CHAPTER – 14

SURFACE AREAS AND VOLUMES

Exercise – 14.1

Page No 14.27:

Question 1: How many balls, each of radius I cm, can be made from a

solid sphere of lead of radius 8 cm?

ANSWER:

We are given a solid sphere with radius R = 8 cm.

From this sphere we have to make spherical balls of radius r= 1 cm.

Let the no. of balls that can be formed be .

We know,

Volume of a sphere = 4

3𝜋𝑟3.

So, volume of the bigger solid sphere = 4

3𝜋(8)3 …… (a)

Volume of one smaller spherical ball = 4

3𝜋(1)3 …… (b)

We know, the volume of the solid sphere should be equal to the sum of

the volumes of the n spherical balls formed.

So, using (a) and (b), we get,

⇒ 𝑛 ×4

3𝜋(1)3 =

4

3𝜋(8)3

⇒ 𝑛 ×4

3𝜋(1)3 =

4

3𝜋(8)3

Therefore, 𝑛 = (8)3

𝑛 = 512

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Hence, the no. of balls of radius r =1 that can be formed out of solid sphere

of radius R = 8 is 512.

Page No 14.27:

Question 2: How many spherical bullets each of 5 cm in diameter can be

cast from a rectangular block of metal 11dm × 1 m × 5 dm?

ANSWER:

We are given a metallic block of dimension 11dm × 1 m × 5 dm

We know that, 1 dm = 10–1 dm

So, the volume of the given metallic block is

= 11 × 10−1 × 1 × 5 × 10−1

= 55 × 10−2 m³

We want to know how many spherical bullets can be formed from this

volume of the metallic block. It is given that the diameter of each bullet

should be 5 cm.

We know,

Volume of a sphere =4

3𝜋(𝑟)3

Here, 𝑟 = 25 × 10−3 m

Let the no. of bullets formed be n.

We know that the sum of the volumes of the bullets formed should be

equal to the volume of the metallic block.

⇒ 55 × 10−2 = 𝑛 ×4

3×

22

7× (25 × 10−3)3

𝑛 =55×3×7×10−2

4×22×25×25×25×10−9

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=21×107

(2×5)3×25

=21×107

103×25

= 8400

Hence the no. of bullets that can be formed is 8400.

Page No 14.27:

Question 3: A spherical ball of radius 3 cm is melted and recast into three

spherical balls. The radii of the two of the balls are 1.5 cm and 2 cm

respectively. Determine the diameter of the third ball.

ANSWER:

We have one spherical ball of radius 3 cm

So, its volume =4

3𝜋(3)3 …… (a)

It is melted and made into 3 balls.

The first ball has radius 1.5 cm

So, its volume =4

3𝜋(1.5)3 …… (b)

The second ball has radius 2 cm

So, its volume =4

3𝜋(2)3 …… (c)

We have to find the radius of the third ball.

Let the radius of the third ball be 𝑟

The volume of this third ball =4

3𝜋(𝑟)3 …… (d)

We know that the sum of the volumes of the 3 balls formed should be

equal to the volume of the given spherical ball.

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Using equations (a), (b), (c) and (d)

4

3𝜋(𝑟)3 +

4

3𝜋(1.5)3 +

4

3𝜋(2)3 =

4

3𝜋(3)3

⇒ (𝑟)3 + (1.5)3 + (2)3 = (3)3

𝑟3 = 27 − 8 −27

8

𝑟3 =7×27−64

8

𝑟3 =125

8

⇒ 𝑟 =5

2= 2.5 cm

Hence the diameter of the third ball should be 5 cm

Page No 14.27:

Question 4: 2.2 cubic dm of brass is to be drawn into a cylindrical wire

0.25 cm in diameter. Find the length of the wire.

ANSWER:

The brass volume that has to be drawn into a cylindrical wire is given is

2.2 dm³ = 2.2 × 10−3 m³

We have to make a cylindrical wire out of it with diameter = 0.25 cm

So the radius of this wire 0.125 × 10−3 m

We have to find the length of this wire.

Let the length of this wire be

We know that the volume of a cylinder = 𝜋𝑟2ℎ.

We know, the volume of the cylinder should be equal to the volume of the

given brass

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⇒ 𝜋(0.125 × 10−2)2 × ℎ = 2.2 × 10−3

ℎ =22×10−3×104

𝜋×.125×.125

=22×7

22×.125×.125

=4×7

.25×.25

= 448

Therefore, h = 448 m

Hence, the length of the cylindrical wire that can be formed is 448 m

Page No 14.28:

Question 5: What length of a solid cylinder 2 cm in diameter must be

taken to recast into a hollow cylinder of length 16 cm, external diameter

20 cm and thickness 2.5 mm?

ANSWER:

We are given a solid cylinder of, diameter = 2 cm

We have to recast it into a hollow cylinder of length = 16 cm

External Diameter = 20 cm and thickness = 2.5 mm = 0.25 cm

We have to find the height of the solid cylinder that can be used to get a

hollow cylinder of the desired dimensions.

Volume of a solid cylinder = 𝜋𝑟2ℎ

So,

The volume of the given solid cylinder = 𝜋(1)2ℎ …… (a)

Here, height ℎ has to be found.

Volume of a hollow cylinder = 𝜋ℎ(𝑅2 − 𝑟2)

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Where is the external radius and is the internal radius

External radius is given. Thickness of the hollow cylinder is also given.

So, we can find the internal radius of the hollow cylinder.

Thickness = R – r

⇒ 0.25 = 10 − 𝑟

⇒ 𝑟 = 9.75 cm

So, the volume of the hollow cylinder = 𝜋 × 16 × (100 − 95.0625)

…… (b)

From (a) and (b) we get,

𝜋(1)2ℎ = 𝜋 × 16 × (100 − 95.0625)

𝜋ℎ = 𝜋 × 16 × (100 − 95.0625)

ℎ = 16 × (4.9375)

ℎ = 79 cm

Hence, the required height of the solid cylinder is h = 79 cm

Page No 14.28:

Question 6: A cylindrical vessel having diameter equal to its height is full

of water which is poured into two identical cylindrical vessels with

diameter 42 cm and height 21 cm which are filled completely. Find the

diameter of the cylindrical vessel.

ANSWER:

A cylindrical vessel whose height is equal to its diameter is given.

It is filled with water.

