 Lesson: Surface Areas and Volumes
Exercise 13.1
Write the correct answer in each of the following:
Question: 1
The radius of a sphere is 2r, and then its volume will
be:
(a)
3
4
3
(b)
3
4 r
(c)
3
8
3
r
(d)
3
32
3
r
Solution:
d
Volume of a sphere of radius
3
4
3
rr
If the radius is 2r then the volume will be
33
4 32
(2 ) .
33
rr

 Question: 2
The total surface area of a cube is 96 cm
2
. The volume
of the cube is:
(a) 8 cm
3
(b) 512 cm
3
(c) 64 cm
3
(d) 27 cm
3
Solution:
c
Surface area of a cube of side
2
6aa
.
Therefore,
cm
22
6 96a
Volume of a cube of side
3
aa
For,
4a
, the volume
3
4 64
cm
3
. Question: 3
A cone is 8.4 cm high and the radius of its base is 2.1
cm. It is melted and recast into a sphere. The radius of
the sphere is:
(a) 4.2 cm
(b) 2.1 cm
(c) 2.4 cm
(d) 1.6 cm
Solution:
b
Volume of a cone of height h and the radius of its base
2
1
.
3
r r h
So,
The volume of the cone of height 8.4 cm and the
radius of its base
cm =
2
1
2.1 (2.1) 8.4
3
As,
The cone is melted and recast into a sphere.
Therefore,
The volume of the cone is equal to the volume of the
sphere.
Now, Volume of a sphere of radius
3
4
.
3
rr
Then,
32
41
(2.1) 8.4
33
πr π
32
(2.1) 2.1r
2. c1 mr
Question: 4
In a cylinder, radius is doubled and height is halved,
curved surface area will be
(a) Halved
(b) Doubled
(c) Same
(d) Four times
Solution:
c
Curved surface area of a cylinder of radius r and
height
2h rh
In a cylinder, radius is doubled and height is halved,
So,
New curved surface area will be 22()
2
2.
π r h
πrh
Question: 5
The total surface area of a cone whose radius is
2
r
and
slant height 2l is
(a)
2πr l r
(b)
4
r
πr l



(c)
()πr l r
(d)
2πrl
Solution:
b
Total surface area of a right circular cone of radius r
and slant height
2
, . ., .l πrl πr i e πr l r
The total surface area of a cone whose radius is
2
r
and
slant height 2l is
()
2
22
π r r
l



4
r
πr l



 Question: 6
The radii of two cylinders are in the ratio of 2:3 and
their heights are in the ratio of 5:3. The ratio of their
volumes is:
(a) 10:17
(b) 20:27
(c) 17:27
(d) 20:37
Solution:
b
Volume of a cylinder of radius r and height
2
πh r h
Therefore,
If the radii of two cylinders are in the ratio of 2:3 and
their heights are in the ratio of 5:3, the ratio of their
volumes is 20:27. Question: 7
The lateral surface area of a cube is 256 m
2
. The
volume of the cube is
(a) 512 m
3
(b) 64 m
3
(c) 216 m
3
(d) 256 m
3
Solution:
a
The lateral surface area of a cube is 256 m
2
.
Let the side of a cube is a m.
So,
2
4 256a
m8a
Thus,
The volume of the cube is
m
33
8
i.e.
m
3
512
. Question: 8
The number of planks of dimensions
(
m cm cm4 50 20
) that can be stored in
a pit which is 16 m long, 12 m wide and 4 m deep is
(a) 1900
(b) 1920
(c) 1800
(d) 1840
Solution:
b
Volume of the plank
m m m4 0.50 0.20
m
3
0.4
Capacity of the pit
m m m16 12 4
m
3
768
Therefore,
Number of planks
768
0.4
1920 Question: 9
The length of the longest pole that can be put in a
room of dimensions (
m m m10 10 5
) is
(a) 15 m
(b) 16 m
(c) 10 m
(d) 12 m
Solution:
a
The length of the longest pole that can be put in a
room of dimensions
m m m10 10 5
will be equal to
the diagonal of the room.
So,
The length of the longest pole is
2 2 2
10 10 5
m i.e.
15 m. Question: 10
The radius of a hemispherical balloon increases from 6
cm to 12 cm as air is being pumped into it. The ratios
of the surface areas of the balloon in the two cases is
(a) 1:4
(b) 1:3
(c) 2:3
(d) 2:1
Solution:
a
Case I:
When the radius of the hemispherical balloon is 6 cm,
The surface area of the balloon is
2
3 (6)
i.e.
108
.
Case II:
When the radius of the hemispherical balloon is 12
cm,
Then,
The surface area of the balloon is
2
3 (12)
i.e.
432
.
Therefore,
The ratio of surface areas of the balloon in both cases
is

