Lesson: Surface Areas And Volumes

Exercise 12.1 (20)

Question: 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.

Solution:

(a)

The part with the pointed end is a cone, and the

rest of the part is a cylinder.

Hence, the correct option is (A).

Question: 2

A surahi is the combination of

a. a sphere and a cylinder

b. a hemisphere and a cylinder

c. two hemispheres

d. a cylinder and a cone

Solution:

(a)

A surahi is formed by combining a sphere and a

cylinder.

Hence, the correct option is (A).

Question: 3

A plumbline (sahul) is the combination of (see given

figure)

a. a cone and a cylinder

b. a hemisphere and a cone

c. frustum of a cone and a cylinder

d. sphere and cylinder

Solution:

(b)

A plumbline is a formed by combining a hemisphere

and a cone.

Hence, the correct option is (B).

Question: 4

The shape of a glass (tumbler) (see given figure) is

usually in the form of

a. a cone

b. frustum of a cone

c. a cylinder

d. a sphere

Solution:

(b)

The shape of a glass is frustum of a cone.

This is because, the radius of the upper circular part

is larger than the lower circular part.

Hence, the correct option is (B).

Question: 5

The shape of a gilli, in the gilli-danda game (see

given figure), is a combination of

a. two cylinders

b. a cone and a cylinder

c. two cones and a cylinder

d. two cylinders and a cone

Solution:

(c)

The shape of gilli, is formed by combining two

cones and a cylinder.

Hence, the correct option is (C).

Question: 6

A shuttle cock used for playing badminton has the

shape of the combination of

a. a cylinder and a sphere

b. a cylinder and a hemisphere

c. a sphere and a cone

d. frustum of a cone and a hemisphere

Solution:

(d)

A shuttle cock is formed by combining a frustum of

a cone and a hemisphere.

Hence, the correct option is (D).

Question: 7

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 a cone

b. cone

c. cylinder

d. sphere

Solution:

(a)

The part that is left after removing the conical part

has the shape of a frustum of a cone.

Hence, the correct option is (A).

Question: 8

A hollow cube of internal edge 22 cm is filled with

spherical marbles of diameter 0.5 cm and it is

assumed that space of the cube remains unfilled.

Then the number of marbles that the cube can

accommodate is

a. 142296

b. 142396

c. 142496

d. 142596

Solution:

(a)

Let the radius of the spherical marble be .

Given: Diameter of the spherical marble

0.5cm

So, radius

 

0.5



10 2







Length of the cube

 

l 22cm

Fig. Exm_12.1_8 (ii)

Let the number of marbles that can fill the cube be

Volume of part of the

x marbles

volume of the cube

 



 



 

 

 

 

 

 

3 3

4 7

x πr l

3 8

  

7 3

x l

8 4πr

  

7 22 22 22 3 7 4 4 4

8 4 22

       



 

x 7 22 11 3 7 4     

x 142296

Hence, the number of marbles, that the cube can

accommodate are 142296 marbles.

Question: 9

A metallic spherical shell of internal and external

diameters 4 cm and 8 cm, respectively is melted and

recast into the form 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

Solution:

(b)

When we recast a shape into another shape, there

is no change in its volume.

Internal diameter of shell

4 cm

So, internal radius

 



2cm

External diameter of shell

8cm

So, external radius

 



4 cm

Diameter of the base of the cone

8cm

Radius

 



4 cm

Now, find the height of the cone.

Volume remains the same during recasting so,

Volume of Volume of the

cone spherical shell

   



   

   

2 3 3

2 1

1 4 4

πr h πr πr

3 3 3

 

 

2 3 3

2 1

1 4

πr h π r r

3 3

 

 

2 3 3

2 1

r h 4 r r 

 

2 3 3

4 h 4 4 2 

 

4 h 64 8  



h 14 cm

Hence, height of the cone is 14 cm.

Question: 10

A solid piece of iron in the form of a cuboid of

dimensions , is moulded to

49cm 33cm 24cm 

form a solid sphere. The radius of the sphere is

a. 21 cm

b. 23 cm

c. 25 cm

d. 19 cm

Solution:

(a)

Length of the cuboid

 

49cm

Breadth of the cuboid

 

33cm

Height of the cuboid

 

24cm

Now, find the radius of the solid sphere.

 

Volume remains the same during moulding from

one shape to another so,

   

Volume of cuboid Volume of sphere

l b h πr

  

4 22

49 33 24 r

3 7

   

49 33 24 7 3

4 22

   





r 49 3 3 7 3    

r 7 7 3 3 7 3     

 

r 21

r 21cm

Hence, radius of the sphere is 21 cm.

Question: 11

A mason constructs a wall of dimensions

with the bricks each of

270cm 300cm 350cm 

size and it is assumed

22.5cm 11.25cm 8.75cm 

that space is covered by the mortar. Then the

number of bricks used to construct the wall is

a. 11100

b. 11200

c. 11000

d. 11300

Solution:

(b)

The space covered by the mortar in making the wall



So,

 

The space covered by the bricks

volu

me of wa1

 

 

 

 

volume of l

wal

Length of brick

 

l 22.5cm

Breadth of brick

 

b 11.25cm

Height of brick

 

h 8.75cm

Now, Length of wall

 

l 270cm

Breadth of wall

 

b 300cm

Height of wall

 

h 350cm

Let the total number of bricks used be .

