimensions 49 cm × 33 cm × 24 cm is
moulded to form a solid sphere. The radius of the
sphere is :
[2011 (T-II)]
(a) 25 cm
(b) 21 cm
(c) 19 cm
(d) 23 cm
Sol. (b) Volume of sphere = Volume of cuboid
G
O
4 3
r = (49 × 33 × 24) cm3 = 38808 cm3
3
r3 =
38808 × 3 × 7
cm 3 = 9261 cm 3
4 × 22
r = 21 cm
x3 =
22
× 7 (10.5 + 7) cm2
7
= 44 × 17.5 cm2 = 770 cm2.
=2×
Q.8. On increasing each of the radius of the
base and the height of a cone by 20%, its volume
will be increased by :
(a) 25%
(b) 40%
(c) 50%
(d) 72.8%
13
Sol. (d) Volume of the original cone =
6r
120r
=
=
5
100
6h
120h
=
=
5
100
cube =
N
PR
O
91 100 3
= 72.8%.
375
Q.9. A sphere and a cube have the same
surface. Show that the ratio of the volume of
BR
Q.11. A solid ball is exactly fitted inside the
cubical box of side a. The volume of the ball is
YA
L
[2011 (T-II)]
Sol. Let the radius of the sphere be r and the edge
4 3
a . Is it true?
3
G
O
of the cube be x. Whole surface area of sphere = 4r2
and whole surface area of cube = 6x2.
r2
x
2
=
6
3
r
=
⇒ =
4π 2π
x
Sol. Diameter of the ball = side of the cube
6x2.
3
2π
4 3
πr
Volume
of
sphere
Now,
= 3 3
Volume of cube
x
3
=
)
= 7 cm.
6: π
=
(
S
ER
TH
91 2
100
r h
1
375
r 2 h
3
sphere to that of the cube is
6: π
2
cm.
3
= Volume of the metal in the spherical shell
32 4
2
= π 53 − 33
π×r ×
3 3
32 2 4
r = (125 − 27)
3
3
3
4
× × 98
r2 =
32 3
49
7
⇒ r = cm
r2 =
4
2
Hence, the diameter of the base of the cylinder
height 10
91 2
r h.
375
Per cent increase in volume
According to question,
π
r cm. Then, volume of the metallic solid cylinder of
=
4r2
6
HA
72 2
1
r h – r2h
3
125
216 125 2
=
r h
375
=
π
=
Q.10. The internal and external radii of a
hollow spherical shell are 3 cm and 5 cm
respectively. If it is melted to form a solid
2
cylinder of height 10 cm, find the diameter of
3
the cylinder.
[2011 (T-II)]
Sol. Let the radius of the base of the cylinder be
2
=
2× 3
Hence, ratio of the volume of sphere to that of
1 6r
6h
New volume = ×
3 5
5
72 2
=
r h.
125
Increase in volume =
4
3
3
2 3
π×
×
=
=
3
2π
2π
2π
AS
New height
=
AK
New radius
1 2
r h
3
Radius of the ball =
a
2
Volume of the ball =
4
a3
a3
×
=
3
8
6
Hence, the statement is false.
Q.12. From a solid cube of side 7 cm, a
conical cavity of height 7 cm and radius 3 cm is
hollowed out. Find the volume of the remaining
solid.
2
4 r
4 r
r
π = π ×
3 x
3 x
x
14
Sol. Volume of the cube = 73 cm3
= 343 cm3
Sol.
1
× 32 × 7 cm3
3
= 66 cm3
Volume of the remaining solid
= (343 – 66) cm3
Volume of the cone =
= 277 cm3.
(R – r) =
88 × 7
=1
28 × 22
HA
AS
AK
S
PR
Capacity of the shape =
TH
⇒
Radius of the hemispherical portion
= 5 cm
= radius of the cone.
Height of the conical portion
= (10 – 5) cm = 5 cm.
ER
be r cm and R cm respectively.
Then, the height of the cylinder = 14 cm.
Inner surface of the cylinder = 2πr × 14 cm2
= 28πr cm2
Outer surface of the cylinder = 2πR × 14 cm2
= 28R cm2
Difference of the two surfaces = (28πR – 28πr)
⇒
88 = 28π (R – r)
N
Q.13. The difference between the outer and
inner curved surface areas of a hollow right
circular cylinder
14
cm long
is
88 cm2. If the volume of metal used in making
cylinder is 176 cm3, find the outer and inner
diameters of the cylinder.
[2010]
Sol. Let the inner and outer radii of the cylinder
1
2750
×
cm3.
6
7
Required volume of the ice cream
O
BR
YA
L
176 × 7
=4
22 × 1 × 14
R+r= 4
O
⇒
(R + r) =
...(ii)
G
r = 1.5 cm
Hence, inner and outer diameters of the cylinder
are 3 cm and 5 cm respectively.
Q.14. An ice cream cone, full of ice cream is
having radius 5 cm and height 10 cm as shown.
Calculate the volume of ice cream provided that
its
2750 2750
=
cm3
67
7
=
Solving (i) and (ii), we have
R = 2.5 cm and
1 2
r (2r + h)
3
1
22
=
×
× 5 × 5 (2 × 5 + 5) cm3
3
7
2750
22 25
=
× 15 cm3 =
cm3.
7
21
=
Space which remains unfilled =
⇒
R–r= 1
...(i)
Volume of the metal used in making the cylinder
= π(R2 – r2) × 14 cm3
.
..
176 = π (R + r) (R – r) × 14
⇒
2 3 1 2
r + r h
3
3
2750 5
cm3 = 327.4 cm3.
7
6
Q.15. A solid toy is in the form of a
hemisphere surmounted by a right-circular cone.
The height of the cone is 4 cm and the diameter
of the base is 8 cm. Determine the volume of the
toy. If a cube circumscribes the toy, then find the
difference of the volumes of cube and the toy.
Also, find the total surface area of the toy.
Sol. Volume of the toy = Volume of the cone +
Volume of the hemisphere
1
part is left unfilled with ice cream.
6
=
15
1 2
2
1
r h + r3 = r2 (h + 2r)
3
3
3
Sol. Capacity of the box
1 22
1408
×
× 4 × 4 (4 + 8) cm3 =
cm3.
3
7
7
= 16 × 8 × 8 cm3 = 1024 cm3
Volume of the 16 glass spheres
4
= 16 × r3
3
4
22
= 16 ×
×
× 2 × 2 × 2 cm3
3
7
11264
=
cm3
21
Volume of water filled in the box
HA
N
=
AS
11264
10240
= 1024
cm3
cm3 =
21
21
A cube circumscribes this toy, hence edge of the
cube = 8 cm.
Volume of the cube = 83 cm3 = 512 cm3
Required difference in the volumes of the toy
and the cube
= 487.61 cm3.
AK
Q.17. A building is in the form of a cylinder
surmounted by a hemispherical valuted dome
PR
and contains 41
1408
= 512
cm3
7
diameter of the dome is equal to its total height
above the floor, find the height of the building.
[2001]
Sol. Let the internal height of the
ER
S
2176
cm3 = 310.86 cm3.
7
Total surface area of the toy
= curved curface area of the cone
+ curved surface area of the hemisphere
=
O
TH
cylindrical part be h and the internal
radius be r.
Then, total height of the building
=h+r
Also, 2r = h + r
h = r.
Now, volume of the building
= Volume of the cylindrical part +
Volume of the hemispherical part
BR
2
2
2
= r h r 2r
2
2
= πr h + r + 2 r
22
4 16 16 2 4 cm2
7
YA
L
=
19
m3 of air. If the internal
21
2
19
= r2h + r3
3
21
88 4
=
7
2
880
= r3 + r3 [ r = h]
3
21
5r 3
880
=
21
3
G
O
22
4 4 2 8 cm2
=
7
2 2 cm2
88 4
=
× 3.41 cm2 = 171.47 cm2.
7
41
880 3 7
=8
21 5 22
r =2
Hence, height of the building = h + r
Q.16. 16 glass spheres each of radius 2 cm
are packed into a cubical box of internal
dimensions 16 cm × 8 cm × 8 cm and then the
box is filled with water. Find the volume of water
filled in the box.
r3 =
= (2 + 2) m = 4 m.
16
Q.18. A godown building is in the form as
shown in the figure. The vertical cross section
parallel to the width side of the building is a
rectangle 7 m × 3 m, mounted by a semicircle of
radius 3.5 m. The inner measurements of the
cuboidal portion of the building are 10 m × 7 m
× 3 m. Find the volume of the godown and the
total interior surface area excluding the floor
1 2
= 2 πr = πr2
2
22
× (3.5) 2 m2 = 38.5 m2
7
Total interior surface area excluding the base floor
= area of the four walls
=
1
(curved surface area of the cylinder)
2
+ 2 (area of the semicircle)
= (102 + 110 + 38.5) m2
+
HA
N
22
(base). π =
.
7
= 250.5 m2.
the bottom and the top of the building is in the form
of half of the cylinder.
YA
L
BR
O
TH
ER
Length of the cuboid = 10 m,
Breadth of the cuboid = 7 m
Height of the cuboid = 3 m
Volume of the cuboid = lbh = 10 × 7 × 3 m3
= 210 m3.
Radius of the cylinder = 3.5 m
Length of the cylinder = 10 m
1 2
Volume of the half of the cylinder =
πr h
2
1 22
= ×
× (3.5)2 × 10 m3
2
7
= 192.5 m3
Volume of the godown = volume of the cuboid
+ volume of the half cylinder
= (210 + 192.5) m3
= 402.5 m3
Interior surface area of the cuboid
= Area of four walls
= 2 (l + b) h
= 2(10 + 7) 3 m2 = 102 m2
Interior curved surface area of half of the cylinder
22
= πrh =
× 3.5 × 10 m2 = 110 m2
7
S
Sol. The godown building consists of cuboid at
PR
AK
AS
Q.19. A tent is in the shape of a cylinder
surmounted by a conical top. If the height and
diameter of the cylindrical part are 2.1 m and 4
m respectively and the slant height of the top is
2.8 m, find the area of canvas used for making
the tent. Find the cost of the canvas of the tent at
the rate of Rs 550 per m2. Also, find the volume
of air enclosed in the tent.
[2008C]
Sol.
G
O
Height of the cone, H
=
2.8 2 22
m
= 7.84 4 m = 1.95 m
Area of canvas required for making the tent
= Curved surface area of the tent
= Curved surface area of the cylindrical part
+ curved surface area of the conical part
= 2rh + rl = r (2h + l )
=
Interior area of two semicircles
17
22
× 2 (2 × 2.1 + 2.8) m2
7
= 36 64 cm
= 10 cm
Total surface area of the remaining solid
= curved surface area of the cylinder
+ area of top + curved surface area of the cone
= 2rh + r2 + rl
= r (2h + r + l)
= 3.14 × 6 (16 + 6 + 10) cm2
= 18.84 × 32 cm2
= 602.88 cm2.
44
× 7 m2 = 44 m2.