We know that the volume of a cylinder = 𝜋𝑟2ℎ

In this particular case,

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Height is equal to the diameter, that is ℎ = 2𝑟,

The volume of cylindrical vessel becomes = 2𝜋𝑟3

The water from this vessel is transferred into two identical cylindrical

vessels of

Diameter = 42 cm and, height h = 21 cm

Volume of each vessel = 𝜋(21)2 × 21

We know that the sum of the volumes of the two identical vessels must be

equal to the volume of the given cylindrical vessel.

⇒ 2𝜋𝑟3 = 2 × (𝜋(21)2 × 21)

𝑟3 = (21)3

Therefore, 𝑟 = 21

The diameter of the given cylinder is 42 cm

Page No 14.28:

Question 7: 50 circular plates each of diameter 14 cm and thickness 0.5

cm are placed one above the other to form a right circular cylinder. Find

its total surface area.

ANSWER:

We have 50 circular plates, each with diameter = 14 cm

That is, radius = 7 cm and thickness = 0.5 cm

These plates are stacked on top of one another.

So, the total thickness = 0.5 × 50 cm = 25 cm

This is clearly a cylindrical arrangement.

We know,

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Total surface area of a cylinder = 2𝜋𝑟ℎ + 2𝜋𝑟2

= 2𝜋𝑟(ℎ + 𝑟)

= 2𝜋 × 7(25 + 7)

= 448𝜋

= 1408

So, the total surface area of the given arrangement is 1408 cm²

Page No 14.28:

Question 8: 25 circular plates, each of radius 10.5 cm and thickness 1.6

cm, are placed one above the other to form a solid circular cylinder. Find

the curved surface area and the volume of the cylinder so formed.

ANSWER:

We have 25 circular plates, each with radius = 10.5 cm and thickness =

1.6 cm

These plates are stacked on top of one another.

So, the total height of the arrangement becomes = 1.6 × 25 = 40 cm

Volume of this arrangement = 𝜋𝑟2ℎ = 𝜋(10.5)2 × 40 = 13860 cm³

Curved surface area = 2𝜋𝑟ℎ = 2𝜋 × 10.5 × 40 = 2640 cm²

Hence volume = 13860 cm³ and C.S.A = 2640 cm².

Page No 14.28:

Question 9: Find the number of metallic circular 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 diameter 4.5 cm.

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ANSWER:

Given the diameter of the base of the circular disc = 1.5 cm

Height = 0.2 cm

Volume of the circular disc = 𝜋𝑟2ℎ = 𝜋 × (1.5

2)

2× 0.2

= 𝜋 × (0.75)2 × 0.2 ...(i)

Height of the cylinder = 10 cm

Diameter = 4.5 cm

Volume of the cylinder = 𝜋𝑅2𝐻 = 𝜋 (4.5

2)

2× 10

= 𝜋 × (2.25)2 × 10 ...(ii)

Now since the circular discs are used to make the cylinder so, let n be

the number of circular discs required.

𝑛 × Volume of circular disc = Volume of cylinder

⇒ Volume of cylinderVolume of circular disc = 𝑛

⇒𝜋×(2.25)2×10

𝜋×(0.75)2×0.2= 𝑛

⇒ 𝑛 = 450

Hence, 450 metallic circular discs need to be melted to form the right

circular cylinder.

Page No 14.28:

Question 10: How many spherical lead shots each of diameter 4.2 cm can

be obtained from a solid rectangular lead piece with dimension

6cm × 42 cm × 21 cm .

ANSWER:

The dimensions of the solid rectangular lead piece is 66 cm × 42 cm ×

21 cm.

Diameter of the spherical lead shots = 4.2 cm

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Let n spherical lead shots be obtained from the rectangular piece.

𝑛 × volume of spherical lead shot = Volume of the rectangular lead piece

⇒ Volume of the rectangular lead piecevolume of spherical lead shot = 𝑛

⇒66×42×21

4

3𝜋𝑟3

= 𝑛

⇒66×42×21

4

3𝜋(

4.2

2)

3 = 𝑛

⇒58212

38.808= 𝑛

⇒ 𝑛 = 1500

Hence, 1500 lead shots can be formed.

DISCLAIMER: There is some error in the question given. Instead of 6

cm, there should be 66 cm.

The result obtained is by taking 66 cm as the dimensions of the

rectangular piece.

Page No 14.28:

Question 11: How many spherical lead shots of diameter 4 cm can be

made out of a solid cube of lead whose edge measures 44 cm.

ANSWER:

Diameter of the spherical lead shots = 4 cm

Edge length of the solid cube (a) = 44 cm.

Let n be the number of spherical lead shots made out of the solid cube.

𝑛 × Volume of the spherical lead shots = Volume of the solid cube

⇒Volume of the solid cube

Volume of the spherical lead shots= 𝑛

⇒𝑎3

4

3𝜋𝑟3

= 𝑛

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⇒443

4

3𝜋×(

4

2)

3 = 𝑛

⇒ 2541 = 𝑛 Hence, 2541 spherical lead shots can be made

Page No 14.28:

Question 12: Three cubes of a metal whose edges are in the ratios 3 : 4 :

5 are melted and converted into a single cube whose diagonal is 12√3 cm.

Find the edges of the three cubes.

ANSWER:

The three cubes of metal are in the ratio 3 : 4 : 5.

Let the edges of the cubes be 3x, 4x and 5x.

Volume of the three cubes will be

𝑉1 = (3𝑥)3

𝑉2 = (4𝑥)3

𝑉3 = (5𝑥)3

Diagonal of the single cube = 12√3 cm

We know diagonal of the cube = 𝑎√3 = 12√3

Hence, the side of the cube = 12 cm

Volume of the bigger cube Vb = (12)3

Volume of the three cubes = Volume of the single

(3𝑥)3 + (4𝑥)3 + (5𝑥)3 = (12)3

⇒ 27𝑥3 + 64𝑥3 + 125𝑥3 = 1728

⇒ 216𝑥3 = 1728

⇒ 𝑥3 =1728

216= 8

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⇒ 𝑥 = 2

Hence, the edges of the three cubes will be 3 × (2), 4 × (2), 5 × (2) =

6,8,10 cm

Page No 14.28:

Question 13: 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.

ANSWER:

Radius of the solid metallic sphere, r = 10.5 cm

Radius of the cone, R = 3.5 cm

Height of the cone, H = 3 cm

Let the number of smaller cones formed be n.

Volume of the metallic sphere, 𝑉𝑆 =4

3𝜋(𝑟)3 =

4

3𝜋(10.5)3

Volume of the cone, 𝑉𝐶 =1

3𝜋(𝑅)2𝐻 =

1

3𝜋(3.5)2 × 3

Let the number of cones thus formed be n.