Now, according to the question,

 

Volume of x bricks olume of wallV

 

 

 

 

1 1 1

x l b h l b h

      

7 270 300 350

8 22.5 11.25 8.75

  



  

7 270 300 350 10 100 100

8 225 1125 875

     



  

x 11200

Hence, the number of bricks used are 11200.

Question: 12

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

a. 4 cm

b. 3 cm

c. 2 cm

d. 6 cm

Solution:

(c)

When we recast a shape into another shape, there

is no change in its volume.

Now, diameter of base of the cylinder

2cm

Radius of base of the cylinder

 

1cm



Height of the cylinder

16cm

Now, find the radius of the sphere.

Volume of Volume of

12spheres cylinder

   



   

   

3 2

12 πR πr h

 

3 2

16 R r h  

r h





1 16





R 1

R 1cm

So, diameter of sphere

2R

2 1 

2cm

Hence, the diameter of each sphere is 2 cm.

Question: 13

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

b. 4951

c. 4952

d. 4953

Solution:

(a)

Radius of bottom

 

r 7cm

Radius of top

 

r 28cm

Slant height

 

l 45cm

 

1 2

Curved surface

πl r r

area of bucket

 

 

 

 

 

45 7 28

  

45 35

  

4950cm

Hence, curved surface area of the bucket is

4950cm

Question: 14

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

a. 0.36

b. 0.35

c. 0.34

d. 0.33

Solution:

(a)

Capsule Cylinder 2Hemsphere 

Diameter of the cylinder is same as the diameter of

the two hemispheres.

Diameter

0.5cm

Radius

 

0.5



0.25cm

Now, find the height of the cylinder.

Total length of thecapsule r r h  

2 0.25 0.25 h  

h 2 0.5 

h 1.5cm

Volume of Volume of

cylind

Volumeof

er Hemispc hapsule ere

   





 



 

 

    



2 3

πr h 2 πr

 

 

 

 

πr h r

 

 

 

 

22 4

0.25 0.25 1.5 0.25

7 3

 

    

 

 

22 25 25 15 4 25

7 100 100 10 3 100

 

    

 

 

22 1 1 3 1

7 4 4 2 3

 

   

 

 

22 1 1 9 2

7 4 4 6



 

  

 

 

22 1 1 11

7 4 4 6

 

  

 

 

121

336



0.36cm

Hence, the capacity of the capsule is .

0.36cm

Question: 15

If two solid hemispheres of same base radius are

joined together along their bases, then curved

surface area of this new solid is

4πr

6πr

3πr

8πr

Solution:

(a)

On joining the base of two hemispheres, having

equal radii, we get a new solid, which has the shape

of a sphere.

Curved surface area of sphere

4πr

Question: 16

A right circular cylinder of radius cm and height

cm just encloses a sphere of diameter

 

h 2r

a. cm

b. cm

c. cm

d. cm

Solution:

(b)

The cylinder encloses the sphere.

So, the diameter of cylinder is equal to the diameter

of the sphere.

Now, radius of cylinder

r

Diameter of cylinder

2r

Hence, diameter of circle is cm.

Question: 17

During conversion of a solid from one shape to

another, the volume of the new shape will

a. increase

b. decrease

c. remain unaltered

d. be doubled

Solution:

(c)

When a solid is converted from one shape to

another, the volume of new solid formed will

remain unaltered.

Question: 18

The diameters of the two circular ends of the

bucket are 44 cm and 24 cm. 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

Solution:

(a)

Bucket has the shape of frustum of a cone.

Diameter of the base of bucket = 24 cm

So, radius

 



12cm

Diameter of the top of bucket = 44 cm

So, radius

 



22cm

Height of the bucket

35cm

 

2 2

1 2 1 2

Volumeof thebucket πh r r rr

  

 

2 2

1 22

35 12 22 12 22

3 7

     

 

22 5

144 484 264



  

 

22 5

892





98120



32706.66

32.7litres

Hence, volume of the bucket is 32.7 litres.

Question: 19

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

Solution:

(a)

If a cross-section of the plane is cut parallel to the

base in a right circular cone, we get a circle.

Question: 20

Volumes of two spheres are in the ratio . The

64:27

ratio of their surface areas is

3:4

4:3

9:16

16:9

Solution:

(d)

Let the volumes of two spheres be .

1 2

V and V

According to the question,

V 64

V 27



πr



r 64

r 27



 



r 4

r 3



Now, find the ratio of the total surface areas of the

two spheres.

1 1

2 2

TSA 4πr



 



 

 

 



 

 



Hence, the ratio of the total surface areas of the

two spheres is .

16:9

Exercise 12.2 (8)

Write ‘True’ or ‘False’ and justify your answer in

the following:

Question: 1

Two identical solid hemispheres of equal base

radius cm are stuck together along their bases.

The total surface area of the combination is .

6πr

Solution:

The given statement is false.

On joining the base of two hemispheres, having

equal radii, we get a new solid, which has the shape

of a sphere.

Total surface area of sphere

4πr

Hence, the given statement is false.

Question: 2

A solid cylinder of radius and height is placed

over other cylinder of same height and radius. The

total surface area of the shape so formed is

4πrh 4πr

Solution:

The given statement is false.

If a solid cylinder of radius and height is placed

over other cylinder of same height and radius, then

the shape formed will also be a cylinder.

Height of new cylinder

2h

Radius of new cylinder

r

 

Total surfacearea

of new formed cylinder

2πr r 2h

 

 



 



2πr 4πrh 

4πrh 2πr 

Hence, the given statement is false.

Question: 3

A solid cone of radius and height is placed over

a solid cylinder having same base radius and height

as that of a cone. The total surface area of the

combined solid is .