7
Cost of canvas = Rs 500 × 44 = Rs 22000.
Volume of the air enclosed in the tent
= Volume of the cylindrical part
+ Volume of the conical part
=
HA
88 8.25 3
m = 34.57 m3.
7
3
AS
=
Q.21. A juice seller serves his
customers using a glass as shown
in the figure. The inner diamater
of the cylindrical glass is 5 cm,
but the bottom of the glass has a
hemispherical portion raised
which reduces the capacity of the
glass. If the height of the glass is
10 cm, find the apparent capacity of the glass
and its actual capacity.
(Use = 3.14)
[2009]
Sol. Radius of the cylindrical glass r = 2.5 cm
ER
Q.20. 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
two places of decimal. Also find the total surface
area of the remaining solid. (Take = 3.14)
[2008, 2011 (T-II)]
AK
1.95
22
× 22 2.1
m3
3
7
PR
=
N
H
1 2
r H = r2 × h
3
3
S
= r2h +
YA
L
BR
O
TH
Sol.
O
Radius of the cylinder
= radius of the cone = 6 cm.
Height of the cylinder
= height of the cone = 8 cm.
Volume of the remaining solid
1
2
= r2h – r2h = r2h
3
3
2
=
× 3.1416 × 36 × 8 cm3
3
= 603.19 cm3
Slant height of the cone, l
Q.22. A cylindrical vessel with internal
diamater 10 cm and height 10.5 cm is full of
water. A solid cone of the diameter
7 cm and height of 6 cm is completely immersed
in water. Find the volume of
(i) water displaced out of the cylindrical
vessel.
(ii) water left in the cylindrical vessel.
G
=
Height of the glass = 10 cm
Apparent capacity of the glass = r2h
= 3.14 × 2.5 × 2.5 × 10 cm3
= 196.25 cm3
Volume of the hemispherical portion
2
2
= r3 =
× 3.14 × 2.5 × 2.5 × 2.5 cm3
3
3
= 32.71 cm3
Actual capacity of the glass
= (196.25 – 32.71) cm3
= 163.54 cm3.
[Take =
r 2 h2
18
22
]
7
[2009]
10 cm, 5 cm and 4 cm. The radius of each of the
conical depressions is 0.5 cm and depth is 2.1
cm. The edge of the cubical depression is 3 cm.
Find the volume of the wood in the entire stand.
Sol. Volume of a cuboid = 10 × 5 × 4 cm3
= 200 cm3.
Volume of the conical depression
1 2
1
22
r h =
×
× (0.5)2 × 2.1 cm3
3
3
7
Volume of 4 conical depressions
HA
N
=
4
22
×
× (0.5)2 × 2.1 cm3
3
7
= 2.2 cm3
Volume of cubical depression
= 33 cm3 = 27 cm3.
Volume of wood in the entire stand
= [200 – (2.2 + 27)] cm3
= 170.8 cm3.
=
AS
Height of the cylinder, h = 10.5 cm
Capacity of the vessel = r2h
22
=
× 5 × 5 × 10.5 cm3 = 825 cm3
7
1
Volume of the cone = r2h
3
1
22
=
×
× 3.5 × 3.5 × 6 cm3 = 77 cm3.
3
7
(i) Water displaced out of the cylinder
= Volume of the cone = 77 cm3
(ii) Water left in the cylindrical vessel
= Capacity of the vessel
– Volume of the cone
= (825 – 77) cm3 = 748 cm3.
AK
Sol. Radius of the cylinder, r = 5 cm
S
PR
Q.23. A pen stand made of wood is in the
shape of a cuboid with four conical depressions
and a cubical depression to hold pens and pins
respectively. The dimensions of the cuboid are
ER
PRACTICE EXERCISE 13.2A
5. The capacity of a cylindrical vessel with a
hemispherical portion raised upward at the bottom
as shown in the figure is :
(a) r2h
O
TH
Choose the correct option (Q 1 – 5) :
1. The surface area of a sphere is 154 cm2. The
volume of the sphere is :
2
1
(a) 179 cm3
(b) 359 cm3
3
2
2 3
1
(c) 1215 cm
(d) 1374 cm3
3
3
2. The ratio of the volumes of two spheres is
8 : 27. The ratio between their surface areas is :
(a) 2 : 3
(b) 4 : 27
(c) 8 : 9
(d) 4 : 9
3. The curved surface area of a cylinder is
264 m2 and its volume is 924 m3. The height of the
cylinder is :
(a) 3 m
(b) 4 m
(c) 6 m
(d) 8 m
4. The radii of the base of a cylinder and a cone
of same height are in the ratio 3 : 4. The ratio between their volumes is :
(a) 9 : 8
(b) 9 : 4
(c) 3 : 1
(d) 27 : 16
G
O
YA
L
BR
r 2
3h 2r
3
r 2
3h 2r
(c)
3
(b)
(d)
πr 3
(3h + 4r )
3
6. Two solid cones A and B are placed in a
cylindrical tube as shown in the figure. The ratio
of their capacities is 2 : 1. Find the heights and
capacities of the cones. Also, find the volume of
the remaining portion of the cylinder.
7. Marbles of diameter 1.4 cm are dropped into
a cylindrical beaker of diameter 7 cm containing
19
some water. Find the number of marbles that should
be dropped into the beaker so that the water level
rises by 5.6 cm.
8. A solid is in the form of a right circular cone
mounted on a hemisphere. The radius of the
hemisphere is 3.5 cm and the height of the cone is
4 cm. The solid is placed in a cylindrical tub, full
of water, in such a way that the whole solid is
submerged in water. If the radius of the cylinder is
5 cm and height 10.5 cm, find the volume of water
left in the cylindrical tub.
9. The largest possible sphere is carved out
from a solid cube of side 7 cm. Find the volume
of the sphere.
10. A cylindrical boiler, 2 m high, is 3.5 m in
diameter. It has a hemispherical lid. Find the
volume of its interior, including the part covered
22
by the lid. π =
7
15. A building is in the form of a cylinder
surmounted by a hemispherical dome as shown in
the figure. The base diameter of the dome is equal
2
of the total height of the building. Find the
3
height of the building, if it contains 67 1 m3 of
27
HA
N
to
[2011 (T-II)]
S
PR
AK
AS
air.
ER
11. An ice cream cone consists of a right
circular cone of height 14 cm and the diameter of
the circular top is 5 cm. It has a hemispherical
scoop of ice cream on the top with the same
diameter as of the circular top of the cone. Find
the volume of ice cream in the cone.
12. A solid toy is in the form of a hemisphere
surmounted by a right circular cone. Height of
the cone is 2 cm and the diameter of the base is
4 cm. If a right circular cylinder circumscribes
the toy, find how much more space it will cover.
[2011 (T-II)]
13. A cylindrical tub of radius 12 cm
contains water to a depth of 20 cm. A spherical
iron ball is dropped into the tub and thus the
level of water is raised by 6.75 cm. What is the
radius of the ball?
14. From a solid cylinder of height 12 cm
and base diameter 10 cm, a conical cavity with
the same height and diameter is carved out. Find
the volume of the remaining solid.
G
O
YA
L
BR
O
TH
16. A heap of rice is in the form of a cone of
diameter 9 m and height 3.5 m. Find the volume of
the rice. How much canvas cloth is required to just
cover the heap?
17. 500 persons are taking a dip into a cuboidal
pond which is 80 m long and 50 m broad. What is
the rise of water level in the pond, if the average
displacement of the water by a person is 0.04 m3.
18. A rocket is in the form of a right circular
cylinder closed at the lower end and surmounted
by a cone with the same radius as that of the
cylinder. The diameter and height of the cylinder
are 6 cm and 12 cm respectively. If the slant height
of the conical portion is 5 cm, find the total surface
area and volume of the rocket. (Take = 3.14)
13.3 CONVERSION OF SOLID FROM ONE SHAPE TO ANOTHER
TEXTBOOK’S EXERCISE 13.3
22
, unless stated otherwise.
7
Q.1. A metallic sphere of radius 4.2 cm is
melted and recast into the shape of a cylinder of
radius 6 cm. Find the height of the cylinder.
Sol. Radius of sphere = 4.2 cm
Take π =
Volume of sphere =
20
4 3 4
r = (4.2)3 cm3
3
3
Volume of cylinder = R2H
= (6)2H cm3
As per condition,
Volume of the sphere = Volume of the cylinder
4
(4.2)3 = (6)2H
3
H =
4 4.2
2
7
Volume of well = r2h = 20 m3
2
= 245 m3
Length of platform (L) = 22 m
Width of platform (B) = 14 m
Let the height of the platform be H.
Then, volume of the platform = LBH
= 22 × 14 × H m3 = 308H m3
As per condition,
Volume of platform = Volume of well
308 H = 245
36
H = 2.74
Hence, the height of the cylinder is 2.74 cm.
N
2
AS
HA
Q.2. Metallic spheres of radii 6 cm, 8 cm
and 10 cm, respectively, are melted to form a
single solid sphere. Find the radius of the
resulting sphere.
4 3
r
3
TH
ER
...(ii)
Q.4. A well of diameter 3 m dug 14 m deep.
The earth taken out of it has been spread evenly
all around it in the shape of a circular ring of
width 4 m to form an embankment. Find the
height of the embankment.
[2011 (T-II)]
Sol. For well :
S
4
(8)3 cm3
3
Volume of sphere of radius 10 cm
4
(10)3 cm3
...(iii)
3
Let the radius of the resulting sphere be R cm.
Then volume of the resulting sphere
Diameter = 3 m
...(iv)
2
63
3
= (14) m3 =
m3
2
2
Width of the embankment = 4 m
O
4 3
4
4
4
R = (6)3 + (8)3 + (10)3
3
3
3
3
3
3
3
3
R = (6) + (8) + (10)
R3 = 1728
Let the height of the embankment be H m.
Radius of the well with embankment, R
G
3
m
2
Depth of well (h) = 14 m
Volume of earth taken out = r2h
Radius of well (r) =
YA
L
As per condition,
4 3 3
R cm
3
BR
O
=
=
245
245 22
H=
H= 2.5
308
308 7
Hence, the height of the platform is 2.5 m.
...(i)
=
H =
PR
Volume of sphere of radius 6 cm
4
= (6)3 cm3
3
Volume of sphere of radius 8 cm
7
m
2
Depth (h) = 20 m
3
Sol. We know that, volume of the sphere =
Radius (r) =
AK
11
3
= + 4 cm =
m
2
2
R = 3 1728
R = 12
Hence, the radius of the resulting sphere is 12 cm.
Volume of earth= Volume of the embankment
R2H – r2H = (R2 – r2) H
Q.3. A 20 m deep well with diameter 7 m is
dug and the earth from digging is evenly spread
out to form a platform 22 m by 14 m. Find the
height of the platform.