108 :432
: i.e. 1:4.
EXERCISE 13.2 Write True or False and justify your answer in each of
the following:
Question: 1
The volume of a sphere is equal to two-third of the
volume of a cylinder whose height and diameter are
equal to the diameter of the sphere.
Solution:
The height and diameter of right circular cylinder are
equal to the diameter of the sphere.
Now,
Suppose that the base radius of the right circular
cylinder is r and the height of that cylinder is 2r
whereas the radius of the sphere is r.
So, 2
11
3
2
4
3
Cylinder
Sphere
V
rh
V
r
2
4
3
r r r
rrr


3
2
Thus,
2
3
Sphere Cylinder
VV
Therefore,
The volume of a sphere is equal to two-third of the
volume of a cylinder whose height and diameter are
equal to the diameter of the sphere.
Hence,
The given statement, “The volume of a sphere is equal
to two-third of the volume of a cylinder whose height
and diameter are equal to the diameter of the sphere”
is true. Question: 2
If the radius of a right circular cone is halved and
height is doubled, the volume will remain unchanged.
Solution:
The radius of a right circular cone is halved and the
height is doubled.
Now,
Suppose that initially base radius of the right circular
cone is 2x and height is y and later base radius of right
circular cone becomes x and height becomes 2y.
So,
2
1 1 1
2
2 2 2
V r h
V r h
22
2
x x y
x x y


2
Thus, 21
1
2
VV
Therefore,
If the radius of a right circular cone is halved and
height is doubled, the new volume is half of the
original volume.
Hence,
The given statement, “If the radius of a right circular
cone is halved and height is doubled, the volume will
remain unchanged” is false.
Question: 3
In a right circular cone, height, radius and slant height
do not always be sides of a right triangle.
Solution:
Since,
For a right circular cone,
2 2 2
r h l
Where,
l
slant height of right circular cone
r
base radius of right circular cone
h
height of right circular cone.
So, In a right circular cone, height, radius and slant height
always be sides of a right triangle.
Hence,
The given statement, “In a right circular cone, height,
radius and slant height do not always be sides of a
right triangle” is false.
Question: 4
If the radius of a cylinder is doubled and its curved
surface area is not changed, the height must be halved.
Solution:
The radius of a cylinder is doubled and its curved
surface area is not changed.
Now,
Suppose that initially the base radius of the right
circular cylinder is x and height of that cylinder is 2y
and later the base radius of the right circular cylinder
becomes 2x whereas the curved surface area remains
unchanged. So,
In both cases as the curved surface areas are same.
Therefore,
Curved surface area for case
I
Curved surface area
for case II
1 1 2 2
22rh r h

2
22x y xh
As a result,
2
hy
Thus,
The height of the right circular cylinder is halved.
Hence,
The given statement, “If the radius of a cylinder is
doubled and its curved surface area is not changed, the
height must be halved” is true.
Question: 5
The volume of the largest right circular cone that can
be fitted in a cube whose edge is 2r equals to the
volume of a hemisphere of radius r. Solution:
The volume of the largest right circular cone that can
be fitted in a cube whose edge is 2r.
So,
The base radius of the right circular cone
2
2
r
r
And,
The height of right circular cone
2r
Therefore,
Volume of right circular cone
2
1
2
3
rr