𝑛 × volume of the cone = volume of the sphere

⇒ volume of the sphere(V𝑠)volume of the cone(V𝑐) = 𝑛

⇒4

3𝜋(10.5)3

1

3𝜋(3.5)2×3

= 𝑛

⇒ 126 = 𝑛

Hence, 126 cones are thus formed.

Page No 14.28:

Question 14: The diameter of a metallic sphere is equal to 9 cm. It is

melted and drawn into a long wire of diameter 2 mm having uniform

cross-section. Find the length of the wire.

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ANSWER:

The radius of the metallic sphere is 9

2= 4.5 cm = 4.5 mm. Therefore, the

volume of the metallic sphere is

𝑉 =4

3𝜋 × (45)3 Cubic mm

The metallic sphere is melted to produce a long wire of uniform cross

section of radius 2

2= 1 mm. Let the length of the wire be l mm. Then, the

volume of the wire is

𝑉1 = 𝜋 × (1)2 × 𝑙 = 𝜋𝑙 Cubic mm

Since, the volume of the metallic sphere is equal to the volume of the wire,

we have

𝑉 = 𝑉1

⇒4

3𝜋 × (45)2 = 𝜋𝑙

⇒ 𝑙 =4

3× (45)2

⇒ = 4 × (45)2 × 15

⇒ = 121500

Hence, the length of the wire is 121500 mm = 12150 cm.

Hence length = 12150 cm

Page No 14.28:

Question 15: 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 1/4

of the 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.

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ANSWER:

Let the radius of the big metallic ball is 4r. Therefore, the volume of the

big metallic ball is

𝑉 =4

3𝜋 × (4𝑟)3

The metallic sphere is melted to produce small balls of radius 4𝑟

4= 𝑟.

Then, the volume of each of the small balls is

𝑉1 =4

3𝜋 × (𝑟)3

Since, the volume of the big metallic ball is equal to the sum of the

volumes of the small balls, we have the number of produced small balls

is

𝑉

𝑉1=

4

3𝜋×(4𝑟)3

4

3𝜋×(𝑟)3

= (4)3

= 64

Hence, the number of small balls is 64

The surface area of the big ball is

𝑆 = 4𝜋 × (4𝑟)2

The surface area of each of the small ball is

𝑆1 = 4𝜋 × (𝑟)2

Therefore, the total surface area of the 64 small balls is

𝑆2 = 64 × 4𝜋 × (𝑟)2

Now, we compute the following ratio

𝑆2

𝑆=

64×4𝜋×(𝑟)2

4𝜋×(4𝑟)2 = 4

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⇒ 𝑆2 = 4𝑆

Hence, the total surface area of the small balls is equal to four times the

surface area of the original big ball.

Page No 14.28:

Question 16: A copper sphere of radius 3 cm is melted and recast into a

right circular cone of height 3 cm. Find the radius of the base of the cone.

ANSWER:

The radius of the copper sphere is 3cm. Therefore, the volume of the

copper sphere is

𝑉 =4

3𝜋 × (4𝑟)3 cm³

The copper sphere is melted to produce a right circular cone. The height

of the right circular cone is 3cm. Let the base-radius of the right circular

cone is r. Then, the volume of the right circular cone is

𝑉1 =4

3𝜋 × (𝑟)2 × 3

= 𝜋 × 𝑟2 cm³

Since, the sphere is melted to recast the cone; the volumes of the sphere

and the cone are equal. Hence, we have

𝑉 = 𝑉1

⇒ 4

3𝜋 × (3)3 = 𝜋 × 𝑟2

⇒ 4 × (3)3 =4

3× (45)2

⇒ 𝑟 = 2 × 3

⇒ 𝑟 = 6

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Hence, the base-radius of the right circular cone is 6 cm.

Page No 14.28:

Question 17: A copper rod of diameter 1 cm and length 8 cm is drawn

into a wire of length 18 m of uniform thickness. Find the thickness of the

wire.

ANSWER:

The radius of the copper rod is 0.5 cm and length is 8 cm. Therefore, the

volume of the copper rod is

𝑉 = 𝜋 × (0.5)2 × 8 cm³

Let the radius of the wire is r cm. The length of the wire is 18 m=1800

cm. Therefore, the volume of the wire is

𝑉1 = 𝜋 × (𝑟)2 × 1800 cm³

Since, the volume of the copper rod is equal to the volume of the wire; we

have

𝑉1 = 𝑉

⇒ 𝜋𝑟2 × 1800 = 𝜋 × (0.5)2 × 8

⇒ 𝑟2 =0.25×8

1800=

1

900

⇒ 𝑟 =1

30= 0.033 cm

Hence, the radius of the wire is 0.033 cm = 0.33 mm.

So, thickness = 0.33 × 2 = 0.66 mm

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Page No 14.28:

Question 18: The diameters of internal and external surfaces of a hollow

spherical shell are 10 cm and 6 cm respectively. If it is melted and recast

into a solid cylinder of length of 223 cm, find the diameter of the cylinder.

ANSWER:

The internal and external radii of the hollow sphere are 3cm and 5cm

respectively. Therefore, the volume of the spherical shell is

𝑉 =4

3𝜋 × {(5)3 − (3)3}

=4

3𝜋 × 98 cm³

The spherical shell is melted to recast a solid cylinder of length 8

3 cm. Let

the radius of the solid cylinder is r cm. Therefore, the volume of the solid

cylinder is

𝑉1 = 𝜋 × (𝑟)2 ×8

3 cm³

Since, the volume of the hollow spherical shell is equal to the volume of

the solid cylinder; we have

𝑉1 = 𝑉

⇒ 𝜋 × (𝑟)2 ×8

3=

4

3𝜋 × 98

⇒ 𝑟2 = 49

⇒ 𝑟 = 7

Hence, the diameter of the solid cylinder is two times its radius, which is

14 cm.

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Page No 14.28:

Question 19: How many coins 1.75 cm in diameter and 2 mm thick must

be melted to form a cuboid 11 cm × 10 cm × 7 cm?

ANSWER:

The dimension of the cuboid is 11 cm × 10 cm × 7 cm. Therefore, the

volume of the cuboid is

𝑉1 = 11 × 10 × 7 = 770 cm³

The radius and thickness of each coin are 1.75

2= 0.875 cm and 2mm =

0.2cm respectively. Therefore, the volume of each coin is

𝑉2 = 𝜋 × (0.875)2 × 0.2 cm³

Since, the total volume of the melted coins is same as the volume of the

cuboid; the number of required coins is

𝑉1

𝑉2=

770

𝜋×(0.875)2×0.2

=770×7

22×(0.875)2×0.2

= 1600

Page No 14.28:

Question 20: 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 (Use π = 22/7).