2 2

πr r h 3r 2h

 

  

 

Solution:

The given statement is false.

Radius of the cone and cylinder

r

Height of the cone and cylinder

h

Curved

Curved surface

surface area

The total surface

area of cone

of cylinder

area of the

combined solid

Area of base

of cylinder

 

 



 

 

 

 



 

 

 

 

 



 

 

TSA πrl 2πrh πr  

 

 

πr l 2h r ...... 1  

Now, slant height

 

2 2

l r h 

Put the equation of slant height in equation .

 

2 2

TSA πr r h 2h r

 

   

 

Hence, the given statement is false.

Question: 4

A solid ball is exactly fitted inside the cubical box of

side . The volume of the ball is .

πa

Solution:

The given statement is false.

If a solid ball is exactly fitted inside the box having

the shape of a cube, then the diameter of the ball

will be equal to the side of cube.

Side of the cube

a

So, diameter of the ball

a

Now, radius of ball



Volume of a sphere

πr



4 a

3 2

 



 

 

πa



Hence, the given statement is incorrect.

Question: 5

The volume of the frustum of a cone is

, where is vertical height of the

2 2

1 2 1 2

πh r r rr

 

 

 

frustum and are the radii of the ends.

1 2

r ,r

Solution:

The given statement is false.

Volume of the frustum of a cone is given by the

formula,

2 2

1 2 1 2

V πh r r rr

 

  

 

Hence, the given statement is incorrect.

Question: 6

The capacity of a cylindrical vessel with a

hemispherical portion raised upward at the bottom

as shown in the given figure is .

 

πr

3h 2r



Solution:

The given statement is true.

Radius of the cylinder and the hemispherical

portion is .

r cm

Height of the cylinder

h cm

Now, find the capacity of vessel.

Capacity of Volume of Volume of

the vessel cylinder hemisphere

     

 

     

     

2 3

πr h πr

 

πr h r

 

 

 

 

 

πr

3h 2r

 

Hence, the given statement is correct.

Question: 7

The curved surface area of a frustum of a cone is

, where , are the

 

1 2

πl r r

 

1 2

l h r r 

1 2

r andr

radii of the two ends of the frustum and is the

vertical height.

Solution:

The given statement is false.

 

 

1 2

Curved surfacearea πl r r 

Slant height

   

1 2

l h r r  

Hence, the statement is incorrect.

Question: 8

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 a cone + area of

circular base + curved surface area of cylinder

Solution:

The given statement is true.

The total surface area of the metallic sheet used for

making the vessel will be equal to the curved

surface area of frustum of a cone and the total

surface area of cylinder excluding the area of its

top.

So,

Curved surface Curved surface

area of frustum area of cylinder

Total surface

area of vessel

Area of thebase

of cylinder

 

   



 

   

 

   

 



 

 

 

 



 

 

Hence, the statement is correct.

Exercise 12.3 (14)

Question: 1

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.

Solution:

According to the question,

Length of the side of first cube

3cm

   

Volume of fir Vst cube 3

27cm

Length of the side of second cube

4 cm

   

Volume of seco Vnd cube 4

64 cm

Length of the side of third cube

5cm

   

Volume of thi Vrd cube 5

125cm

Let the side of the new cube be .

 

1 2 3

Volume of the new cu e V Vb V V  

 

a 27 64 125  

a 216

3 3

a 6

Hence, the side of the new cube is 6 cm.

Question: 2

How many shots each having diameter 3 cm can be

made from a cuboidal lead solid of dimensions

9cm 11cm 12cm 

Solution:

Length of the cuboidal lead

 

l 12cm

Breadth of the cuboidal lead

 

b 11cm

Height of the cuboidal lead

 

h 9cm

Diameter of the shots

3cm

So, radius of shots



1.5cm

The spherical shots are made from the cuboidal

lead.

Let the number of shots made be .

   

Volume of x shots Volume of cuboidal lead

x πr l b h

   

 

4 22

x 1.5 12 11 9

3 7

     

12 11 9 3 7

4 22 1.5 1.5 1.5

   



   

3 9 3 7 10 10 10

2 15 15 15

     



  

x 3 7 2 2   

x 84

Hence, the total number of lead shots which can be

made are 84.

Question: 3

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.

Solution:

Let the radius of bottom and top of the bucket be

.and respectively and let the height of the bucket

be .

Now, according to the question,

r 21cm

r 28cm

Volume of the bucket = 28490

Now, find the volume of the frustum of a cone.

2 2

1 2 1 2

V πh r r rr

 

  

 

2 2

1 22

28490 h 21 28 21 28

3 7

 

     

 

 

1 22

28490 h 441 784 588

3 7

    

 

1 22

28490 h 1813

3 7

  

28490 3 7

1813 22

 





h 15

Hence, the height of the bucket is 15 cm.

Question: 4

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.

Solution:

Let the original cone be OAB.

Radius of AOB = 8 cm

 

Height of AOB = 12 cm

 

Now, from the figure,

New cone formed is DEO.

Let the radius of DEO be .

Height of DEO

 



6cm

New frustum formed is DABE.

Radius of its top is cm

 

r 4

Radius of its bottom cm

 

r 8

Height = 6 cm

 

Now, (By similarity criterion)

ΔOAB||ΔODE

So,

1 1

2 2

r h



8 h



r 1

8 2



2r 8

r 4 cm

Now, find the required ratio.