3
2
3
r
Volume of the hemisphere of radius r
Hence,
The given statement, “The volume of the largest right
circular cone that can be fitted in a cube whose edge is
2r equals to the volume of a hemisphere of radius r is
true.
Question: 6
A cylinder and a right circular cone are having the
same base and same height. The volume of the cylinder is three times the volume
of the cone.
Solution:
A cylinder and a right circular cone have the same
base and same height.
Now,
Suppose that the base radius of the right circular
cylinder is r and the height of that cylinder is h
whereas the base radius of the right circular cone is r
and the height of the cone is h.
So,
2
2
1
3
Cylinder
Cone
V
rh
V
rh
Thus,
3
Cylinder cone
VV
Therefore, The volume of the right circular cylinder is three times
the volume of the right circular cone.
Hence,
The given statement, “If a cylinder and a right circular
cone are having the same base and same height, the
volume of the cylinder is three times the volume of the
cone” is true.
Question: 7
A cone, a hemisphere and a cylinder stand on equal
bases and have the same height. The ratio of their
volumes is 1 : 2 : 3.
Solution:
A cone, a hemisphere and a cylinder stand on equal
bases and have the same height.
Now,
Suppose that the base radius of the right circular
cylinder is r and the height of that cylinder is r
whereas the base radius of the right circular cone is r
and the height of the cone is r while the radius of the
hemisphere is r. So,
Volume of cone: Volume of hemisphere : Volume of
cylinder
2 3 2
12
::
33
r h r r h
2 3 2
12
::
33
r r r r r
3 3 3
12
::
33
r r r
1:2:3
Therefore,
The ratio of volumes of right circular cone,
hemisphere and right circular cylinder is 1 : 2 : 3.
Hence,
The given statement, “If a cone, a hemisphere and a
cylinder stand on equal bases and have the same
height, the ratio of their volumes is 1 : 2 : 3” is true. Question: 8
If the length of the diagonal of a cube is
63
cm, then
the length of the edge of the cube is 3 cm.
Solution:
The length of the diagonal of a cube is
63
cm.
If the side of a cube is a, the length of the diagonal of
that cube is
3.a
So,
3 6 3a
Therefore,
6a
Thus,
The length of the edge of the cube is 6 cm.
Hence,
The given statement, “If the length of the diagonal of a
cube is
63
cm, then the length of the edge of the
cube is 3 cm” is false. Question: 9
If a sphere is inscribed in a cube, then the ratio of the
volume of the cube to the volume of the sphere will be
6:
.
Solution:
A sphere is inscribed in a cube.
Now,
Suppose that the side of a cube is 2r.
Then,
The radius of the sphere that is inscribed in it will be r.
So,
3
3
2
4
3
Cube
Sphere
r
V
V
r
3
3
2
4
3
r
r 3
3
8
4
3
r
r
6
Therefore,
The ratio of the volume of the cube to the volume of
the sphere is
6: .
Hence,
The given statement, “If a sphere is inscribed in a
cube, then the ratio of the volume of the cube to the
volume of the sphere will be
6:
” is true.
Question: 10
If the radius of a cylinder is doubled and height is
halved, the volume will be doubled.
Solution:
The radius of a right circular cylinder is doubled and
height of that cylinder is halved,
Now,
Suppose that initially the base radius of the right
circular cylinder is r and the height of that cylinder is
2y while later the base radius of right circular cylinder
becomes 2r and the height of that cylinder becomes y. So,
2
1 1 1
2
2 2 2
V r h
V r h
2
22
r r y
r r y


1
2
Therefore,
21
2VV
Thus,
The volume of the right circular cylinder is doubled.
Hence,
The given statement, “If the radius of a cylinder is
doubled and height is halved, the volume will be
doubled” is true.
EXERCISE 13.3 Question: 1
Metal spheres, each of radius 2 cm, are packed into a
rectangular box of internal dimensions 16 cm × 8 cm ×
8 cm. When 16 spheres are packed the box is filled
with preservative liquid. Find the volume of this
liquid. Give your answer to the nearest integer.
[Use
3.14]
Solution:
Volume of the preservative liquid Volume of the
Cuboid Volume of the 16 spheres.
Let r be the radius of the sphere. Its volume will be
3
4
3
r
Let l, b and h respectively be the length, breadth and
height of the rectangular box. Its volume will be
.l ×b×h
Volume of the preservative liquid
 Volume of the Cuboid Volume of the 16 spheres. 3
4
16
3
l b h πr
3
64
16 8 8 3.14 2
3
3.14
64 8 2
3