[2011 (T-II)]
Sol. Diameter of well = 7 m
2
2
121 9
11
3
− H m3
= π − H m3 = π
4
4
2
2
= 28H m3
21
As per condition,
Sol. Diameter of silver coin = 1.75 cm
63
9
63
28H =
H =
H=
2 28
8
2
OR, H = 1.125
Radius of silver coin (r) =
7
1.75
cm = cm
8
2
2
Thickness of silver coin (h) = 2 mm =
cm
10
1
=
cm
5
Hence, the height of the embankment is 1.125 m.
Q.5. A container shaped like a right dircular
cylinder having diameter 12 cm and height 15
cm is full of ice cream. The ice cream is filled
into cones of height 12 cm and diameter 6 cm,
having a hemispherical shape on the top. Find
the number of such cones which can be filled
with ice cream.
[2008, 2011(T-II)]
Sol. Diameter of cylinder = 12 cm
Diameter of cone
Radius of cone (R) =
N
HA
AS
AK
ER
6
cm = 3 cm
2
Height of cone (H) = 12 cm
1
Volume of cone = R2H
3
1
(3)2 (12) cm3 = cm3
3
Radius of hemisphere = 3 cm
BR
2
(3)3 cm3
3
= 18 cm3
Volume of ice cream in one cone-volume of
cone + volume of hemisphere
= (36 + 18) cm3 = 54 cm3
Let n cones be filled with ice cream.
Then, volume of n cones = n (54) cm3
As per condition
n (54) = 540
G
O
YA
L
Volume of hemisphere =
540π
n = 10
54π
Hence, 10 cones can be filled with ice cream.
n =
1925
385
cm3 =
cm3
10
2
As per condition,
=
n×
O
=
49
cm3
320
Length of cuboid (L)= 5.5 cm
Breadth of cuboid (B)= 10 cm
Height of cuboid (H)= 3.5 cm
Volume of the cuboid
= Length × Breadth × Height
= (5.5 × 10 × 3.5) cm3 = 192.5 cm3
TH
Then, volume of coins = n ×
PR
Height of cylinder (h)
Volume of cylinder
12
=
cm = 6 cm
2
= 15 cm
= r2h = (6)2 (15) cm3
= 540 cm3
= 6 cm
S
Radius of cylinder (r)
2
7 1
Volume of a silver coin = r2h = cm3
8 5
49
cm3
=
320
Let n coins be melted.
49
385
=
320
2
n =
385 320
2 49
385 320 7
2
49 22
n = 400
Hence, 400 coins must be melted to form the
required cuboid.
n =
Q.7. A cylindrical bucket, 32 cm high and
with radius of base 18 cm, is filled with sand.
This bucket is emptied on the ground and a
conical heap of sand is formed. If the height of
the conical heap is 24 cm, find the radius and
slant height of the heap.
Sol. For cylindrical bucket :
Radius of cylindrical bucket (r)= 18 cm
Height of cylindrical bucket (h) = 32 cm
Volume of cylindrical bucket = r2h
= (18)2 (32) cm3 = 10368 cm3
Q.6. How many silver coins, 1.75 cm in
diameter and of thickness 2 mm, must be melted
to form a cuboid of dimension 5.5 cm × 10 cm
× 3.5 cm?
22
The area it will irrigate in 30 minutes with 8 cm
of standing water
Height of cone (H) = 24 cm
Let the radius of conical heap be R cm.
1
Then,volume of conical heap = R2H
3
1 2
= R (24) cm3= 8R2 cm3
3
As per condition,
8R2 = 10368
8R2 = 10368
=
= 562500 m2 =
1
hectare]
10000
= 56.25 hectare.
N
HA
R2 =
Q.9. A farmer connects a pipe of internal
diameter 20 cm from a canal into a cylindrical
tank in her field, which is 10 m in diameter and
2 m deep. If water flows through the pipe at the
rate of 3 km/h, in how much time will the tank be
filled?
[2011(T-II)]
Sol. Diameter of cylindrical tank = 10 m
AS
36 2 24 2
R 2 + H2
cm =
PR
Slant height (l ) =
AK
R = 1296
R = 36
Hence, the radius of the conical heap is 36 cm.
=
562500
hectare
10000
[1 hectare = 10000 m2, 1 m2 =
10368
8
R2 = 1296
45000
4500000 2
m2 =
m
8
8
100
1296 576 cm
ER
S
= 1872 cm = 12 12 13 cm = 12 13 cm
Hence, the slant height of the conical heap is
12 13 cm.
10
m=5m
2
Depth of cylindrical tank (h) = 2 m
Radius of cylindrical tank (r) =
TH
Q.8. Water in a canal 6 m wide and 1.5 m
deep is flowing with a speed of 10 km/h. How
much area will it irrigate in 30 minutes, if 8 cm
of standing water is needed?
[2011 (T-II)]
Sol. Width of canal = 6 m
Volume of cylindrical tank
= r2h = (5)2 (2) cm3 = 50 m3
O
Rate of flow of water = 3 km/h = 3000 m/h
BR
=
Internal diameter of pipe = 20 cm
20
cm = 10 cm
2
10
=
m = 0.1 m
100
Volume of water that flow per minute in pipe
Internal radius of pipe (R) =
O
YA
L
15
3
Depth of canal = 1.5 m =
m= m
10
2
Speed of flowing water
= 10 km/h = 10 × 1000 m/h
= 10000 m/h
10000
=
m per min
60
500
=
m/min
3
500 30
=
m per 30 minutes
3
= 5000 m per 30 minutes
Volume of water that flows in 30 minutes
3000
m/min = 50 m/min
60
G
= r2 (50) m3 = (0.1)2 (50) m3
= 0.5 m3 =
Required time =
5 3 3
m =
m
10
2
50
= 100 minutes
2
Hence, the tank will filled in 100 minutes.
3
=6×
× 5000 m3 = 45000 m3
2
23
OTHER IMPORTANT QUESTIONS
Q.4. A spherical iron ball is melted and
recast into 8 identical balls. The radius of each
1
th the radius of the original ball. Is
8
N
new ball is
it true?
capacity of cubical brick
=
capacity of an ice cream cone
22 22 3
223
=
= 363
1 22 2
4
2 7
3 7
4 3
R
3
R3
8
AK
r3 =
R
2
Hence, radius of each smaller ball is half of the
original ball.
Thus, the statement is false.
r =
PR
Q.5. A solid metallic sphere of radius
10.5 cm is melted and recast into a number of
smaller cones, each of radius 3.5 cm and height
3 cm. Find the number of cones so formed.
Sol. Volume of the solid metallic sphere
TH
ER
Q.2. A metallic spherical shell of internal
and external diameters 4 cm and 8 cm
respectively is melted and recast into the form of
a cone of base diameter 8 cm. The height of the
cone is :
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm
Sol. (b) Internal and external radii of the ball are
4 3
r =
3
AS
Sol. We have, 8 ×
S
=
16 3
=1 r=1
12 4
Hence, diameter of each sphere = 2 cm.
r3 =
HA
Q.1. A cubical ice cream brick of edge 22 cm
is to be distributed among some children by
filling ice cream cones of radius 2 cm and height
7 cm up to its brim. How many children will get
the ice cream cones?
(a) 163
(b) 263
(c) 363
(d) 463
Sol. (c) Required number of children
O
2 cm and 4 cm respectively.
Volume of metal used in the spherical ball
4
224
(43 – 23) cm3 =
cm3
3
3
Let the height of the cone be h. Then,
BR
=
1
× (3.5)2 × 3 cm3
3
Number of cones so formed
Volume of the sphere
=
Volume of a cone
4
3
10.5
3
=
= 126.
1
2
3.5 3
3
1
224
× 42 × h =
3
3
O
224
= 14 cm.
16
G
h=
Q.3. Twelve solid spheres of the same size
are made by melting a solid metallic cylinder of
base diameter 2 cm and height 16 cm. The
diameter of each sphere is :
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm
Sol. (a) Let the radius of each sphere be r.
Then, 12 ×
4 3
r = × 12 × 16
3
4
× (10.5)3 cm3
3
Volume of a cone =
YA
L
=
Q.6. The circumference of the circular end of
a hemispherical bowl is 132 cm. Find the
capacity of the bowl.
[2011 (T-II)]
Sol. Let r be radius of the edge of a
hemispherical bowl.
Its circumference = 2r
132 × 7
2r = 132 r =
= 21 cm
2 × 22
24
Now, capacity of the bowl
2 3 2 22
πr = ×
× 21 × 21 × 21 = 44 × 441
3
3 7
= 19404 cm3
=
Q.7. Three cubes of a metal whose edges are
in the ratio 3 : 4 : 5 are melted and converted
into a single cube whose diagonal is
Q.9. A spherical ball of lead 3 cm in
diameter is melted and recast into three spherical
balls. The diameters of two of these balls are
1 cm and 1.5 cm. Find the diameter of the third
ball.
Volume of the bigger ball
=
Sol. Let the edges of the three cubes be 3x, 4x
1
1.5
cm and
cm
2
2
= 0.5 cm and 0.75 cm.
AS
Radii of two smaller balls are
.
. . Total volume of the two smaller balls
=
=
Q.8. If the diameter of the cross-section of a
wire is decreased by 5%, how much per cent will
the length be increased so that the volume
remains the same?
Sol. Let the diameter and length of the cross-
...(ii)
S
ER
TH
4
.
. . Volume of the third ball = r3 cm3 ...(iii)
3
From (i), (ii) and (iii), we have
O
4
4
4
(1.5)3 =
[(0.5)3 + (0.75)3] + r3
3
3
3
(1.5)3 = (0.5)3 + (0.75)3 + r3
BR
YA
L
O
G
400 h
39 h
.
. . Increase in length =
–h=
361
361
39 h
100
39 100
% increase = 361
h
361
4
[(0.5)3 + (0.75)3] cm3
3
Let the radius of the third ball be r cm.
section of the wire by 2r and h respectively.
Then, volume of the wire = r2h
2r × 5
New diameter = 2r –
100
r
19 r
= 2r –
10
10
19r
New radius =
20
To keep the volume same, let the new length be H.
2
4
3 4
3
3
3 π (0.5) + 3 π (0.75) cm
PR
6 x 3 = 12 3 x = 2
Edge of the three cubes are 3 × 2 cm, 4 × 2 cm
and 5 × 2 cm or 6 cm, 8 cm and 10 cm.
AK
Diagonal of the new cube = a 3 = 6 x 3
19r
361 2
2
..
r × H = r2h
× H = r h
20
400
400 h
H=
361
...(i)
HA
and 5x. Then,
Total volume of these three cubes
= 27x3 + 64x3 + 125x3 = 216x3.
Let the edge of the new cube formed be a. Then,
a3 = 216x3 a = 6x
4
(1.5)3 cm3
3
N
12 3 cm. Find the edges of the three cubes.
.
3
cm = 1.5 cm.
2
Sol. Radius of the bigger ball =
r3 = (1.5)3 – (0.5)3 – (0.75)3
r3 = 3.375 – 0.125 – 0.421875
r3 = 2.828125
.
. . Diameter of the third ball
r = 1.41
= 2 × 1.41 cm = 2.82 cm.