2 2

2 1 2 1 2

1 1

πh r r rr

Volumeof frustum

Volumeof cone

πr h

 

 

 



2 2

6 4 8 4 8

4 6

 

   

 





 

16 64 32

 



112



Hence, the required ratio is .

7:1

Question: 5

Two identical cubes each of volume 64 are

joined together end to end. What is the surface area

of the resulting cuboid?

Solution:

Let the edge of two identical cubes be .

Volume of the cube

64 cm

So,

a 64

 

a 4

Now, according to the question,

Length of the cuboid

2a

Breadth of cuboid

a

Height of cuboid

a

Now, find the surface area of the cuboid.

 

Surface area 2 lb lh bh  

 

2 2a a 2a a a a     

2 2 2

2 2a 2a a

 

  

 

 

2a 2 2 1  

 

2a 5

   

2 4 5 

160cm

Hence, the surface area of cuboid formed is 160

Question: 6

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.

Solution:

Let the side of the cube be .

a 7cm

Height of the conical cavity

 

h 7cm

Radius of the conical cavity

 

r 3cm

Now, find the volume of the remaining solid.

Volume of Volume of Volume of

–

remaining solid cube conical cavity

     



     

     

3 2

a πr h

 

3 2

1 22

7 3 7

3 7

    

343 66 

277cm

Hence, volume of the remaining solid is 277 .

Question: 7

Two cones with same base radius 8 cm and height

15 cm are joined together along their bases. Find

the surface area of the shape so formed.

Solution:

Radius of the two cones

 

r 8cm

Height of the two cones

 

h 15cm

 

2 2

Slant heig l h rht  

2 2

15 8 

225 64 

289

17cm

On joining two identical cones base to base, the

total surface area of new shape formed, will be

equal to the sum of curved surface areas of both the

cones.

Total surface area Curved surface

of new shape area of cone

   

 

   

   

 

TSA 2 πrl 

2 8 17

 

   

 

 

5984



854.857cm

Hence, surface area of the new solid is 854.857 .

Question: 8

Two solid cones A and B are placed in a cylindrical

tube as shown in the given figure. The ratio of their

capacities is . Find the heights and capacities of

2:1

cones. Also, find the volume of the remaining

portion of the cylinder.

Solution:

Let the heights of two cones be

1 2

h andh

respectively.

Now, both the cones have same diameter, which is

equal to 6 cm.

So, radii

 



3cm

According to the question,

The ratio of the capacities of two cones is .

2:1

So,

2 2

πr h



 

3 h



2 h

1 h



1 2

h 2h

Now, the length of the cylinder is 21 cm.

So,

 

1 2

h h 21 ...... 1 

Now,

2 2

2h h 21 

3h 21



h 7cm

Put the value of in equation .

 

1 2

h h 21 

h 7 21 

h 21 7 

h 14 cm

Now, find the volume of both the cones separately.

 

1 1

Volume of first

V πc

e ron h

 

1 22

3 14

3 7

   

22 3 2  

132cm

 

2 2 2

Volume of second

V πc

e ron h

 

1 22

3 7

   

22 3 

66cm

Volume of cylinder πr h

3 21

  

22 3 

594 cm

Now, find the capacity of the remaining portion of

the tube.

Volume of Volume of Sum of volume

remaining portion cylinder of two cones

     

 

     

     

 

594 132 66  

594 198 

396cm

Hence, the volume of remaining portion is 396 .

Question: 9

An ice cream cone full of ice cream having radius 5

cm and height 10 cm as shown in the given figure.

Calculate the volume of ice cream, provided that its

part is left unfilled with ice cream.

Solution:

An ice-cream cone is a combination of a cone and a

hemisphere.

Radii of both the hemisphere and the cone are same

and is equal to 5 cm.

Total height of the ice-cream is 10 cm.

Height of cone

10 5 

5cm

According to the question,

part is left unfilled with ice cream.

 

Volume of ice-cr

eam

 

 

 

 

 

Volumeof cone Volumeof hemisphere

 

Volume of cone πr h



1 22

5 5

3 7

   

1 22

125

3 7

  

130.952cm

Volume of hemisphere πr



2 22

3 7

  

2 22

125

3 7

  

261.904 cm

Volume of ice-cream

 

130.952 261.904

 

392.856

 

327.38cm

Hence, the volume of ice-cream is 327.4 .

Question: 10

Marbles of diameter 1.4 cm are dropped into a

cylindrical beaker of diameter 7 cm containing

some water. Find the number of marbles that

should be dropped into the beaker so that the water

level rises by 5.6 cm.

Solution:

Let the number of marbles dropped in the beaker be

Now, diameter of the cylinder is 7 cm.

So, radius

 



3.5cm

Height of water raised

 

h 5.6cm

Volume of water raised πr h

22 7

5.6

7 2

 

  

 

 

11 7

  

215.6cm

Volume of spherical marbles

πR



4 22 1.4

3 7 2

 

  

 

 

4 22 14

3 7 2 10

 

  

 



 

4 22 14 14 14

3 7 2 10 2 10 2 10

 

  

    

22 14 14

3 10 10 10



 

 

1.437cm

So,

   

Volume of water raised Volume of x marbles

 

215.6 x 1.437

215.6

1.437



x 150

Hence, the total number of marbles dropped is 150.

Question: 11

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.