ANSWER:

The surface area of the metallic sphere is 616 square cm. Let the radius of

the metallic sphere is r. Therefore, we have

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4𝜋𝑟2 = 616

⇒ 𝑟2 =616×7

4×22

⇒ 𝑟2 = 7 × 7

⇒ 𝑟 = 7

Therefore, the radius of the metallic sphere is 7 cm and the volume of the

sphere is

𝑉1 =4

3𝜋 × (7)3 cm³

The sphere is melted to recast a cone of height 28 cm. Let the radius of

the cone is R cm. Therefore, the volume of the cone is

𝑉2 =1

3𝜋 × (𝑅)2 × 28 cm³

Since, the volumes of the sphere and the cone are same; we have

𝑉1 = 𝑉2

⇒ 4

3𝜋 × (7)3 =

1

3𝜋 × (𝑅)2 × 28

⇒ 𝑅2 =4×(7)3

28

⇒ 𝑅2 = 72

⇒ 𝑅2 = 7

Hence, the diameter of the base of the cone so formed is two times its

radius, which is 14 cm.

Page No 14.29:

Question 21: A cylindrical bucket, 32 cm high and with radius of base 18

cm, is filled with sand. This bucket is emptied out on the ground and a

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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.

ANSWER:

Let the radius of the cone by r

Now, Volume cylindrical bucket = Volume of conical heap of sand

⇒ 𝜋(18)2(32) =1

3𝜋𝑟2(24)

⇒ (18)2(32) = 8𝑟2

⇒ 𝑟2 = 18 × 18 × 4

⇒ 𝑟2 = 1296

⇒ 𝑟 = 36 cm

Let the slant height of the cone be l.

Thus , the slant height is given by

𝑙 = √(24)2 + (36)2

= √576 + 1296

= √1872

= 12√13 cm

Disclaimer: The answer given in the book for the slant height is not

correct.

Page No 14.29:

Question 22: A solid metallic sphere of radius 5.6 cm is melted and solid

cones each of radius 2.8 cm and height 3.2 cm are made. Find the number

of such cones formed. [CBSE 2014]

ANSWER:

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Let the number of such cones formed be n

Now, Volume of solid metallic sphere = Volume of n solid cones

⇒4

3×

22

7× (5.6)3 = 𝑛 ×

1

3×

22

7× (2.8)2 × 3.2

⇒ 4 × (5.6)3 = 𝑛 × (2.8)2 × 3.2

⇒ 𝑛 = 28

Page No 14.29:

Question 23: A solid cuboid of iron with dimensions 53 cm ⨯ 40 cm ⨯

15 cm is melted and recast into a cylindrical pipe. The outer and inner

diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe.

[CBSE 2015]

ANSWER:

Volume of solid cuboid of iron = Volume of cylindrical pipe

⇒ 𝑙𝑏ℎ = 𝜋ℎ(𝑅2 − 𝑟2)

⇒ 53 × 40 × 15 =22

7× ℎ [(

8

2)

2− (

7

2)

2]

⇒ 53 × 40 × 15 =22

7× ℎ[42 − (3.5)2]

⇒ 53 × 40 × 15 =22

7× ℎ × 3.75

⇒ ℎ = 2698.18 cm

Disclaimer: The answer given in the book is not correct.

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Page No 14.29:

Question 24: The diameters of the internal and external surfaces of a

hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and

recast into a solid cylinder of diameter 14 cm. find the height of the

cylinder.

ANSWER:

The internal and external radii of the hollow spherical shell are 3cm and

5cm respectively. Therefore, the volume of the hollow spherical shell is

𝑉 =4

3𝜋 × {(5)3 − (3)3} cm³

The hollow spherical shell is melted to recast a cylinder of radius 7cm.

Let, the height of the solid cylinder is h. Therefore, the volume of the solid

cylinder is

𝑉1 = 𝜋 × (7)2 × ℎ cm³

Since, the volume of the solid cylinder is same as the volume of the hollow

spherical shell, we have

𝑉1 = 𝑉

⇒ 𝜋 × (7)2 × ℎ =4

3𝜋 × {(5)3 − (3)3}

⇒ 49 × ℎ =4

3× 98

⇒ ℎ =4×98

3×49

⇒ =8

3

Therefore, the height of the solid cylinder is 8

3 cm.

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Page No 14.29:

Question 25: A hollow sphere of internal and external diameters 4 cm

and 8 cm respectively is melted into a cone of base diameter 8 cm.

Calculate the height of the cone.

ANSWER:

The internal and external radii of the hollow sphere are 2cm and 4cm

respectively. Therefore, the volume of the hollow sphere is

𝑉 =4

3𝜋 × {(4)3 − (2)3} cm³

The hollow spherical shell is melted to recast a cone of base- radius 4cm.

Let, the height of the cone is h. Therefore, the volume of the cone is

𝑉1 =1

3𝜋 × (4)2 × ℎ cm³

Since, the volume of the cone is same as the volume of the hollow sphere,

we have

𝑉1 = 𝑉

⇒1

3𝜋 × (4)2 × ℎ =

4

3𝜋 × {(4)3 − (2)3}

⇒ 16 × ℎ = 4 × 56

⇒ ℎ =4×56

16

⇒ = 14

Therefore, the height of the cone is 14 cm.

Page No 14.29:

Question 26: A hollow sphere of internal and external radii 2 cm and 4

cm respectively is melted into a cone of base radius 4 cm. Find the height

and slant height of the cone.

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ANSWER:

The internal and external radii of the hollow sphere are 2cm and 4cm

respectively. Therefore, the volume of the hollow sphere is

𝑉 =4

3𝜋 × {(4)3 − (2)3}

=4

3×

22

7× 56

=32×22

3

The hollow sphere is melted to produce a right circular cone of base-radius

4cm. Let, the height and slant height of the cone be h cm and l cm

respectively. Then, we have

𝑙2 = (4)2 + ℎ2

⇒ 𝑙2 = 16 + ℎ2

The volume of the cone is

𝑉1 =1

3𝜋𝑟1

2ℎ1

=1

3×

22

7× (4)2 × ℎ

Since, the volume of the cone and hollow sphere are same, we have

𝑉1 = 𝑉

⇒1

3×

22

7× (4)2 × ℎ =

32×22

3

⇒ 1

7× (4)2 × ℎ = 32

⇒ ℎ =32×7

16

⇒ = 14

Then, we have

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𝑙2 = 16 + (14)2

⇒ = 212

⇒ 𝑙 = 14.56

Therefore, the height and the slant height of the cone are 14 cm and 14.56

cm respectively.