Q.10. Find the number of metallic discs with
1.5 cm base diameter and of height 0.2 cm, to be
melted to form a right circular cylinder of height
10 cm and diamater 4.5 cm.
2
cm3
4.5
Sol. Volume of the cylinder = × × 10
2
Volume of 1 circular disc
2
1.5
=×
× 0.2 cm3
2
Let n circular discs are needed.
10.80%.
25
Then,
Clearly, the water column forms a cylinder of
radius.
2
1.5
4.5
n××
× 0.2 = ×
× 10
2
2
20
1
cm = 10 cm =
m and
2
10
h = height (length) = 3000x metres
Volume of the water that flows in the cistern
r=
4.5 4.5 10
= 450.
0.2 1.5 1.5
Hence, 450 circular discs are needed.
n=
14
7
cm =
m
2
100
Length = h = 5000x metres
Volume of the water flowing through the
cylindrical pipe in x hours
22
7
7
= πr 2 h =
×
×
× 5000 x m 3 = 77 x m 3
7 100 100
Also,
Volume of the water that falls into the tank in
N
HA
Since the cistern is filled in x hours.
Volume of the water that flows in the cistern in
x hours = Volume of the cistern.
AS
will rise by 7 cm in x hours. Since the water is flowing
at the rate of 5 km/hr. Therefore, length of the water
column in x hours = 5x km = 5000x metres.
Clearly, the water column forms a cylinder whose
radius
22 1
1
in x hours = πr 2 h =
×
×
× 3000 x m 3
7 10 10
Also,
22
× 5 × 5 × 2 m3
Volume of the cistern =
7
[ r = 5 m, h = 2m]
22 1
1
22
×
×
× 3000 x =
× 5×5×2
7 10 10
7
PR
Q.11. Water is flowing at the rate of 5 km/hr
through a pipe of diameter 14 cm into a
rectangular tank, which is 50 m long and 44 m
wide. Determine the time in which the level of
water in the tank will rise by 7 cm. [2011 (T-II)]
Sol. Suppose the level of the water in the tank
AK
2
r=
= 1 hour 40 minutes
O
TH
ER
S
5
5 × 5 × 2 × 10 × 10
x =
hrs = 3 hrs
3000
Let the required height of the water level be h.
Then, 11 × 6 × 5 = × (3.5)2 × h
11 6 5 7
= 8.6 m.
22 3.5 3.5
Hence, height of water level in the tank = 8.6 m.
O
YA
L
BR
7
m3 = 154 m3
100
But, volume of the water flowing through the
cylindrical pipe in x hours = Volume of the water that
falls in the tank in x hours
154
77x = 154 x =
=2
77
x hours = 50 × 44 ×
Q.13. A rectangular water tank of base 11 m
× 6 m contains water up to a height of 5 m. If the
water in the tank is transferred to a cylindrical
tank of radius 3.5 m, find the height of the water
level in the tank.
Sol. Volume of water in the tank = 11 × 6 × 5 m3.
G
Hence, the level of the water in the tank will
rise by 7 cm in 2 hours.
Q.12. Water is flowing at the rate of 3 km/hr
through a circular pipe of 20 cm internal
diameter into a cylindrical cistern of diameter
10 m and depth 2 m. In how much time will the
22
cistern be filled? Take π = [2011 (T-II)]
7
Sol. Suppose the cistern is filled in x hours. Since
h=
Q.14. How many coins 1.75 cm in diameter
and 2 mm thick must be melted to form a cuboid
of dimensions 11 cm × 10 cm × 7 cm?
22
[2011 (T-II)]
Take π =
7
Sol. Here each coin is a cylinder with
r=
water is flowing at the rate of 3 km/hr. Therefore,
length of the water column in x hours = 3x km
= 3000x metres.
26
1.75
cm = 8.75 mm and h = 2 mm
2
Volume of each coin
= r2h = (8.75)2 × 2 cu mm
Volume of the cuboid = 11 × 10 × 7 cu cm
= 770 cu cm = 770000 cu mm
Required number of coins
770000
=
= 1600
π × 8.75 × 8.75 × 2
n=
YA
L
BR
Internal radius of the pipe = 30 cm = 0.3 m.
Thickness of pipe = 5 cm.
External radius of the pipe
= 35 cm = 0.35 m.
Let the length of the pipe be h m.
Then, 4.4 × 2.6 × 1 = × (0.352 – 0.32) × h
G
O
4.4 × 2.6 × 1 × 7
m = 112 m.
22 × 0.65 × 0.05
Hence, length of the pipe = 112 m.
h=
N
AS
HA
Q.18. A solid right circular cone of diameter
14 cm and height 8 cm is melted to form a hollow
sphere. If the external diameter of the sphere is
10 cm, find the internal diameter of the sphere.
[2007]
AK
1
× 72 × 8 cm3
3
Let the internal radius of the sphere be r.
Sol. Volume of the cone =
Then,
1
4
× 72 × 8 =
× × (53 – r3)
3
3
49 8
= 98
4
r3 = 125 – 98 = 27
r =3
Hence, internal diameter of the sphere = 6 cm.
53 – r3 =
Q.19. An iron spherical ball has been melted
and recast into smaller balls of equal size. If the
radius of each of the smaller balls is
1
of the
4
radius of the original ball, how many such balls
are made? Compare the surface area of all the
smaller balls combined together with that of the
original ball?
[2007]
Sol. Let the radius of the bigger ball be r.
Q.17. A hemispherical bowl of internal
radius 9 cm is full of a liquid. The liquid is to be
filled into cylindrical shaped bottles each of
radius 1.5 cm and height 4 cm. How many such
bottles are needed to empty the bowl? [2005]
Sol. Capacity of the hemispherical bowl
=
1458
= 54
1.5 1.5 4 3
Hence, 54 cylindrical bottles are required.
ER
O
TH
Q.16. A solid iron cuboidal block of
dimensions 4.4 m × 2.6 m × 1 m is recast into a
hollow cylindrical pipe of internal radius 30 cm
and thickness 5 cm. Find the length of the pipe.
Sol. Volume of the cuboid = 4.4 × 2.6 × 1 m3.
1458
3
PR
0.5
× 1000 cm3.
2
Capacity of the conical vessel
1
= × × (20)2 × 24 cm3
3
Time taken by the pipe to fill the conical vessel
2
20 24
3
=
min
2
0.5
1000
2
= 51.2 min = 51 min 12 sec.
pipe in 1 minute = ×
n × × (1.5)2 × 4 =
S
Q.15. Water flows at the rate of 10 m/min
through a cylindrical pipe 5 mm in diameter.
How long would it take to fill a conical vessel
whose diamater at the base is 40 cm and depth
24 cm?
Sol. Volume of water which flows out through the
Capacity of 1 cylindrical bottle
= × (1.5)2 × 4 cm3
Let n cylindrical bottles are required.
Then, the radius of each smaller ball =
r
.
4
Let n smaller balls are formed.
3
4 r
3 4
n = 4 × 4 × 4 = 64
Hence, 64 smaller balls are formed.
Also, surface area of the bigger ball = 4r2.
Then,
2
1458
× × 93 cm3 =
cm3
3
3
27
4 3
r = n ×
3
2
r
= 64 × 4 = 16r2
4
Surface area of the 64 smaller balls together
Surface area of the original ball
16r 2
=
= 4 : 1.
4r 2
4
22
×
× 7 × 7 × 7 cm3
3
7
Let R be the radius and h be the height of the
cone.
=
Then, volume of sphere =
1
22
×
× r2 × 28 cm3
3
7
But, according to the question,
=
N
Q.20. The rain water collected on the roof of
a building of dimensions 22 m × 20 m, is drained
into a cylindrical vessel having base diameter
2 m and height 3.5 m. If the vessel is full up to
the brim, find the height of rain water on the
HA
4
22
1
22
×
×7×7×7=
×
× r2 × 28
3
7
3
7
r2 =
22
roof. π = 7
[2010, 2011 (T-II)]
4×7×7×7
=7×7
28
AK
r=7
Hence, base diameter of the cone
= 2 × 7 cm = 14 cm.
S
PR
Q.22. A hemispherical bowl of internal
diameter 36 cm is full of liquid. This liquid is to
be filled in cylindrical bottles of radius 3 cm and
height 6 cm. How many such bottles are required
to empty the bowl?
[2011 (T-II)]
Sol. Diameter of hemispherical bowl = 36 cm
TH
ER
Sol. Let the height of rain water on the roof be
h m. Then,
Volume of water collected on the roof of the
building = (20 × 20 × h) m3.
Radius of the cylindrical vessel = 1 m
Height of the cylindrical vessel = 3.5 m.
Capacity of the cylindrical vessel
= × 12 × 3.5 m3.
But, according to the question
20 × 20 × h = × 12 × 3.5
22
1
1
11
h=
× 3.5 ×
×
=
m
7
20
20
400
11
=
× 100 cm = 2.75 cm.
400
Radius of hemispherical bowl = 18 cm
Volume of hemispherical bowl
O
2 3 2 22
πr = ×
× 18 × 18 × 18 cu cm
3
3 7
Height of bottle = 6 cm, radius of bottle = 3 cm
BR
=
Q.21. The surface area of a solid metallic
sphere is 616 cm2. It is melted and recast into a
cone of height 28 cm. Find the diameter of the
base of the cone so formed. [2010, 2011 (T-II)]
Sol. Let r be the radius of the sphere.
YA
L
Volume of bottle = r2h =
22
× 3 × 3 × 6 cu cm
7
No. of bottles required
O
=
G
Then, 4r2 = 616
616 × 7
r2 =
= 49 r = 7
4 × 22
Volume of the sphere =
1 2
r h
3
AS
And, surface area of the 64 smaller balls
Volume of liquid in the bottle
Volume of 1 bottle
2 22
×
× 18 × 18 × 18
= 3 7
= 72 bottles
22
×3×3×6
7
4 3
r
3
PRACTICE EXERCISE 13.3A
Choose the correct option (Q 1 – 5) :
1. If a marble of radius 2.1 cm is put into a
cylindrical cup full of water of radius
5 cm and height 6 cm, then how much water
flows out of the cylindrical cup?
28
(a) 38.8 cm3
(b) 55.4 cm3
(c) 19.4 cm3
(d) 471.4 cm3
BR
O
TH
N
HA
AS
AK
ER
5. A cubical block having edge 44 cm is melted
and then recast into spherical bullets, each of radius 2 cm. Number bullets thus formed is :
(a) 2000 (b) 2541 (c) 2451 (d) 2145
6. Three metallic solid cubes whose edges are
3 cm, 4 cm and 5 cm are melted and formed into a
single cube. Find the edge of the cube so formed.
7. How many spherical bullets each having
diameter 3 cm can be made from a cuboidal solid
lead of dimensions 9 cm × 11 cm × 12 cm?