Solution:

Length of the cuboid

 

l 66cm

Breadth of the cuboid

 

b 42cm

Height of the cuboid

 

h 21cm

Let the number of spherical lead shots obtained be

Diameter of the sphere

4.2cm

So, radius of the sphere

 

4.2



2.1cm

Now, according to the question,

   

Volumeof cuboid Volumeof x spherical shot

l b h x πr

   

 

4 22

66 42 21 x 2.1

3 7

     

66 42 21 3 7

4 22 2.1 2.1 2.1

   



   

66 42 21 3 7 10 10 10

4 22 21 21 21

      



   

x 3 5 10 10   

x 1500

Hence, the number of shots obtained are 1500.

Question: 12

How many spherical lead shots of diameter 4 cm

can be made out of a solid cube of lead whose edge

measures 44 cm.

Solution:

Let the number of spherical shots casted be .

Length of the solid cube

 

a 44 cm

Diameter of lead shots

4 cm

So, radius

 



2cm

Now, according to the question,

   

Volume of cube Volume of x spheres

3 3

a x πr

 

   

3 3

4 22

44 x 2

3 7

   

44 44 44 3 7

4 22 2 2 2

   



   

x 11 11 3 7   

x 2541

Hence, the total number of spherical lead shots are

2541.

Question: 13

A wall 24 m long, 0.4 m thick and 6 m high is

constructed with the bricks each of dimensions

. If the mortar occupies of

25cm 16cm 10cm 

the volume of the wall, then find the number of

bricks used in constructing the wall.

Solution:

The space covered by the mortar in making the wall



So, the space covered by the bricks

 

 

 

 

volume of

wall

Length of brick

 

l 25cm 0.25m 

Breadth of brick

 

b 16cm 0.16m 

Height of brick

 

h 10cm 0.10m 

Now, length of wall

 

l 24m

Breadth of wall

 

b 0.4m

Height of wall

 

h 6m

Let the total number of bricks used be .

Now, according to the question,

 

Volume of xbricks Volume of wall

 

 

 

 

1 1 1

x l b h l b h

      

x 0.25 0.16 0.10 24 0.4 6

      

x 0.004 57.6

  

x 0.004 57.6

  

9 57.6

10 0.004







9 576 10

 



x 12960

Hence, the number of bricks used are 12960.

Question: 14

Find the number of metallic circular disc 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

diameter 4.5 cm.

Solution:

Let the number of discs to be melted be .

Diameter of disc

1.5cm

So, radius

 

1.5



Height of disc

 

H 0.2cm

Now, diameter of cylinder

4.5cm

So, radius

 

4.5



Height of disc

 

h 10cm

Now, according to the question,

   

Volume of x discs Volume of cylinder

2 2

x πR H πr h 

πr h

πR H



r h

R H



0.2

 



 

 



 



 

 

9 9 10 10

3 3 2

  



 

x 9 5 10  

x 450

Hence, the number of metallic circular discs to be

melted are 450.

Exercise 12.4 (20)

Question: 1

A solid metallic hemisphere of radius 8 cm is melted

and recasted into a right circular cone of base

radius 6 cm. Determine the height of the cone.

Solution:

Let the height of the cone be .

Radius of hemisphere

 

R 8cm

Radius of cone

 

r 6cm

Now, find the height of cone.

Volume of Volume

hemisphere of cone

   



   

   

2 3

1 2

πr h πR

3 3



   

2 3

6 h 2 8 

2 8 8 8

6 6

  





2 2 8 8

3 3

  





h 28.44 cm

Hence, the height of the recasted right circular cone

is 28.44 cm.

Question: 2

A rectangular water tank of base

11m 6m

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.

Solution:

Let the height of the water level in cylindrical tank

be .

Length of rectangular water tank

 

l 11m

Breadth of rectangular water tank

 

b 6m

Height of water level in rectangular tank

 

H 5m

Radius of cylindrical tank

 

r 3.5m

Now, according to the question, water from the

rectangular tank is transferred to the cylindrical

tank.

Volume of water Volume of water

in cylindrical tank in rectangular tank

   



   

   

lr h bπ H  

 

3.5 h 11 6 5

    

330 7

22 3.5 3.5





 

330 7 10 10

22 35 35

  



 

6 10





h 8.571m

Hence, the height of the water level in the tank is

8.58 m approx.

Question: 3

How many cubic centimetres of iron is required to

construct an open box whose external dimensions

are and provided the thickness

36cm,

25cm

16.5cm

of the iron is 1.5 cm. If one cubic cm of iron weighs

7.5 g, find the weight of the box.

Solution:

Given: External dimensions of the box are as

follows.

Length

 

l 36cm

Breadth

 

b 25cm

Height

 

h 16.5cm

Now, calculate the internal dimensions of the box.

Length

 

l 36 1.5 1.5  

33cm

Breadth

 

b 25 1.5 1.5  

22cm

Height

 

h 16.5 1.5 

15cm

Now, according to the question,

1 1 1 2 2 2

Volume of iron

in the open

l bh

box

l b h

 

 

 

 

 

36 25 16.5 33 22 15     

14850 10890 

3960cm

Weight of one cubic cm of iron = 7.5 g

So, weight of , iron

3960cm

7.5 3960 

29700

29700

1000



29.7Kg

Hence, the open box weighs 29.7 kg.

Question: 4

The barrel of a fountain pen, cylindrical in shape, is

7 cm long and 5 mm in diameter. A full barrel of ink

in the pen is used up on writing 3300 words on an

average. How many words can be written in a bottle

of ink containing one fifth of a litre?

Solution:

Let the barrel of ink be filled times.