512
2.86
3

3
1464.32
cm
3
3
488.11cm
Question: 2
A storage tank is in the form of a cube. When it is full
of water, the volume of water is 15.625 m
3
. If the
present depth of water is 1.3 m, find the volume of
water already used from the tank.
Solution: Let a be the side of the cube. The volume of the cube
will be a
3
.
The volume of the cube is 15. 625 m
3
.
Therefore,
3
15.625a
3
15.625a
2.5ma
The volume of the water already used from the tank
=aa
height of the tank not filled with water
2.5 2.5 1.2
3
7.5m
Question: 3
Find the amount of water displaced by a solid
spherical ball of diameter 4.2 cm, when it is
completely immersed in water.
Solution:
Let r be the radius of the sphere. As given,
Diameter 4.2
2.1
22
r
The volume of water displaced
The volume of the sphere
3
4
3
πr
4 22
2.1 2.1 2.1
37
3
38.808cm
Question: 4
How many square metres of canvas is required for a
conical tent whose height is 3.5 m and the radius of
the base is 12 m?
Solution:
l
2
= h
2
+ r
2 l
2
= 3.5
2
+ 12
2
= 156.25
l = 12.5 m
The curved surface area of the canvas (conical shape)
of radius r and slant height
l πrl
22
12 12.5
7
3300
7
3
471.43m
Question: 5
Two solid spheres made of the same metal have
weights 5920 g and 740 g, respectively. Determine the
radius of the larger sphere, if the diameter of the
smaller one is 5 cm.
Solution: Volume of a sphere of radius
3
4
3
r πr
Let r
1
and r
2
be the radii and V
1
and V
2
be the volumes
of the large and the small sphere respectively.
Since both the spheres are made up of the same metal,
their weights will be in proportion to their volumes.
Therefore,
1
2
5920
740
V
V
3
1
3
2
4
296
3
4
37
3
πr
πr
3
1
296 5 5 5
37 2 2 2
r
33
3
1
5r
Radius of the larger sphere = 5 cm.
Question: 6 A school provides milk to the students daily in a
cylindrical glass of diameter 7 cm. If the glass is filled
with milk up to a height of 12 cm, find how many
litres of milk is needed to serve 1600 students.
Solution:
Milk required for 1600 students
 the1600 capacity of glass
1600 volumeof aglass
Volume of a cylinder of radius r and height h
2
πr h
Therefore,
Milk required for 1600 students
2
1600 πr h
22 7 7
1600 12
7 2 2
3
739200cm
739200
1000
l 739.2l
Question: 7
A cylindrical roller 2.5 m in length, 1.75 m in radius
when rolled on a road was found to cover the area of
5500 m
2
. How many revolutions did it make?
Solution:
Curved surface area of a cylinder of radius r and
height h
2πrh
Here,
1.75mr
2.5mh
The area covered by the cylinder = 5500 m
2
The area covered in 1 revolution
= The curved surface area of the cylinder
2πrh
22
2 1.75 2.5
7
2
27.5m
The number of revolution made by the cylinder
=
Area covered by thecylinder
Curved surfaceareaof thecylinder 5500
27.5
200
Question: 8
A small village, having a population of 5000, requires
75 litres of water per head per day. The village has got
an overhead tank of measurement 40 m × 25 m × 15 m.
For how many days will the water of this tank last?
Solution:
Water required per day
=Population of the village
×
Water requirement per
5000 75
375000 l
3
375000
m
1000
The volume of a rectangular tank of length l, breadth b
and height h
l b h
40 25 15
3
15000m
Required number of days tankVolumeof the water in the
Volumeof water requirement per day
15000
375
40
Question: 9
A shopkeeper has one spherical laddoo of radius 5 cm.
With the same amount of material, how many laddoos
of radius 2.5 cm can be made?
Solution:
Volume of a sphere of radius r
3
4
3
πr
Let r
1
and r
2
be the radii of the bigger and the smaller
Number of laddoos that can be made from the bigger
Volumeof the bigger laddoo 3
1
3
2
4
3
4
3
πr
πr
555
2.5 2.5 2.5