8. The radii of the internal and external surfaces
of a hollow spherical shell are 3 cm and 5 cm
respectively. If it is melted and recast into a solid
PR
3. During conversion of a solid from one
shape to another, the volume of the new shape
will :
(a) increase
(b) decrease
(c) remains unaltered (d) be doubled
4. A solid sphere of radius 3 cm is melted and
then cast into small spherical balls each of diameter 0.6 cm. The number of small balls thus obtained is :
(a) 50
(b) 100
(c) 500
(d) 1000
10. A piece of metal in the form of a cone of
radius 3 cm and height 7 cm is melted and recast
into a cube. Find the side of the cube.
11. The diameter of a copper sphere is 6 cm.
The sphere is melted and drawn into a long wire
of uniform cross section. Find the radius of the
wire if its length is 36 cm.
12. A vessel in the form of a hemispherical
bowl is full of water. The contents are emptied
into a cylinder. The internal radii of the bowl and
cylinder are 6 cm and 4 cm respectively. Find the
height of water level in the cylinder.
13. How many lead balls each of radius
1 cm, can be made from a sphere of radius
8 cm?
14. A metallic toy in the form of a cone of
radius 11 cm and height 62 cm mounted on a
hemisphere of the same radius is melted and
recast into a solid cube. Find the surface area of
the cube thus formed.
15. A canal is 3 m wide and 1.2 m deep. The
water in the canal is flowing with a speed of
20 km/h. How much area will it irrigate in
20 minutes if 8 cm of standing water is required?
16. How many spherical lead shots each of
diameter 4.2 cm can be obtained from a solid
rectangular lead piece with dimensions 66 cm,
42 cm and 21 cm?
17. A solid metallic hemisphere of radius 8 cm
is melted and recast into a right circular cone of
base radius 6 cm. Determine the height of the cone.
18. Water flows through a cylindrical pipe,
whose inner radius is 1 cm, at the rate of
80 cm/sec in an empty cylindrical tank, the radius
of whose base is 40 cm. What is the rise of water
level in the tank in half an hour?
[2007, 2011 (T-II)]
S
2. A solid piece of aluminium in the form of
a cuboid of dimensions 49 cm × 33 cm
× 24 cm is moulded to form a solid sphere. The
radius of the sphere is :
(a) 21 cm (b) 23 cm (c) 25 cm (d) 19 cm
8
cm, find the diameter of the
3
YA
L
cylinder of height
G
O
cylinder.
[2011 (T-II)]
9. A right circular cone is 3.6 cm high and
has base radius 1.6 cm. It is melted and recast
into a right circular cone with radius of its base
as 1.2 cm, find its height.
13.4 FRUSTUM OF A CONE
1. When a cone is cut by a plane parallel to
the base of the cone, then the portion between
the plane and the base is called the frustum of
the cone.
(a) Slant height of the frustum,
l=
h2 R r
2
(b) Volume of
frustum of the cone
=
the
h
[R2 + r2 + Rr]
3
(c) Lateral surface area
of the frustum of the cone
= l (R + r)
29
= [R2 + r2 + l (R + r)]
(d) Total surface area of the frustum of the
cone = (area of the base) + (area of the top)
+ (lateral surface area)
= [R2 + r2 + l (R + r)].
TEXTBOOK’S EXERCISE 13.4
Q.1. A drinking glass is in the shape of a
frustum of a cone of height 14 cm. The diameters
of its two circular ends are 4 cm and 2 cm. Find
the capacity of the glass.
N
2
cm = 1 cm
2
HA
Sol. Let radius of lower end (r1) =
Sol.
4
cm = 2 cm
2
Height of the glass (h) = 14 cm
Capacity of the glass = Volume of the frustum
AS
Radius of upper end (r2) =
AK
Radius of upper end, r1 = 10 cm
Radius of lower end, r2 = 4 cm
1
= h (r12 + r22 + r1r2)
3
PR
1
22
×
× 14 [(2)2 + (1)2 + (2) (1] cm3
3
7
1
22
308 3
=
×
× 14 × 7 cm3 =
cm
3
7
3
2
= 102 cm2
3
2
Hence, capacity of the glass is 102 cm2.
3
Slant height of frustum, l = 15 cm
Surface area of the frustum
= (r1 + r2) l + r2 2
TH
ER
S
=
O
= 600 cm2 +
BR
YA
L
O
G
Q.3. A fez, the cap used by the Turks, is
shaped like the frustum of a cone (see figure). If
its radius on the open side is 10 cm, radius at the
upper base is 4 cm and its slant height is 15 cm,
find the area of material used for making it.
22
(10 + 4) (15) cm2 + (4)2 cm2
7
= 660 cm2 + 16 cm2 = 600 cm2 + 16 ×
Q.2. The slant height of a frustum of a cone
is 4 cm and the perimeters (circumference) of its
circular ends are 18 cm and 6 cm. Find the
curved surface area of the frustum.
Sol. Let r1 be the radius of upper end and r2 be the
radius of lower end.
Then, slant height l = 4 cm
Circumference of upper end, 2r1 = 18 cm
r1 = 9 cm
Circumference of lower end, 2r2 = 6 cm
r2 = 3 cm
Curved surface area of the frustum
= (r1 + r2) l = (r1 + r2) l
= (9 + 3) 4 = 48 cm2.
=
=
22
cm2
7
352 2
4620 352
cm =
cm2
7
7
4972
2
cm2 = 710 cm2
7
7
Hence, the area of material used for making it is
710
2
cm2.
7
Q.4. A container, opened from the top is
made up of a metal sheet, is in the form of a
frustum of a cone of height 16 cm with radii of
its lower and upper ends as 8 cm and 20 cm,
respectively. Find the cost of the milk which can
completely fill the container, at the rate of Rs 20
per litre. Also, find the cost of metal sheet used
to make the container, if it costs Rs 8 per
100 cm2. (Take = 3.14).
Sol. Height of container (h) = 16 cm
30
1
(3.14) (16) (624) cm3
3
= (3.14) (16) (208) cm3 = 10449.92 cm3
= 10.44992 litres
Cost of the milk = Rs 10.44992 × 20
= Rs 208.9984 Rs 209
Total surface area
=
r2
10
r2 =
cm
10
3
r1
20
In EOF, tan 30° =
r1
20
r1 =
cm
20
3
3
Height of the frustum = h = 10 cm
1
=
AS
20 2 10 2 20 10
cm3
3 3 3 3
+ (3.14) (8)2] cm2
PR
1
22
400 100 200
×
× 10 ×
cm3
3
3
3
7
3
1
22
700
22000
×
× 10 ×
cm3 =
cm3
3
7
3
9
S
=
BR
O
TH
= [(3.14) (28) 256 144 + (3.14) (64)] cm2
= [(3.14) (28) (20) + 200.96] cm2 = [1758.4 + 200.96] cm2
= 1959.36 cm2
Area of the metal sheet used = 1959.36 cm2
8
Cost of metal sheet = Rs 1959.36 ×
100
= Rs 156.7488 = Rs 156.75
O
YA
L
Q.5. A metallic right circular cone 20 cm
high and whose vertical angle is 60° is cut into
two parts at the middle of its height by a plane
parallel to its base. If the frustum so obtained be
drawn into a wire of diameter
1
h (r12 + r22 + r1r2)
3
AK
1
22
= ×
× 10 ×
7
3
N
3
=
=
ER
16 2 20 82
1
Volume of the frustum =
= (r1 + r2) h 2 (r1 r2 )2 + r2 2
= [(3.14) (20 + 8)
r2
10
In ODB, tan 30° =
HA
Radius of upper end, r1 = 20 cm
Radius of lower end, r2 = 8 cm
Volume of the container
= Capacity of the container
1
= h (r12 + r22 + r1r2)
3
1
= (3.14) (16) {(20)2 + (8)2 + (20) (8)} cm3
3
1
=
(3.14) (16) (400 + 64 + 160) cm3
3
1
cm, find the
16
Diameter of the wire =
1
cm
16
Radius of the wire r
1
1
1
×
cm =
cm
32
2 16
Let length of the wire be x cm.
Then, volume of the wire = r2 x
=
2
22 1
11x
× x =
cm3
7
3584
32
According to the question,
=
G
length of the wire.
Sol.
11x
22000
=
3584
9
x =
22000 3584
2000 3584
x=
11 9
9
7168000
9
x = 796444.44 cm
x = 7964.4 m
Hence, the length of the wire is 7964.4 m.
x=
31
OTHER IMPORTANT QUESTIONS
the cone.
TH
O
=
16 2 + (20 − 8)2 cm
=
256 + 144 cm
=
400 cm = 20 cm
G
Q.5. The radii of the ends of the frustum of
a cone of height h are R and r. The volume of the
frustum of the cone is :
(a)
h
(R2 + r2 + Rr)
3
(b)
h
(R2 + r2 – Rr)
3
(d)
(R2 + r2 – Rr)
3
(R2 + r2 + Rrhh)
3
h
Sol. (a) Required volume =
(R2 + r2 + Rr)
3
(c)
N
AS
Q.7. The diameters of two circular ends of a
bucket are 44 cm and 24 cm. If the height of the
bucket is 35 cm, the capacity of the bucket is :
(a) 32.7 litres
(b) 33.7 litres
(c) 34.7 litres
(d) 31.7 litres
Sol. (a) Capacity of the bucket
=
h
[R2 + r2 + Rr]
3
22
35
×
[222 + 122 + 22 × 12] cm3
7
3
= 32706.6 cm3 = 32.7 litres.
=
Q.8. The total surface area of a frustum of a
cone is [R2 + r2 + l (R + r)],
O
BR
h 2 + (R − r )2
YA
L
Sol. (c) Slant height =
22
× 45 (28 + 7) cm2 = 4950 cm2.
7
HA
=
ER
Q.3. In a right circular cone, the cross-section made by a plane parallel to the base is a:
(a) circle
(b) frustum of a cone
(c) sphere
(d) hemisphere
Sol. (a) The desired cross-section is a circle.
Q.4. A frustum of a right circular cone of
height 16 cm with radii of its circular ends as 8
cm and 20 cm has its slant height equal to :
[2011 (T-II)]
(a) 18 cm
(b) 16 cm
(c) 20 cm
(d) 24 cm
= l (R + r)
AK
Q.2. A cone is cut through a plane parallel to
its base and then the cone that is formed on one
side of that plane is removed. The new part that
is left over on the other side of the plane is
called :
(a) a frustum of the cone (b) cone
(c) cylinder
(d) sphere
Sol. (a) The desired shape is called a frustum of
PR
in the form of frustum of a cone.
Q.6. The radii of the top and bottom of a
bucket of slant height 45 cm are 28 cm and 7 cm
respectively. The curved surface area of the
bucket is :
(a) 4950 cm2
(b) 4951 cm2
(c) 4952 cm2
(d) 4953 cm2
Sol. (a) Curved surface area of the bucket
S
Q.1. The shape of a glass (tumbler) is usually in the form of :
(a) cone
(b) frustum of a cone
(c) a cylinder
(d) a sphere
Sol. (b) The shape of a glass (tumbler) is usually
where l =
h 2 R r , R, and r are the
2
radii of the two ends of the frustum and h is the
height. Is it true?