Page No 14.29:

Question 27: A spherical ball of radius 3 cm is melted and recast into

three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm.

Find the diameter of the third ball.

ANSWER:

The radius of the big spherical ball is 3cm. Therefore, the volume of the

big spherical ball is

𝑉 =4

3𝜋 × (3)3 cubic cm

The radii of the 1st and 2nd small spherical balls are 1.5 cm and 2 cm

respectively. Therefore, the volumes of the 1st and 2nd spherical balls are

respectively

𝑉1 =4

3𝜋 × (1.5)3 cubic cm,

𝑉2 =4

3𝜋 × (2)3 cubic cm

Let, the radius of the 3rd small spherical ball is r cm. Then, its volume is

𝑉3 =4

3𝜋 × (𝑟)3 cubic cm

Since, the big spherical ball is melted to produce the three small spherical

balls; the volume of the big spherical ball is same as the sum of the

volumes of the three small spherical balls. Therefore, we have

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𝑉 = 𝑉1 + 𝑉2 + 𝑉3

⇒4

3𝜋 × (3)3 =

4

3𝜋 × (1.5)3 +

4

3𝜋 × (2)3 +

4

3𝜋 × (𝑟)3

⇒ (3)3 = (1.5)3 + (2)3 + (𝑟)3

⇒ (𝑟)3 = (3)3 − (1.5)3 − (2)3

⇒ = 15.625

⇒ = 2.5

Therefore, the diameter of the 3rd ball is 2𝑟 = 5 cm

Page No 14.29:

Question 28: A path 2 m wide surrounds a circular pond of diameter 40

m. How many cubic metres of gravel are required to grave the path to a

depth of 20 cm?

ANSWER:

Diameter of the circular pond is given = 40 m

So, the radius of this pond is 20 m

There is a path surrounding the pond. We are given the thickness of this

path as 2 m

We have to grave this path with gravel. The depth of the path is also given

20 cm=0.2 m

This circular path can be viewed as a hollow cylinder of thickness 0.2 m

and depth 0.2 m

We know,

Volume of a hollow cylinder = 𝜋ℎ(𝑅2 − 𝑟2)

So the volume of the circular path with height 0.2 m

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= 𝜋 × 0.2(222 − 202)

= 𝜋 × 0.2(484 − 400)

= 𝜋 × 0.2 × 84

= 52.77 m³

Hence, the volume of gravel required is 52.77 m³

Page No 14.29:

Question 29: A 16 m deep well with diameter 3.5 m is dug up and the

earth from it is spread evenly to form a platform 27.5 m by 7 m. Find the

height of the platform.

ANSWER:

Assume the well as a solid right circular cylinder. Then, the radius of the

solid right circular cylinder is

𝑟 =3.5

2= 1.75 m

The well is 16m deep. Thus, the height of the solid right circular cylinder

is ℎ = 16 m.

Therefore, the volume of the solid right circular cylinder is

𝑉1 = 𝜋𝑟2ℎ

=22

7× (1.75)2 × 16 cubic meters

Let the height of the platform formed be x m. The length and the breadth

of the platform are l = 27.5 m and b = 7 m respectively. Therefore, the

volume of the platform is

𝑉2 = 𝑙𝑏𝑥 = 27.5 × 7 × 𝑥 = 192.5𝑥 cubic meters

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Since, the well is spread to form the platform; the volume of the well is

equal to the volume of the platform. Hence, we have

𝑉1 = 𝑉

⇒22

7× (1.75)2 × 16 = 192.5𝑥

⇒ 𝑥 =22

7×192.5× (1.75)2 × 16

⇒ 𝑥 =22×3.0625×16

7×192.5

⇒ = 0.8

Hence, the height of the platform is 0.8 m = 80 cm.

Page No 14.29:

Question 30: A well of diameter 2 m is dug 14 m deep. The earth taken

out of it is spread evenly all around it to form an embankment of height

40 cm. Find the width of the embankment.

ANSWER:

Assume the well as a solid right circular cylinder. Then, the radius of the

solid right circular cylinder is

𝑟 =2

2= 1

The well is 14m deep. Thus, the height of the solid right circular cylinder

is ℎ = 14 m.

Therefore, the volume of the solid right circular cylinder is

𝑉1 = 𝜋𝑟2ℎ =22

7× (1)2 × 14 = 44 cubic meters

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Since, the embankment is to form around the right circular cylinder. Let

the width of the embankment be x m. The height of the embankment is h =

40 cm = 0.4 m. Therefore, the volume of the platform is

𝑉2 = 𝜋((1 + 𝑥)2 − 12) × .4

Since, the well is spread to form the platform; the volume of the well is

equal to the volume of the platform. Hence, we have

𝑉2 = 𝑉1

⇒ 𝜋((1 + 𝑥)2 − 12) × .4 = 44

⇒ 𝑥2 + 2𝑥 − 35 = 0

⇒ (𝑥 − 5)(𝑥 + 7) = 0

⇒ 𝑥 = 5 or 𝑥 = −7

Hence, x = 5

Hence, width = 5 m

Page No 14.29:

Question 31: A well with inner radius 4 m is dug 14 m deep. Earth taken

out of it has been spread evenly all around a width of 3 m it to form an

embankment. Find the height of the embankment.

ANSWER:

The inner radius of the well is 4m and the height is 14m. Therefore, the

volume of the Earth taken out of it is

𝑉1 = 𝜋 × (4)2 × 14 m²

The inner and outer radii of the embankment are 4m and 4 + 3 = 7 m

respectively. Let the height of the embankment be h. Therefore, the

volume of the embankment is

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𝑉2 = 𝜋 × {(7)2 − (4)2} × ℎ m³

Since, the volume of the well is same as the volume of the embankment;

we have

𝑉1 = 𝑉2

⇒ 𝜋 × (4)2 × 14 = 𝜋 × {(7)2 − (4)2} × ℎ

⇒ ℎ =(4)2×14

3

⇒ ℎ = 6.78 m

Hence, the height of the embankment is 6.78 m

Page No 14.29:

Question 32:

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has

been spread evenly all around it to a width of 4 m to form an embankment.