Height of the cylindrical barrel

 

h 7cm

Diameter of the cylindrical barrel

5mm

So, radius

 

2 10







Now, according to the question,

   

x Volumeof barrel Volumeof ink used 

 

x πr h onelitre

  

22 1 1

x 7 1000cm

7 4 5

 

    

 

 

x 22 200

 

  

 

 

200 4 4

 



100 4 4

 



1600



A full barrel of ink in the pen can write 3300 words.

So, find the number of words can be written by

number of barrels.

number of barrels can write words

3300 x 

1600

3300

 

480000

Hence, the number of words that can be written are

480000.

Question: 5

Water flows at the rate of 10 m/minute through a

cylindrical pipe 5 mm in diameter. How long would

it take to fill a conical vessel whose diameter at the

base is 40 cm and depth 24 cm?

Solution:

Water flows through a cylindrical pipe into the

conical vessel.

Diameter of the pipe = 5 mm

So, radius of pipe

 

r mm



2000



Rate of flow of water

 

v 10m / min

m/sec



= m/sec

 

Now, volume of flowing water V A v t  

Here, is the area of the base and let time be .

Now, diameter of conical vessel = 40 cm

So, radius

 

R cm



200



Height of conical vessel

 

h 24 cm

100



According to the question,

Volume of water Volume of

in conical vessel flowing water

   



   

   

πR h A v t

  

 

2 2

πR h πr v t

  

2 2

R h r v t

  

2 2

1 1 24 5 1

3 5 100 2000 6

   

   

   

   

2 2

1 8 5 1

5 100 2000 6

   

   

   

   

1 1 8 2000 2000 6

5 5 100 5 5

    



   

8 2000 20 6

5 5 5 5

  



  

t 8 16 4 6   

8 16 4 6

t minutes

  



8 16 4

t minutes

 



512

t minutes



t 51.2minutes

Now,

t 51minutes 0.2minutes 

t 51minutes 0.2 60seconds  

t 51minutes 12seconds 

Hence, it will take 51 minutes and 12 seconds to fill

the conical vessel.

Question: 6

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?

Solution:

According to the question,

Heap of rice has the shape of a cone.

Diameter of cone = 9 m

So, radius

 

r m



Height of the cone

 

h 3.5m

Now, find the volume of cone.

Volumeof cone πr h



1 22 9

3.5

3 7 2

 

   

 

 

1 22 9 9 35

3 7 2 2 10

    

3 9

2 2

  

74.25m

Now, find the quantity of canvas required to cover

the heap of rice.

It is equal to the curved surface area of the conical

heap of rice.

 

2 2

Slant height l r h 

 

3.5

 

 

 

 

   

2 2

4.5 3.5 

20.25 12.25 

32.5

5.7m

 

Curved surfacearea CSA πrl

22 9

5.7

7 2

  

9 5.7

  

80.61m

Hence, the area of canvas cloth required to cover

the heap is .

80.61m

Question: 7

A factory manufactures 120000 pencils daily. The

pencils are cylindrical in shape each of length 25 cm

and circumference of base as 1.5 cm. Determine the

cost of colouring the curved surfaces of the pencils

manufactured in one day at Rs 0.05 per .

Solution:

According to the question,

The pencils manufactured are cylindrical in shape.

Height of the cylinder = 25 cm

Circumference of the base = 1.5 cm

 

2πr

Now, find the curved surface area of one pencil.

 

Curved surface CSR ar a he 2πr

 

2πr h

1.5 25 

37.5cm

Curved surface area

of 12000

120000 37.5

0pencils

 

 

 

 

4500000

4500000

100



45000dm

Cost of colouring 1 curved surfaces of the

pencils = Rs 0.05

So, cost of colouring, 45000 curved surfaces of

the pencils

45000 0.05 

450 5 

2250

Hence, cost of colouring the pencils per day is Rs

2250.

Question: 8

Water is flowing at the rate of 15 km/h through a

pipe of diameter 14 cm into a cuboidal pond which

is 50 m long and 44 m wide. In what time will the

level of water in pond rise by 21 cm?

Solution:

Water flows through a cylindrical pipe into the

cuboidal pond.

Diameter of the pipe = 14 cm

So, radius of pipe

 



7cm

0.07m

Rate of flow of water

 

v 15km / hour

15000m / hour

 

Volume of flowin Vg water A v t  

Here, is the area of the base and let time be .

Now, length of cuboidal pond

 

l 50m

Breadth of cuboidal pond

 

b 44m

Height of cuboidal pond

 

h 21m

According to the question,

   

Volume of water in pond Volume of flowing water

l b h A v t    

 

l b h πr v t    

 

50 44 21 0.07 15000 t

     

50 44 0.21 7

22 0.07 0.07 15000

  



  

50 44 21 7 100 100

22 7 7 15000 100

    



   

50 44 21

22 7 15 10

 



  

5 2 3

 



t 2hours

Hence, the level of water in pond will rise in 2

hours.

Question: 9

A solid iron cuboidal block of dimensions

is recast into a hollow cylindrical

4.4m 2.6m 1m 

pipe of internal radius 30 cm and thickness 5 cm.

Find the length of the pipe.

Solution:

Length of the cuboidal block

 

l 4.4m

Breadth of the cuboidal block

 

b 2.6m

Height of the cuboidal block

 

h 1m

Now, let the height of pipe be .

Internal radius of the pipe

 

r 30cm

0.3m

Thickness of the pipe

 

t 5cm

0.05m

So, external radius

 

r 0.3 0.05 

0.35m

Now, find the height of the cylindrical pipe.