Sol. Yes, the statement is true.
Q.9. The slant height of the frustion of a cone
is 5 cm. If the difference between the radii of its
two circular ends is 4 cm, write the height of the
frustum.
[2010]
Sol. We have, l =
h2 + (r1 − r2 )2
5 = h 2 + 42 h2 = 25 – 16 = 9
h = 3 cm.
Q.10. A cone of radius 8 cm and height 12
cm is divided into two parts by a plane through
the mid-point of its axis parallel to its base. Find
the ratio of the volumes of two parts.
[2011 (T-II)]
32
28490 21
= 15
22 1813
Hence, height of the bucket = 15 cm.
Sol.
h=
HA
N
Q.12. A milk container of height 16 cm is
made of metal sheet in the form of a frustum of
a cone with radii of its lower and upper ends as
8 cm and 20 cm respectively. Find the cost of
milk at the rate of Rs 22 per litre which the container can hold.
Sol. Here, h = 16 cm, r = 8 cm, R = 20 cm
Capacity of the container
h
=
(R2 + r2 + Rr)
3
22
16
=
×
(202 + 82 + 20 × 8) cm3
7
3
1
16
=
7
112
TH
=
O
1
h
42
3
2
= h
2
2
8 4 8 4
3 2
AK
PR
S
ER
Volume of the smaller cone
Volume of the frustum of the cone
73216
352
× 624 cm3 =
cm3
7
21
= 10.46 litres
Cost of milk which this container can hold
= Rs 22 × 10.46 = Rs 230.12.
=
Q.13. A bucket is in the form of a frustum of
a cone of height 30 cm with radii of its lower and
upper ends as 10 cm and 20 cm respectively.
Find the capacity and the surface area of the
bucket.
[2011 (T-II)]
Sol. Here, h = 30 cm, R = 20 cm, r = 10 cm
h
Capacity of the bucket =
(R2 + r2 + Rr)
3
22
30
=
×
(202 + 102 + 20 × 10) cm3
7
3
220
=
× 700 cm3 = 22000 cm3 = 22 litres.
7
BR
AS
Let h be the height of the given cone. On dividing
the cone through the mid-point of its axis and parallel
to its base into two parts, we get the shapes as shown
in the figure.
OAB ~ OCD
OB
AB
h
8
=
=
h
OD
CD
r
2
8
2=
r = 4 cm.
r
Hence, required ratio = 1 : 7.
G
O
YA
L
Q.11. A bucket is in the form of a frustum of
a cone and holds 28.490 litres of water. The radii
of the top and bottom are 28 cm and 21 cm,
respectively. Find the height of the bucket.
[2008C]
Sol. Here, R = 28 cm, r = 21 cm
Now,
Volume = 28.490 litres = 28490 cm3.
Volume of the bucket
1
= h (R2 + r2 + Rr)
3
1 22
28490 = ×
× h × (282 + 212 + 28 × 21)
3 7
28490 21
= h × 1813
22
l =
h 2 + (R − r )2
=
302 (20 10) 2 cm
=
900 100 cm
= 1000 cm = 10 10 cm
Total surface area of the bucket
= Curved surface area of the bucket
+ surface area of the bottom
[l (R + r) + r2]
33
=
=
=
⇒
⇒
1
27
2
30 R 2
30 30
⇒
h
=
27 r 2
27 h
h=
3
⇒
3
h
1
30
= ⇔h=
= 10 cm.
30 3
3
.
. . Height of the frustum
= 30 cm – 10 cm = 20 cm.
AS
PR
AK
Q.15. A tent is made in the form of a frustum
of a cone surmounted by another cone. The diameters of the base and the top of the frustum are
20 m and 6 m respectively and the height is 24 m.
If the height of the tent is 28 m and the radius of
the conical part is equal to the radius of the top
of the frustum, find the quantity of canvas required.
[2011 (T-II)]
Sol. Diameter of the base of the frustum = 20 m
O
TH
ER
S
Let its radius be R cm
1
h
1
⇒ =
=
30
27 3
1
the base. If its volume be
th of the volume of
27
the section made?
[2011 (T-II)]
Sol. Height of the cone = 30 cm.
30 R
2
[From (iii)]
Q.14. The height of a cone is 30 cm. A small
cone is cut off at the top by a plane parallel to
the given cone, at which height above the base is
r2 h
N
=
22
[ 10 10 (20 + 10) + 100] cm2
7
22
(300 10 + 100) cm2
7
22
× 100 (9.49 + 1) cm2
7
22
× 100 × 10.49 cm2
7
3295.86 cm2.
HA
=
Diameter of the top of the frustum = 6 m
Height of the frustum = 24 m
Let l be the slant height of the frustum. Then
l 2 = (r1 – r2)2 + h2
BR
⇒
Let the height of the small cone
= h and its radius = r
l 2 = (10 – 3)2 + (24)2
= (7)2 + (24)2 = 49 + 576 = 625
YA
L
⇒
l = 25 m
G
O
Volume of the given cone
π 2 π 2
= R H = R (30) cm3 ...(i)
3
3
π 2
Volume of the smaller cone =
rh
...(ii)
3
In similar OAB and OCD, we have
R 30
=
... (iii)
r
h
As per the question,
1
Volume of smaller cone
=
27
Volume of the given cone
⇒
2
r h
3
2
R (30)
3
1
27
Curved surface of the frustum = π (r1 + r2) l
22
=
× (10 + 3) × 25 cm2
7
[From (i) and (ii)]
34
=
22
× 13 × 25 cm2
7
.
.. l =
= 1021.43 m2
Diameter of the base of the cone = 6 m.
Height of the cone = 4 m
Let l be the slant height of the cone.
l2 = (3)2 + (4)2 = 9 + 16 = 25
⇒
l = 5
Curved surface of the cone
22
= πrl =
× 3 × 5 m2 = 47.14 m2
7
Area of the canvas required
= curved surface area of frustum
+ curved surface area of cone
= (1021.43 + 47.14) m2
= 1068.57 m2.
(6)2 (2.5 1)2 cm
36 2.25 cm
=
38.25 cm = 6.2 cm (approx).
.
HA
N
. . CSA of frustum part = πl (r1 + r2)
= π × 6.2 × (2.5 + 1) cm2
= π × 6.2 × 3.5 cm2 = 21.7 π cm2
CSA of hemispherical part with radius 1 cm
= 2π r12 = 2π(1)2 cm2 = 2π cm2
AS
.
. . Total surface area = (21.7π + 2π) cm2
= 23.7π cm2
AK
22
cm2 = 74.49 cm2
7
≅ 74.5 cm2 (approx).
PR
= 23.7 ×
S
Q.17. A hollow cone is cut by a plane parallel to the base and the upper portion is removed.
If the curved surface of the remainder is 8/9 of
the curved surface of the whole cone, find the
ratio of line segments into which the altitude of
the cone is divided by the plane.
[2004]
Sol. Let L be the slant height of the given cone
and R be its base radius.
Then, the smaller cone as shown in the figure has
been removed.
G
O
YA
L
BR
O
TH
ER
Q.16. A shuttle cock used for playing badminton has the shape of a frustum of a cone
mounted on a hemisphere (see figure). The external diameters of the frustum are 5 cm and 2 cm
and the height of the entire shuttle cock is 7 cm.
Find its external surface area.
[2011 (T-II)]
h2 + (r2 − r1 )2
Sol. For the upper portion of the shuttle cock :
5
cm = 2.5 cm
2
and h = (7 – 1) cm = 6 cm.
r1 = 1 cm,
r2 =
Let l and r be the slant height and radius respectively of the smaller cone.
Then,
OAB ~ OCD
35
AB
OB
OA
=
=
CD
OD
OC
...(i)
L
R
Rl
=
r=
...(ii)
l
r
L
Now, curved surface area of the remainder
8
of the curved surface area of the whole cone
9
Curved surface area of the smaller cone
=
Sol. We have OA = r, OO = h, OA =
h2
r2
9
1
RL
9
1
Rl
l = RL
L
9
9l2 = L2
1
l
=
3
L
HA
rl =
[From (ii)]
AS
N
1
of the curved surface area of the whole cone
9
AK
=
r2
. Show that
9
13r 2 h
the volume of the frustum is
.
27
h2
the base of the frustum is
Now, from (i)
PR
.... (iii)
Now, in OOA, we have
OA2
= OO2 + OA2
ER
BR
O
OC
1
=
AC
2
Hence, required ratio = 1 : 2.
2OC = AC
TH
OA
OC + AC
OC =
=
3
3
Choose the correct option (Q 1 – 5) :
1. The shape of a bucket is usually in the
form of :
(a) a cone
(b) a cylinder
(c) sphere
(d) frustum of a cone
2. Slant height of frustum of cone of a height
h and radii of its ends as R and r is :
R2 + r2
(c)
h2 + ( R − r )
2
OA 2 OO 2
r
r2
h2 =
3
9
Volume of the frustum
h
=
(OA2 + OA2 + OA × OA)
3
h 2 r 2
r
= 3 r 9 r 3
=
=
h2
2
2
2
h 9r r 3r
3
9
=
13r 2 h
Proved.
27
PRACTICE EXERCISE 13.4A
G
O
YA
L
Q.18. The radius of base of a right circular
cone is r. It is cut by a plane parallel to the base
at a height h from the base. The distance of the
boundary of the upper surface from the centre of
(a)
OA =
S
OB
OA
=
OD
OC
OA
L
3
=
=
[From (iii)]
OC
1
l
(
(b)
h + R +r
(d)
h2 − ( R + r )
2
2
2
3. Curved surface area of a frustum of a
cone of height h and radii of its ends as R1 and
R2 is :
(a) h (R1 + R2)
(b) (R1 + R2) {h2 + (R1 – R2)2}
(c) (R1 + R2)
)
(d) (R1 + R2)
2
36
h 2 + ( R1 − R 2 )
2
S
PR
AK
AS
HA
N
of 4 cm. Find the area of interior surface of cone
not in contact with water. (Use π = 3.14)
12. The radii of the circular ends of a conical
curd container, 6 cm high are 16 cm and
24 cm. Find the volume of the container in litres.
(Take π = 3.14)
13. An open metallic bucket is in the shape
of a frustum of a cone mounted on a hollow
cylindrical base made of metallic sheet. If the
diameters of the two ends of the bucket are 45
cm and 25 cm, the total vertical height of the
bucket is 30 cm and that of the cylindrical
portion is 6 cm, find the area of the metallic sheet
used to make the bucket. Also find the volume of
22
water it can hold Take
. [2011 (T-II)]
7
G
O
YA
L
BR
O
TH
ER
4. If the radii of the ends of a frustum of a cone,
which is 45 cm high are 28 cm and 7 cm, then the
volume of the frustum of the cone is :
(a) 48510 cm3
(b) 45810 cm3
3
(c) 41510 cm
(d) 48501 cm3
5. The volume of a frustum of a cone is
616 cm3. If the radii of its ends are 10 cm and 6 cm,
then the height of frustum of the cone is :
(a) 3 cm
(b) 4 cm
(c) 5 cm
(d) 7 cm
6. An open metallic bucket is in the shape of
a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to
curved surface area of frustum of the cone + area
of circular base + curved surface area of the cylinder. Is this statement true?