Find the height of the embankment.

ANSWER:

The inner radius of the well is 3

2 m and the height is 14m. Therefore, the

volume of the Earth taken out of it is

𝑉1 = 𝜋 × (3

2)

2× 14 m³

The inner and outer radii of the embankment are 3

2 m and 4 +

3

2=

11

2 m

respectively. Let the height of the embankment be h. Therefore, the

volume of the embankment is

𝑉2 = 𝜋 × {(11

2)

2− (

3

2)

2} × ℎ m³

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Since, the volume of the well is same as the volume of the embankment;

we have

𝑉1 = 𝑉2

⇒ 𝜋 × (3

2)

2× 14 = 𝜋 × {(

11

2)

2− (

3

2)

2} × ℎ

⇒ ℎ =9×14

112

⇒ ℎ =9

8 m

Hence, the height of the embankment is ℎ =9

8 m

Page No 14.29:

Question 33: Find the volume of the largest right circular cone that can

be cut out of a cube whose edge is 9 cm.

ANSWER:

We have the following figure

The length of each side of the cube is 9 cm. We have to find the volume

of the largest right circular cone contained in the cube.

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The diameter of the base circle is same as the length of the side of the

cube. Thus, the diameter of the base circle of the right circular cone is 9

cm. Therefore, the radius of the base of the right circular cone is 𝑟 = 4.5

cm.

From the right angled triangle ∆𝐴𝑂𝐵 we have

h = 9 cm

Therefore, the volume of the solid right circular cone is

𝑉 =1

3𝜋𝑟2ℎ

=1

3×

22

7× (4.5)2 × 9

= 190.93

Hence largest volume of cone is 190.93 cm³

Page No 14.29:

Question 34: A cylindrical bucket, 32 cm high and 18 cm of radius of the

base, 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.

ANSWER:

The height and radius of the cylindrical bucket are ℎ = 32 cm and 𝑟 = 18

cm respectively. Therefore, the volume of the cylindrical bucket is

𝑉 = 𝜋𝑟2ℎ

=22

7× (18)2 × 32

The bucket is full of sand and is emptied in the ground to form a conical

heap of sand of height ℎ1 = 24 cm. Let, the radius and slant height of the

conical heap be 𝑟1 cm and 𝑙1cm respectively. Then, we have

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𝑙12 = 𝑟1

2 + ℎ12

⇒ 𝑟12 = 𝑙1

2 − ℎ12

⇒ 𝑟12 = 𝑙1

2 − (24)2

The volume of the conical heap is

𝑉1 =1

3𝜋𝑟2ℎ

=1

3×

22

7× 𝑟1

2 × 24

=22

7× 𝑟1

2 × 8

Since, the volume of the cylindrical bucket and conical hear are same, we

have

𝑉1 = 𝑉

⇒22

7× 𝑟1

2 × 8 =22

7× (18)2 × 32

⇒ 𝑟12 = (18)2 × 4

⇒ 𝑟1 = 18 × 2

⇒ 𝑟1 = 36

Then, we have

𝑙12 = 𝑟1

2 + ℎ12

⇒ 𝑙12 = (36)2 − (24)2

⇒ 𝑙1 = 43.27

Therefore, the radius and the slant height of the conical heap are 36 cm

and 43.27 cm respectively.

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Page No 14.29:

Question 35: Rain water, which falls on a flat rectangular surface of

length 6 m and breadth 4 m is transferred into a cylindrical vessel of

internal radius 20 cm. What will be the height of water in the cylindrical

vessel if a rainfall of 1 cm has fallen?

ANSWER:

The fallen rains are in the form of a cuboid of height 1 cm, length 6 m =

600 cm and breadth 4 m = 400 cm. Therefore, the volume of the fallen

rains is

𝑉 = 600 × 400 × 1 = 240000 cm³

The fallen rains are transferred into a cylindrical vessel of internal

radius r1 = 20 cm. Let, the height of the water in the cylindrical vessel

is h1 cm. Then, the volume of the water in the cylinder is

𝑉1 = 𝜋𝑟12ℎ1 =

22

7× (20)2 × ℎ1

Since, the volume of the water in the cylinder is same as the volume of

the rainfalls, we have

𝑉1 = 𝑉

⇒22

7× (20)2 × ℎ1 = 240000

⇒ ℎ1 =240000×7

(20)2×22

⇒ = 190.9

Therefore, the height of the water in the cylinder is 190.9 cm.

Page No 14.29:

Question 36:

The rain water from a roof of dimensions 22 m × 20 m drains into a

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

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water collected from the roof just fills the cylindrical vessel, then find the

rainfall in cm.

ANSWER:

The dimension of the roof is 22 m × 20 m.

Diameter of the cylinderical vessel = 2 m

Radius of the cylinderical vessel, R = 1 m

Height of the cylinderical vessel, H = 3.5 m

Let the height of the roof be h.

Volume of water thus collected on the roof = 22 × 20 × ℎ

Volume of the cylinderical vessel

= 𝜋(𝑅)2𝐻 = 𝜋 (2

2)

2× 3.5 = 𝜋 × 1 × 3.5 = 3.5𝜋

Volume of water collected on the roof = Volume of the cylinderical

vessel

22 × 20 × ℎ = 3.5𝜋

⇒ ℎ =3.5𝜋

22×20= 0.025 m

⇒ ℎ = 2.5 cm

Page No 14.29:

Question 37: A conical flask is full of water. The flask has base-

radius r and height h. The water is poured into a cylindrical flask of base-

radius Mr. Find the height of water in the cylindrical flask.

ANSWER:

The base-radius and height of the conical flask are r and h respectively.

Let, the slant height of the conical flask is l. Therefore, the volume of the

water in the conical flask is

𝑉 =1

3× 𝜋 × 𝑟2 × ℎ

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The water in the conical flask is poured into a cylindrical flask of base-

radius mr. Let, the height of the water in the cylindrical flaks is h1. Then,

the volume of the water in the cylindrical flaks is

𝑉1 = 𝜋 × (𝑚𝑟)2 × ℎ1

Since, the volume of the water in the cylindrical flaks is same as the

volume of the water in the conical flaks, we have

𝑉1 = 𝑉

⇒ 𝜋 × (𝑚𝑟)2 × ℎ1 =1

3× 𝜋 × 𝑟2 × ℎ

⇒ 𝑚2 × ℎ1 =1

3× ℎ

⇒ ℎ1 =ℎ

3𝑚2

Therefore, the height of the water in the cylinder is ℎ

3𝑚2

Page No 14.30:

Question 38: A rectangular tank 15 m long and 11 m broad is required to

receive entire liquid contents from a fully cylindrical tank of internal

diameter 21 m and length 5 m. Find the least height of the tank that will

serve the purpose.