According to the question,

   

Volume of cuboid Volume of cylindrical pipe

 

2 2

2 1

l b h πH r r   

   

 

2 2

4.4 2.6 1 H 0.35 0.30

     

   

 

2 2

4.4 2.6 7

22 0.35 0.30

 



 

 

44 26 7

22 0.1225 0.09 10 10

 



   

2 26 7

0.0325 10 10

 



 

2 26 7 10000

325 10 10

  



 

2 26 7 100

325

  



H 112m

Hence, the length of the hollow cylindrical pipe is

112 m.

Question: 10

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 ?

0.04m

Solution:

The pond is cuboidal in shape.

So, length of the pond

 

l 80m

Breadth of the pond

 

b 50m

Now, let the rise of water level be .

According to the question,

Average displacement of the water by one person

0.04m

Total number of persons taking a dip into the pond

= 500

Now, total volume of water displaced

500 0.04 

500 4

100





20m

Volume of Total volume of

the pond water displaced

   



   

   

l b h 20  

80 50 h 20  

50 80





0.005

0.005 100 

0.5cm

Hence, the water level rises by 0.5 cm.

Question: 11

16 glass spheres each of radius 2 cm are packed into

a cuboidal box of internal dimensions

and then the box is filled with

16cm 8cm 8cm 

water. Find the volume of water filled in the box.

Solution:

Radius of one sphere = 2 cm

 

Length of cuboidal box

 

l 16cm

Breadth of cuboidal box

 

b 8cm

Height of cuboidal box

 

h 8cm

Now, according to the question,

Volume of Volume Volume of

water filled of cuboid 16 spheres

     

 

     

     

l b h 16 πr

 

    

 

 

 

4 22

16 8 8 16 2

3 7

 

      

 

 

4 22

16 8 8 16 8

3 7

 

      

 

 

16 8 8 1

 

   

 

 

21 11

1024



 



 

 

1024

 



 

 

487.6cm

Hence, the volume of water filled in the box is

487.6cm

Question: 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.

Solution:

It is given that the milk container has the shape of a

frustum of a cone.

So, height of frustum

 

h 16cm

Lower radius of frustum

 

r 8cm

Upper radius of frustum

 

r 20cm

Volume of milk Volume of

in a container frustum

   



   

   

2 2

1 2 1 2

πh r r rr

 

  

 

2 2

1 22

16 8 20 8 20

3 7

 

     

 

 

1 22

16 64 400 160

3 7

    

1 22

16 624

3 7

   

16 208

  

10459.4 cm

10.459litre

Cost of 1 litre of milk = Rs. 22

So, cost of milk

10.459litres

22 10.459 

230.098

Hence, the cost of milk which the container can

hold is Rs. 230.098.

Question: 13

A cylindrical bucket of height 32 cm and base radius

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.

Solution:

According to the question,

A cylindrical bucket is emptied, and the heap of

sand forms a cone.

Height of the cylinder

 

H 32cm

Radius of cylinder

 

R 18cm

Let the radius and slant height of the cone be

respectively.

r andl

Height of cone

 

h 24 cm

   

Volume of cylinder Volume of cone

2 2

h hπ πr

r 

2 2

π 18 32 πr 24

   

2 2

18 32 r 24

  

18 18 32 3

  



18 18 32

 



r 18 18 4  

r 18 18 2 2   

 

r 18 2 

r 36cm

Now, find the value of slant height.

 

2 2

Slant heig l r hht  

2 2

l 36 24 

1296 576 

1872

43.266cm

Hence, the radius and slant height of the conical

heap are 36 cm and 43.266 cm respectively.

Question: 14

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 [Use = 3.14].

Solution:

The shape of the rocket is the combination of a

cylinder and a cone.

Let the height of the cone be .

Diameter of cone and cylinder is same and is equal

to 6 cm.

So, radius

 



3cm

Height of the cylinder

 

H 12cm

Now, slant height of cone

 

l 5cm

 

2 2

Slant height l r h 

2 2

5 3 h 

 

2 2

5 3 h 

25 9 h 

h 25 9 

h 16

h 4

Now, find the volume of rocket.

Volume Volume of Volume

of rocket cylinder of cone

     

 

     

     

2 2

πr H πr h

 

πr H h

 

 

 

 

3.14 3 12 4

 

   

 

 

3.14 3 12

 

  

 

 

36 4

3.14 3



 

 

 

 

3.14 3

 

 

 

 

3.14 3 40  

376.8cm

Now, find the total surface area of the rocket.

 

Curved Surface

Area of cone

Total Surface

Area of the

Curved Surface

base of

rocket TSA

Area of cylinder

cylinder

 

 

 

 

 



 

 

 

 

 

 

 

 

 

 

 

πrl 2πrH πr  

 

πr l 2H r  

 

3.14 3 5 2 12 3    

 

3.14 3 5 24 3   

 

3.14 3 32 

3.14 96 

301.44 cm

Hence, the volume and surface area of rocket are

and respectively.

376.8cm

301.44 cm

Question: 15

A building is in the form of a cylinder surmounted

by a hemispherical vaulted dome and contains

of air. If the internal diameter of dome is

41 m

equal to its total height above the floor, find the

height of the building?

Solution:

According to the question,

Building is formed by the combination of a cylinder

and a hemisphere.

Volume of air in the building =

41 m

So, Volume of building = Volume of air

Let the radii of both the cylinder and hemisphere be

Also, let the heights of the cylinder and the building

be respectively.

h andH

Now, diameter of the dome

2r

It is given that internal diameter of dome is equal to

total height of building.