222
8
Question: 10
A right triangle with sides 6 cm, 8 cm and 10 cm is
revolved about the side 8 cm. Find the volume and the
curved surface of the solid so formed.
Solution:
The solid so formed is a cone.
Volume of a cone of radius r and height
2
1
3
h πr h
Curved surface area of a cone of radius r and slant
height
l πrl Since, the right triangle with sides 6 cm, 8 cm and 10
cm is revolved about the side 8 cm, the radius of the
cone will be 6 cm and the height will be 8 cm.
The slant height of the cone will be
22
8 6 10cm 
Therefore,
The volume of the right circular cone
2
1
3
πr h
1 22
668
37
2112
7
3
301.71cm
The curved surface area
22
6 10
7 1320
7
2
188.57cm
EXERCISE 13.4
Question: 1
A cylindrical tube opened at both the ends is made of
iron sheet which is 2 cm thick. If the outer diameter is
16 cm and its length is 100 cm, find how many cubic
centimeters of iron has been used in making the tube?
Solution:
Volume of iron used
22
πh R r
22
22
100 8 6
7 2200
8 6 8 6
7
2200
2 14
7
3
8800cm
Question: 2
A semi-circular sheet of metal of diameter 28 cm is
bent to form an open conical cup. Find the capacity of
the cup.
Solution:
The length of the arc
πr
22
14
7
 44 cm
The circumference of base of the cone
= Length of the arc
= 44 cm
Therefore, the circumference of the cone is
2 44πr
22
2 44
7
r
7cmr
The slant height (l) of the cone = 14 cm
Height of the cone,
22
= h l r
22
= 14 7
2 2 2
= 7 2 1
=7 4 1
=7 3 cm
The volume of the conical vessel
1 22
= 7 7 7 3
37
3
1078 3
cm
3
Question: 3
A cloth having an area of 165 m
2
is shaped into the
form of a conical tent of radius 5 m. (i) How many students can sit in the tent if a student,
on an average, occupies
2
5
m
7
on the ground?
(ii) Find the volume of the cone.
Solution:
The area of the cloth
=The curved surface area of the tent
So,
165πrl
22
5 165
7
l
21
m
2
l
The number of students who can sit inside
Areaof the base
Areaoccupied by astudent
2
5
7
πr
22 7
55
75 110
Height of the tent
22
h l r
2
2
21
5
2




441 100
4
341
2
Volume of air in the tent
2
1
3
πr h
1 22 341
55
3 7 2
3
275
341m
21
3
241.82m
Question: 4
The water for a factory is stored in a hemispherical
tank whose internal diameter is 14 m. The tank
contains 50 kilolitres of water. Water is pumped into
the tank to fill to its capacity. Calculate the volume of
water pumped into the tank. Solution:
The capacity of the tank
3
2
3
πr
2 22
777
37
3
2156
m
3

3
2156
kl 1m 1kl
3
Water available into the tank = 50 kl
The volume of water required to be pumped in to fill
the tank
=Capacity of the tank − Water available in the tank
2156
50
3

2156 150
3
2006
3
668.67 kl Question: 5
The volumes of the two spheres are in the ratio 64 :
27. Find the ratio of their surface areas.
Solution:
Let the volume of the first sphere be V
1
and volume of
the second sphere be V
2
.
So,
1
2
64
27
V
V
3
3
1
3
2
4
4
3
4
3
3
πr
πr




1
2
4
....... 1
3
r
r

The ratios of their surface area
2
1
2
2
4
4
πr
πr 2
1
2
r
r



2
4
3



16
9
Question: 6
A cube of side 4 cm contains a sphere touching its
sides. Find the volume of the gap in between.
Solution:
The gap can be found by subtracting the volume of the
sphere from the volume of the cube.
Volume of the cube = 4
3
Volume of the sphere
4 22
222
37
Difference in the volume
3
4 22
4 2 2 2
37 3
11
41
21




21 11
64
21



64 10
21
640
21
3
30.48cm
Question: 7
A sphere and a right circular cylinder of the same
radius have equal volumes. By what percentage does
the diameter of the cylinder exceed its height?
Solution:
Let the radius, diameter and the height of the right
circular cylinder be r, d and h respectively.
Since the volume and the radius of the sphere and the
right circular cylinder are same, therefore 32
4
3

r r h
4
3
hr
2
2
3
hr
2
3
hd
Difference between the height and the diameter
2
33
d
d h d d
Percentage by which the diameter of the cylinder
exceed its height
Difference between the height and the diameter
Height
100
3
100
2
3
50
d
d

Question: 8
30 circular plates, each of radius 14 cm and thickness
3 cm are placed one above the another to form a
cylindrical solid. Find:
(i) The total surface area
(ii) Volume of the cylinder so formed. Solution:
Let r be the radius and h be the height of the 30
combined plates.
So,
cm14 , 3 30 90r cm h
The total surface area of the combined 30 circular
plates
2 ( )r r h

22
2 14 (14 90)
7
88 104
cm
2
9152
The volume of the cylinder so formed
2
rh
22
14 14 90
7
cm
3
55440