ANSWER:

Suppose height of the rectangular tank is equal to h.

Length of the tank = 15 m

Breadth of the tank = 11 m

Further,

length of cylindrical tank = 5 m

Radius of cylindrical tank = 21

2 m

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To find out the least height of the tank, equate the volumes of two tanks.

15 × 11 × ℎ = 𝜋 (21

2)

2× 5

⇒ ℎ =22

7×

21

2×

21

2×

5

15×

1

11

⇒ ℎ =21

2

⇒ ℎ = 10.5

Hence, the least height of the tank is equal to 10.5.

Page No 14.30:

Question 39: A hemispherical bowl of internal radius 9 cm is full of

liquid. The liquid is to be filled into cylindrical shaped small bottles each

of diameter 3 cm and height 4 cm. How many bottles are necessary to

empty the bowl?

ANSWER:

The internal radius of the hemispherical bowl is 9cm. Therefore, the

volume of the water in the hemispherical bowl is

𝑉 =2

3𝜋 × (9)3 cm³

The water in the hemispherical bowl is required to transfer into the

cylindrical bottles each of radius 3

2 cm and height 4cm. Therefore, the

volume of each of the cylindrical bottle is

𝑉1 = 𝜋 × (3

2)

2× 4 cm³

Therefore, the required number of cylindrical bottles is

𝑉

𝑉1=

2

3𝜋×(9)3

𝜋×(3

2)

2×4

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=2×(9)3×(2)2

3×(3)2×4

= 54

Hence No. of bottle = 54

Page No 14.30:

Question 40: A cylindrical tub of radius 12 cm contains water to a depth

of 20 cm. A spherical ball dropped into the tub and the level of the water

is raised by 6.75 cm. Find the radius of the ball.

ANSWER:

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball

into the tub, the height of the raised water is 6.75cm. Therefore, the

volume of the raised water is

𝑉 = 𝜋 × (12)2 × 6.75 cm³

Let, the radius of the spherical ball is r. Therefore, the volume of the

spherical ball is

𝑉1 =4

3𝜋 × 𝑟3 cm³

Since, the volume of the raised water is same as the volume of the

spherical ball, we have

𝑉1 = 𝑉

⇒4

3𝜋 × 𝑟3 = 𝜋 × (12)2 × 6.75

⇒ 𝑟3 =(12)2×6.75×3

4

⇒ = 12 × 3 × 6.75 × 3

⇒ = 9

Therefore, the radius of the spherical ball is 9 cm.

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Page No 14.30:

Question 41: 500 persons have to dip in a rectangular tank which is 80 m

long and 50 m broad. What is the rise in the level of water in the tank, if

the average displacement of water by a person is 0.04 m3?

ANSWER:

The average displacement of water by a person is 0.04 cubic m. Hence,

the total displacement of water in the rectangular tank by 500 persons is

𝑉 = 500 × 0.04 = 20 Cubic m.

The length and width of the rectangular tank are 80m and 50m

respectively. Upon dipping in the tank, let the height of the raised water

is be h m. Therefore, the volume of the raised water is

𝑉1 = 80 × 50 × ℎ

= 4000ℎ cubic m

Since, the volume of the raised water is same as the volume of the water

displaced by 500 persons, we have

𝑉1 = 𝑉

⇒ 4000ℎ = 50

⇒ ℎ =20

4000

⇒ = 0.005

Therefore, the water will be raised by 0.005 m or 0.5 cm.

Page No 14.30:

Question 42: A cylindrical jar of radius 6 cm contains oil. Iron spheres

each of radius 1.5 cm are immersed in the oil. How many spheres are

necessary to raise the level of the oil by two centimetres?

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ANSWER:

The radius of the cylindrical jar is 6cm. The volume of the oil of height

2cm contained in the jar is

𝑉 = 𝜋 × (6)2 × 2 cubic cm

The radius of each small sphere is 1.5cm. Therefore, the volume of each

small sphere is

𝑉1 =4

3× 𝜋 × (1.5)3 cubic cm

Since, the volume of the raised water is same as the sum of the volumes

of the immersed iron spheres, we have the number of immersed sphere is

𝑉

𝑉1=

𝜋×(6)2×24

3×𝜋×(1.5)3

=3×36×2×1000

4×15×15×15

= 16

Therefore, the number of iron spheres is 16

Page No 14.30:

Question 43: A cylindrical tub of radius 12 cm contains water to a depth

of 20 cm. A spherical from ball of radius 9 cm is dropped into the tub and

thus the level of water is raised by h cm. What is the value of h?

ANSWER:

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball

of radius 9cm into the tub, the height of the raised water is h cm.

Therefore, the volume of the raised water is

𝑉 = 𝜋 × (12)2 × ℎ cubic cm

The volume of the spherical ball is

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𝑉1 =4

3𝜋 × (9)3 cubic cm

Since, the volume of the raised water is same as the volume of the

spherical ball, we have

𝑉1 = 𝑉

⇒4

3𝜋 × (9)3 = 𝜋 × (12)2 × ℎ

⇒ ℎ =4×(9)3

3×(12)2

⇒ =27

4

⇒ = 6.75

Therefore, the height of the raised water is ℎ = 6.75 cm

Page No 14.30:

Question 44: Metal spheres, each of the radius 2 cm, are packed into a

rectangular box of internal dimension 16 cm × 8 cm × 8 cm when 16

spheres are packed the box is filled with preservative liquid. Find the

volume of this liquid.

ANSWER:

The radius of each of the metallic sphere is 2cm. Therefore, the volume

of each metallic sphere is

𝑉 =4

3𝜋 × (2)3 cm³

The total volume of the 16 spheres is

𝑉1 = 16 ×4

3𝜋 × (2)3 cm³

The internal dimension of the rectangular box is 16cm × 8cm × 8cm.

Therefore, the volume of the rectangular box is

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𝑉2 = 16 × 8 × 8 cm³

Therefore, the volume of the liquid is

𝑉2 − 𝑉1 = 16 × 8 × 8 − 16 −4

3𝜋 × (2)3

= 1024 − 536.03

= 488

Hence volume of liquid is 488 cm³

Page No 14.30:

Question 45: A vessel in the shape of a cuboid contains some water. If

three identical spheres immersed in the water, the level of water is

increased by 2 cm. If the area of the base of the cuboid is 160 cm2 and its

height 12 cm, determine the radius of any of the spheres.