So,

H 2r

Height of Height of Height of

–

cylinder building hemisphere

     



     

     

h H r 

2r r 

r

Volume of Volume of Volume of

building cylinder hemisphere

     

 

     

     

2 3

19 2

41 πr h πr

21 3

 

880 2

πr h r

21 3

 

 

 

 

 

880 22 2

r r r h r

21 7 3

 

  

 

 

880 22 5

21 7 3

 



 

 

880 7 3

21 22 5

 



 

r 8

r 2

Now, height of the building

 

H 2r

2 2 

4m

Hence, height of the building is 4 m.

Question: 16

A hemispherical bowl of internal radius 9 cm is full

of liquid. The liquid is to be filled into cylindrical

shaped bottles each of radius 1.5 cm and height 4

cm. How many bottles are needed to empty the

bowl?

Solution:

Let the number of bottles required to empty the

bowl be .

Radius of the bowl

 

R 9cm

Now, radius of cylindrical bottle

 

r 1.5cm

Height of the bottle

 

h 4 cm

Now, according to the question,

Volume of Volume of x

bowl number of bottles

   



   

   

3 2

πR x πr h

 

π R

πr h





3r h



 

2 9

3 1.5 4





 

3 9 9

1.5 1.5 2

 



 

3 9 9 10 10

15 15 2

   



 

x 54

Hence, the total number of bottles required are 54.

Question: 17

A solid right circular cone of height 120 cm and

radius 60 cm is placed in a right circular cylinder

full of water of height 180 cm such that it touches

the bottom. Find the volume of water left in the

cylinder, if the radius of the cylinder is equal to the

radius of the cone.

Solution:

Height of the cone

 

h 120cm

Radius of the cone

 

r 60cm

Now, radius of cylinder is same as that of cone.

So, radius of cylinder

 

r 60cm

Height of cylinder

 

H 180cm

According to the question,

Cone is placed inside the cylinder having water up

to the brim. So, volume of water displaced by the

cylinder will be equal to the volume of cone.

Now, find the volume of water left in the cylinder.

Volume of Volume of Volume

–

water left cylinder of cone

     



     

     

2 2

πr H πr h

 

πr H h

 

 

 

 

   

22 1

60 180 120

7 3

 

  

 

 

 

60 60 180 40

   

60 60 140

   

22 60 60 20   

1584000cm

1584000



1.584m

Hence, volume of water left in the cylinder is

1.584m

Question: 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 tank in half an

hour?

Solution:

Water flows through a cylindrical pipe into the

cylindrical tank.

Inner radius of pipe

 

r 1cm

Rate of flow of water

 

v 80cm / sec

Now, area of the base of cylinder

 

A πr

Time

 

t hour



60 60

  

1800seconds

Now, let the height of the cylindrical tank be .

Radius of the bases

 

R 40cm

According to the question,

Volume of Volume of

water in tank flowing water

   



   

   

πR h A v t  

 

2 2

πR h πr v t  

2 2

R h r v t  

   

2 2

40 h 1 80 1800  

80 1800

40 40







2 180





h 90cm

Hence, in half an hour, the water level rises by 90

cm in the tank.

Question: 19

The rain water from a roof of dimensions

drains into a cylindrical vessel having

22m 20m

diameter of base 2 m and height 3.5 m. If the rain

water collected from the roof just fill the cylindrical

vessel, then find the rainfall in cm.

Solution:

The roof has the shape of a cuboid.

Let the height of cuboid be , which is equal to

h cm

the rainfall.

Length of cuboid

 

l 22m

2200cm

Breadth of cuboid

 

b 20m

2000cm

Now, water drains into a cylindrical vessel.

Diameter of cylinder = 2 m

So, its radius

 



1m

100cm

Height of cylinder

 

H 3.5m

350cm

   

Volume of cuboid Volume of cylinder

l b h πr H  

 

2200 2000 h 100 350

    

22 100 100 350

7 2200 2000

  



 

22 35

7 22 2





 



h 2.5cm

Hence, the measurement of rainfall is 2.5 cm.

Question: 20

A pen stand made of wood is in the shape of a

cuboid with four conical depressions and a cubical

depression to hold the pens and pins, respectively.

The dimension of the cuboid are 10 cm, 5 cm and 4

cm. The radius of each of the conical depressions is

0.5 cm and the depth is 2.1 cm. The edge of the

cubical depression is 3 cm. Find the volume of the

wood in the entire stand.

Solution:

The pen stand has the shape of a cuboid with four

conical depressions and a cubical depression.

Length of the cuboid

 

l 10cm

Breadth of the cuboid

 

b 5cm

Height of the cuboid

 

h 4 cm

 

Volume of cub V lo d bi h  

10 5 4  

200cm

Height of conical depression

 

h 2.1cm

Radius of conical depression

 

r 0.5cm

 

Volume of conical depres

sio hn V πr



 

1 22

0.5 2.1

3 7

   

1 22 5 5 21

3 7 10 10 10

    

2 10







Edge of cubical depression

 

a 3cm

 

Volume of cubical depressio Vn a

 

3

27cm

Now, according to the question,

Volume Volume of 4

–

of cuboid conical depression

Volume

of wood

Volume of cubical

–

depression

 

   

 

   

 

   

 



 

 

 

 

 

 

1 2 3

V 4 V V  

200 4 27

   

200 27

  

1000 11 135

 



854



170.8cm

Hence, the volume of the wood in the entire stand

is .

170.8cm