7. A cone of radius 4 cm is divided into two
parts by drawing a plane through the mid-point of
its axis and parallel to its base. Compare the volumes of the two parts.
8. A bucket of height 8 cm and made up of
copper sheet is in the form of a frustum of a right
circular cone with radii of its lower and upper ends
as 3 cm and 9 cm respectively. Calculate :
(i) the height of the cone of which the bucket
is a part.
(ii) the volume of water which can be filled
in the bucket.
(iii) the area of copper sheet required to
make the bucket.
[2003]
9. If the radii of the ends of a bucket 24 cm
high are 5 cm and 15 cm, find the surface area of
the bucket.
10. The radii of the top and bottom of a 12 cm
deep tub are 20 cm and 10 cm. Find its volume and
cost of tin sheet used for making the tub at the rate
of Rs 1.20 per dm2.
11. An inverted cone of vertical height 12 cm
and radius of base 9 cm contains water to a depth
14. The height of a cone is 42 cm. A small
cone is cut off at the top by a plane parallel to the
base. If its volume is
1
of the volume of the
27
given cone, at what height above the base, the
section has been made?
15. A bucket made up of metal sheet is in the
form of a frustum of a cone. Its depth is
24 cm and the diameters of the top and bottom are
30 cm and 10 cm respectively. Find the cost of milk
which can completely fill the bucket at the rate of
Rs 20 per litre and the cost of the metal sheet used,
if it costs Rs 10 per 100 cm2 (use π = 3.14).
[2006, 2011 (T-II)]
TEXTBOOK’S EXERCISE 13.5 (OPTIONAL)
Q.1. A copper wire, 3 mm in diameter, is
wound about a cylinder whose length is 12 cm,
and diameter 10 cm, so as to cover the curved
surface of the cylinder. Find the length and mass
of the wire, assuming the density of copper to be
8.88 g per cm3.
37
In ABC (By Pythagorus Theorem)
(BC)2 = (AB)2 + (AC)2
BC2 = (4)2 + (3)2 = 16 + 9 = 25
BC= 5 cm
3
Diameter
=
cm
20
2
Height of the cylinder = 12 cm
Radius of cylinder = 5 cm
Length of wire used in one turn
= Circumference of cylinder.
Length of wire used in one turn
157
= 2r = 2 × 3.14 × 5 =
cm
5
Number of turns used
Height of the cylinder
=
Diameter of wire
Radius of copper wire =
HA
AS
ER
157
cm = 1256 cm = 12.56 m
5
TH
O
BR
Mass of 1 cm3 = 8.88 g [Given]
Mass of 88.81 cm3 = 88.81 × 8.88 g
YA
L
= 789 g (approx.)
22 12
22 12
×
× 4 cm2
= 3 cm2 +
7
5
7
5
O
Q.2. A right triangle, whose side are 3 cm
and 4 cm (other than hypotenuse is made to
revolve about it a hypotenuse. Find the volume
and surface area of the double cone so formed.
(Choose value of as found appropriate).
Sol. Let ABC be the triangle, right angled at A.
1056
6
= 30
cm3
35
35
= 30.1714 cm3.
Surface area of double cone
= (Surface area of cone ABA)
+ (Surface area of cone ACA)
= × OA × AB + × AO × AC
=
22
3
3
= r2h =
×
×
× 1256 cm3
7
20
20
G
1
12
× AO × 5 = 6 AO =
2
5
Volume of double cone
= (Volume of cone ABA)
+ (Volume of cone ACA)
1
1
= OA2OB + OA2OC
3
3
1
= OA2 (OB + OC)
3
1
22 12 12 9 16
=
×
×
×
×
3
7
5
5
5 5
1
22 12 12
25
=
×
×
×
×
3
7
5
5
5
S
× Length of wire used in one turn
= 88.81 cm3.
1
× BC × AO = 6 cm2
2
AK
= Number of turns
Volume of wire used
Also, area of ABC =
PR
Total length of wire used
= 40 ×
1
× 3 × 4 cm2 = 6 cm2
2
120
12
=
= 40
3
3
10
=
Now, area of ABC =
N
Sol. Diameter of copper wire = 3 mm
=
22 12
264
×
× 7 cm2 =
cm2 = 52.8 cm2
7
5
5
Q.3. A cistern, internally measuring 15 cm ×
120 cm × 110 cm, has 129600 cm3 of water in it.
Porous bricks are placed in the water until the
cistern is full to the brim. Each brick absorbs
one-seventeenth of its own volume of water. How
many bricks can be put in without overflowing
the water, each brick being 22.5 cm × 7.5 cm
× 6.5 cm?
Sol. Volume of 1 brick = 22.5 × 7.5 × 6.5 cm3
= 1096.875 cm3
38
Volume of cistern = 150 × 120 × 10 cm3
= 1980000 cm3
Volume of water = 129600 cm3
Volume of cistern to be filled
= (1980000 – 129600) cm3
= 1850400 cm3
Let n bricks be needed.
Then,water absorbed by n bricks
N
HA
1850400 + n ×
1096.875
cm3
17
Sol. Slant height of the frustum of the cone, l
1096.875
= n (1096.875)
17
=
1850400 17
16 1096.875
n =
n = 1792.4102
=
PR
h 2 r1 r2
ER
S
= 1792 (approx.)
Hence, number of bricks used = 1792.
TH
Q.4. In one fortnight of a given month, there
was a rainfall of 10 cm in a river valley. If the
area of the valley is 97280 km2, check whether
the total rainfall was approximately equivalent
to the addition to the normal water of three
rivers each 1072 km long, 75 m wide and 3 m
deep.
Sol. Volume of total rainfall in the valley
2
18 8
cm
2 2
=
12 2 5 2
=
144 25 cm =
cm
169 cm = 13 cm
Area of metal sheet
= C.S.A. of cylinder + C.S.A. of frustum
= 2 (4) (10) cm2 + (4 + 9) 13 cm2
O
= (80 + 169) cm2 = 249 cm2
= 249 ×
BR
YA
L
2
22 10 2
AK
16n
× 1096.875 = 1850400
17
AS
=n×
=
22
cm2
7
5478 2
4
cm = 782 cm2.
7
7
10
1000 100
= 9.728 km3
Volume of water of three rivers
Q.6. Derive the formula for the curved
surface area and total surface area of the
frustum of a cone.
Sol. Let the frustum be ABCD. The frustum can be
75
3
×
= 0.7236 km3
1000 1000
Hence, the two are not approximately equivalent.
viewed as a difference of two right circular cones OAB
and OCD.
G
O
= 97280 ×
= 3 × 1072 ×
Q.5. An oil funnel made of tin sheet consists
of a 10 cm long cylindrical portion attached to a
frustum of a cone. If the total height is 22 cm,
diameter of the cylindrical portion is 8 cm and
the diameter of the top of the funnel is
18 cm, find the area of the tin sheet required to
make the funnel (see figure).
Let OA = OB = l1, OP = h1,
PB = r1, OC = OD = l2,
OQ = h2 and QD = r2
Let height (PQ) of the frustum be h and slant height
(BD or AC) of the frustum be l.
Then
l2 = l1 – l
....(i)
In DBL,
39
DB2 = DL2 + BL2
N
l2 = h2 + (r1 – r2)2
2
HA
h 2 r1 r2
...(ii)
AS
l=
Let OA = OB = l1, OP = h1, PB = r1
OC = OD = l2, OQ = h2 and QD = r2
Let height (PQ) of the frustum be h.
Then
h = h1 – h2
....(i)
Curved surface area of the frustum
= r1l1 – r2l2
..(iii)
Now, in similar triangles OPB and OQD, we have
AK
In similar triangles OPB and OQD, we have
OP
PB
r1
h
=
1 =
OQ
QD
r
h2
2
l1 =
r1l
r1 r2
...(iv)
TH
BR
O
Also, from (i) and (ii), we have,
r l r l r2l
rl
l2 = 1
–l= 1 1
r1 r2
r1 r2
O
.r12 l
r22 l
r1 r2 r1 r2 l
=
r1 r2 r1 r2
r1 r2
G
r1h
...(ii)
r1 r2
Also, from (i) and (ii), we have
r h r1h r2 h
rh
h2 = 1
–h= 1
r1 r2
r1 r2
h1 =
YA
L
l2 =
h1 =
r2 h
....(iii)
r1 r2
Now, volume of the frustum of the cone
= Volume of the cone OAB
– Volume of the cone OCD.
1
1
= r12h1 – r22h2
3
3
r2 l
....(v)
r1 r2
Substituting the values of l1 and l2 from (iv) and (v)
in (iii), we have curved surface area of the frustum of
the cone
r1 ( h1 − h )
r1h2
=
r2
r2
r2h1 = r1h1 – r1h h1(r1 – r2) = r1h
S
r1 l1 l
r1
r1l2
l1
=
l1 =
=
r2
r2
r2
l2
l1r2 = r1l1 – r1l l1(r1 – r2) = r1l
ER
PR
PB
OB
=
QD
OD
= l (r1 + r2) Proved.
h2 =
1
=
3
Now, total surface area of the frustum
= curved surface area + area of the circular base
+ area of the circular top
= l (r1 + r2) + r12 + r22 Proved.
r13 h
r 3h
2
r1 r2 r1 r2
1
= h
3
r13 r23
r1 r2
r1 r2 r12 r22 r1r2
1
= h
3
r1 r2
Q.7. Derive the formula for volume of the
frustum of a cone.
Sol. Let the frustum be ABCD. The frustum can be
=
seen as a difference of two right circular cones OAB and
OCD.
40
[From (i) and (iii)]
1
h (r12 + r22 + r1r2) Proved.
3
B. FORMATIVE ASSESSMENT
Activity 1
Objective : To compare the curved surface areas and the total surface areas of two right circular
cylinders which are formed from rectangular sheets of paper having same dimensions.
Procedure :
Figure 1
BR
O
TH
ER
S
PR
AK
AS
1. Take two rectangular sheets ABCD of dimensions 22 cm × 11cm.
HA
N
Materials Required : Rectangular sheets of paper, a pair of scissors, sellotape, geometry
box, etc.
G
O
YA
L
2. Fold one of the sheets ABCD along AB such that AD and BC coincide. Use sellotape to get
a cylinder as shown below.
Figure 2
41
AS
HA
N
3. Put the cylinder on a white sheet of paper and draw the outline of its base. Using a pair of scissors
cut out the circular region. Make one more such circular cut out.
AK
Figure 3
BR
O
TH
ER
S
PR
4. Using sellotape, fix these circular cut outs on the top and bottom
of the hollow cylinder as shown in the figure.
Figure 4
G
O
YA
L
5. Take another sheet ABCD and fold it along AD such that AB and DC coincide. Use sellotape to
get a cylinder as shown.
Figure 5
6. Repeat steps 3 and 4 for the cylinder obtained above.
Figure 6
42
Observations :
1. Let r1 be the base radius of the cylinder in figure 2.
Then circumference of its base = 2 r1 = 22 r1 =
22 × 7
cm = 3.5 cm.
2 × 22
Also, height, h1 of the cylinder = 11 cm.
22
× 3.5 × 11 cm2
7
N
2. So, curved surface area of the cylinder in figure 2 = 2 r1h1 = 2 ×
4. Let r2 be the base radius of the cylinder in figure 5.
11 × 7
cm = 1.75 cm.
2 × 22
PR
Then circumference of its base = 2 r2 = 11 r2 =
AS
22
× 3.5 × 3.5) cm2 = 319 cm2
7
AK
= (242 + 2 ×
HA
= 242 cm2.
3. Total surface area of the cylinder in figure 4 = curved surface area + 2 r12
S
Also, height, h2 of the cylinder = 22 cm.
5. So, curved surface area of the cylinder in figure 5
22
× 1.75 × 22 cm2 = 242 cm2.
7
ER
= 2 r2h2 = 2 ×
22
× 1.75 × 1.75) cm2 = 261.25 cm2.
7
O
= (242 + 2 ×
TH
6. Total surface area of the cylinder in figure 6 = curved surface area + 2 r22
BR
7. From 2 and 5, we see that the curved surface area of both the cylinders is same.
8. From 3 and 6, we see that the total surface area of the two cylinders is different.
YA
L
Conclusion :
O
1. The curved surface areas of two cylinders formed by folding the rectangular sheets having
same dimensions is same.
G
2. The total surface areas of two cylinders formed by folding the rectangular sheets having same
dimensions are not same.
Activity 2
Objective : To compare the volumes of two right circular cylinders which are formed from
rectangular sheets of paper having same dimensions.
Materials Required : Rectangular sheets of paper, a pair of scissors, sellotape, geometry
box, etc.
Procedure :
1. Take two rectangular sheets ABCD of dimensions 22 cm × 11 cm.
43
N
HA
AS
Figure 7
BR
O
TH
ER
S
PR
AK
2. Fold one of the sheets ABCD along AB such that AD and BC coincide. Use sellotape to get
a cylinder as shown below.
Figure 8
G
O
YA
L
3. Take another sheet ABCD and fold it along AD such that AB and DC coincide. Use sellotape
to get a cylinder as shown.
Figure 9
Observations :
1. Let r1 be the radius of the base of the cylinder in figure 8.
Then, circumfrence of its base = 2r1 = 22 r1 =
Also, height h1 of the cylinder = 11 cm.
2. Volume of this cylinder = r12h1 =
22 × 7
cm = 3.5 cm.
2 × 22
22
× 3.5 × 3.5 × 11 cm3 = 423.50 cm3
7
44
3. Let r2 be the radius of the base of the cylinder in figure 9.
Then circumference of its base = 2r2 = 11 r2 =
11 × 7
cm = 1.75 cm.
2 × 22
Also, height, h2 of the cylinder = 22 cm.
4. Volume of this cylinder = r22h2 =
22
× 1.75 × 1.75 × 22 cm3 = 211.75 cm3.
7
5. From 2 and 4 above, we see that the two volumes are different.
HA
N
Conclusion : The volumes of two cylinders formed by folding the rectangular sheets having same
dimensions are different.
PR
Materials Required : White sheets of paper, a pair of
scissors, colour pencils, sellotape, geometry box, etc.
AK
Objective : To make a cone of given slant height l and
base circumference.
AS
Activity 3
Procedure : Let us make a cone of slant height, l = 7 cm
and base circumference = 11 cm
ER
S
1. On a white sheet of paper, draw a circle of radius 7
cm and centre O.
YA
L
BR
O
TH
Figure 10
Figure 11
G
O
2. Mark a sector OAB such that AOB = 90°. Cut the sector OAB and bring the radii OA and
OB together. Use sellotape to get a cone as shown above.
Observations :
1. The radius of the circle in figure 11 becomes slant height of the cone.
or, l = 7 cm
2. AOB = 90°
πr θ°
So, length of the arc AB =
, where = 90° =
180°
22
× 7 × 90°
7
cm = 11cm
180°
Hence, base circumference of the cone = length of the arc AB = 11 cm
Conclusion : From the above activity, we see that a cone of given slant height and base
circumference can be made from a sector of a circle.
45
Activity 4
Objective : To give a suggestive
demonstration of the formula for the lateral
surface area of a cone.
N
Materials Required : White sheets of
paper, a pair of scissors, colour pencils,
geometry box, sellotape, gluestick, etc.
HA
Procedure :
1. On a white sheet of paper, draw a circle
of any convenient radius and with centre O.
Mark a sector OAB on the circle and cut it
out. From the previous activity, we know that
this sector can be folded to make a cone.
PR
2. Fold the sector such that OA and OB coincide.
Again fold it as shown below. Press it to make creases.
AK
AS
Figure 12
Figure 13
BR
O
TH
ER
S
3. Unfold the cut out and draw lines along the creases. Now, cut the
sector along the creases to get four smaller sectors as shown below.
Figure 14
YA
L
4. Paste the four smaller sectors on a white sheet of paper to get
an approximate parallelogram as shown alongside.
Observations :
G
O
1. From figure, slant height of the cone = l and base
circumference of the cone = 2r, where r is the base radius of the
cone.
Figure 15
2. Figure 15 is an approximate parallelogram whose base is half of the base circumference of
the cone and height is approximately the slant height of the cone.
i.e, base of the parallelogram =
1
× 2r = r
2
Height of the parallelogram = l
Area of the parallelogram = base × height = r × l = rl
Thus, curved surface area of the cone = area of the parallelogram = rl
Conclusion : The curved (lateral) surface area of a cone of base radius r and slant height l is
equal to rl.
46
Activity 5
Objective : To give a suggestive demonstration of the formula for the volume of a right circular
cone.
Materials Required : Cones and cylinders of same base radius and height, sand, etc.
Procedure :
1. Take a set of cone and cylinder having same base radius (r) and height (h).
3. Pour the sand from the cone into the cylinder
AS
4. Repeat step 3 until the cylinder gets completely filled with sand.
HA
N
2. Fill the cone with sand
5. Repeat steps 2 to 4 for another set of cone and cylinder having same base radius and same height.
AK
Observations :
PR
1. We see that after filling the cone with sand and pouring the sand into the cylinder, it needs
three pourings to fill the cylinder completely.
So, volume of the cylinder = 3 times the volume of the cone
S
1
volume of the cylinder..
3
2. Volume of the cylinder = r2h
1 2
r h
3
TH
3. So, volume of the cone =
ER
or, volume of the cone =
1 2
r h.
3
Activity 6
YA
L
BR
and height h is given by
O
Conclusion : From the above activity, it is verified that the volume of a cone of base radius r
Objective : To give a suggestive demonstration of the formula for the volume of a sphere in
terms of its radius.
G
O
Materials Required : A hollow sphere and two cylinders whose base diameter and height are
equal to the diameter of the sphere, sand, etc.
Procedure :
1. Take a hollow sphere and two cylinders whone base diameter and height are equal to the
diameter of the sphere.
2. Fill the sphere with sand and pour it into one of the cylinders.
3. Fill the sphere with sand second time and pour it into the same cylinder as in step 2, till it
gets completely filled.
4. Pour the remaining sand into the other cylinder.
5. Again fill the sphere with sand and pour it into the semi-filled cylinder to fill the cylinder with
sand completely.
47
Observations :
1. We see that the total sand poured in three pourings completely fill the two cylinders.
2. So, three times the volume of the sphere = two times the volume of the cylinder
= 2r2h = 2r2(2r)
[ h = 2 r for each cylinder]
3 × volume of the sphere = 4r3
Volume of the sphere =
4 3
r
3
4 3
r .
3
HA
given by
N
Conclusion : From the above activity, it is verified that the volume of a sphere of radius r is
AS
Angles in a Cube
PR
AK
Find the angle between the two dotted lines drawn on the surface of a cube.
S
Cutting Cubes
G
O
YA
L
BR
O
TH
ER
A cube of side 4 cm painted red, yellow and green on opposite faces. It was cut into cubical
blocks each of the side 1 cm.
1.
2.
3.
4.
How
How
How
How
many
many
many
many
cubes
cubes
cubes
cubes
are
are
are
are
obtained?
there whose only one face is painted?
there whose no face is painted?
there whose two faces are painted?
48
5.
6.
7.
8.
9.
How many cubes are there whose three faces are painted?
How many cubes are there whose two faces are painted with the same colour?
How many cubes are there whose three faces are painted all with different colours.
How many cubes are there whose all faces are painted?
At most how many painted faces a cube can have?
ANSWERS
HA
N
A. SUMMATIVE ASSESSMENT
(d)
(b)
17.6 cm2
Rs 855.36
AK
3.
7.
11.
15.
4.
8.
12.
16.
(b)
No
163.86 cm2
18480 cm2
(d)
3. (c)
4. (d)
14 cm, 7 cm, 132 cm3, 66 cm3, 396 cm3
7. 150
179.66 cm3
10. 30.48 cm3
11. 124.4 cm3
3 cm
14. 628.57 cm3
15. 6 m
17. 0.5 m
18. 301.44 cm2, 377.1 cm3
TH
Practice Exercise 13.2A
1. (a)
2.
5. (b)
6.
8. 688.83 cm3
9.
3
12. 8 cm
13.
16. 74.25 m3, 80.61 m2
(b)
(b)
No
350 cm2
PR
2.
6.
10.
14.
S
(b)
(a)
No
Rs 99,000
ER
1.
5.
9.
13.
AS
Practice Exercise 13.1A
Practice Exercise 13.3A
1. (a)
2. (a)
5. (b)
6. 6 cm
9. 6.4 cm
10. 4.04 cm (approx.)
13. 512
14. 2904 cm2
17. 28.44 cm
18. 90 cm
YA
L
BR
O
3.
7.
11.
15.
G
O
Practice Exercise 13.4A
1. (d)
2. (c)
3.
5. (a)
6. Yes
7.
3
2
8. (i) 12 cm (ii) 312 cm (iii) 405.43 cm
9.
11. 376.8 cm2
12. 7.64 litres
13.
14. 28 cm
15. Rs 163.28, Rs 17.13
(c)
84
1 cm
30 hectares
4.
8.
12.
16.
(d)
14 cm
9 cm
1500
(c)
4. (a)
1:7
2420 cm2
10. 8800 cm3, Rs 21.44
2
3822.5 cm , 26.6 litres
B. FORMATIVE ASSESSMENT
Angles in a cube
Complete the triangle. The sides of this triangle are diagonals of the faces of the cube. So, the
triangle is equilateral. So, the required angle is 60°.
Cutting cubes
1. 84
2. 24
3. 8
4. 24
5. 8
49
6. 0
7. 8
8. 0
9. 3
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