ANSWER:

The area of the base of the cuboid is 160 cm2. After immersing three

identical spheres the level of the water is increased by 2 cm. Therefore,

the volume of the increased water is

𝑉 = 160 × 2 cm³

Let the radius of each of the spheres is r cm. Then, the volume of each of

the sphere is

𝑉1 =4

3𝜋 × (𝑟)3 cm³

The total volume of the three spheres is

𝑉2 = 3 ×4

3𝜋 × (𝑟)3

= 4𝜋 × (𝑟)3 cm³

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Since, the volume of the increased water is equal to the total volume of

the three spheres; we have

𝑉2 = 𝑉

⇒ 4𝜋 × (𝑟)3 = 160 × 2

⇒ 𝑟3 =320×7

4×22

⇒ 𝑟3 = 25.45

⇒ 𝑟 = 2.94

Hence, the radius of each of the sphere is 2.94 cm.

Page No 14.30:

Question 46: 150 spherical marbles, each of diameter 1.4 cm are dropped

in a cylindrical vessel of diameter 7 cm containing some water, which are

completely immersed in water. Find the rise in the level of water in the

vessel. [CBSE 2014]

ANSWER:

Let the rise in the level of water in the vessel be h cm.

Now, Volume of 150 spherical marbles = Volume of water displaced in

the vessel

⇒ 150 ×4

3×

22

7× (

1.4

2)

3=

22

7× (

7

2)

2× ℎ

⇒ 200 × (0.7)3 = (7

2)

2× ℎ

⇒ ℎ = 5.6 cm

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Page No 14.30:

Question 47: Sushant has a vessel, of the form of an inverted cone, open

at the top, of height 11 cm and radius of top as 2.5 cm and is full of water.

Metallic spherical balls each of diameter 0.5 cm are put in the vessel due

to which (2

5)

th of the water in the vessel flows out. Find how many balls

were put in the vessel. Sushant made the arrangement so that the water

that flows out irrigates the flower beds. What value has been shown by

Sushant? [CBSE 2014]

ANSWER:

Let the number of the balls be n.

Volume of water flows out = Volume of n spherical bolls

⇒ 2

5×

1

3𝜋𝑅2ℎ = 𝑛 ×

4

3𝜋𝑟3

⇒2

5× (2.5)2 × 11 = 𝑛 × 4 (

0.5

8)

3

⇒ 27.5 =0.5

8𝑛

⇒ 𝑛 = 440

Page No 14.30:

Question 48: 16 glass spheres each of radius 2 cm are packed into a

cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the

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

ANSWER:

Radius of the glass spheres, r = 2 cm

Dimensions of the cuboidal box = 16cm × 8cm × 8cm

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volume of the spheres = 𝑉𝑠 =4

3𝜋(𝑟)3 =

4

3𝜋(2)3

volume of the cuboidal box = 𝑉𝑐 = 16 × 8 × 8 = 1024

Volume of water in the cuboidal box = Volume of the cuboidal box −

Volume of the 16 glass spheres

= 1024 − 16 ×4

3𝜋(2)3

= 1024 − 536.6

= 487.6 cm3

Hence, the volume of the water in the cuboidal box = 487.6 cm3

Page No 14.30:

Question 49: 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?

ANSWER:

The inner radius of the cylindrical pipe r =1 cm.

Rate of flow of water = 80 cm/sec

volume of the water that flows through pipe in 1sec is 𝜋𝑟2 × 80 =

80𝜋 cm3

volume of the water that flows through pipe in half an hour 80𝜋 × 30 ×

60 = 144000𝜋 cm3

radius of the base of the cylindrical tank is R = 40 cm

let the water level in the cylinderical tank after half an hour be h.

volume of the raised water = 𝜋(𝑅)2ℎ = 𝜋(40)2ℎ

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volume of the raised water in tank = volume of the water that flows

through pipe

⇒ 𝜋(40)2ℎ = 144000𝜋

⇒ ℎ =144000

1600= 90 cm

Thus water level will rise by 90 cm in half an hour.

Page No 14.30:

Question 50: Water in a canal 1.5 m wide and 6 m deep is flowing with

a speed of 10km/hr. How much area will it irrigate in 30 minutes if 8 cm

of standing water is desired?

ANSWER:

The canal is 1.5 m wide and 6 m deep. The water is flowing in the canal

at 10 km/hr. Hence, in 30 minutes, the length of the flowing standing

water is

= 10 ×30

60 km

= 5 km

= 5000 km

Therefore, the volume of the flowing water in 30 min is

𝑉1 = 5000 × 1.5 × 6 m³

Thus, the irrigated area in 30 min of 8 cm = 0.08 m standing water is

=5000×1.5×6

0.08

= 562500 m²

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Page No 14.30:

Question 51: A farmer runs a pipe of internal diameter 20 cm from the

canal into a cylindrical tank in his 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?

ANSWER:

The internal radius of the pipe is 10 cm = 0.1 m. The water is flowing in

the pipe at 3km/hr = 3000m/hr. Let the cylindrical tank will be filled

in t hours. Therefore, the length of the flowing water in t hours is =

3000 × 𝑡 meter

Therefore, the volume of the flowing water is

𝑉1 = 𝜋 × (0.1)2 × 3000 × 𝑡 m³

The radius of the cylindrical tank is 5 m and the height is 2 m. Therefore,

the volume of the cylindrical tank is

𝑉2 = 𝜋 × (5)2 × 2 m³

Since, we have considered that the tank will be filled in t hours; therefore,

the volume of

the flowing water in t hours is same as the volume of the cylindrical tank.

Hence, we have

𝑉2 = 𝑉1

⇒ 𝜋 × (5)2 × 2 = 𝜋 × (0.1)2 × 3000 × 𝑡

⇒ 𝑡 =(5)2×2

(0.1)2×3000

⇒ 𝑡 =5

3 hours

⇒ 𝑡 =5×60

3= 100 minutes

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Hence, the tank will be filled in 1 hour 40 minutes.

Page No 14.31:

Question 52: A cylindrical tank full of water is emptied by a pipe at the

rate of 225 litres per minute. How much time will it take to empty half the

tank, if the diameter of its base is 3 m and its height is 3.5 m?

[Use π =22

7] [CBSE 2014]

ANSWER:

Volume of cylindrical tank = 22

7× (

3

2)

2× 3.52 = 24.75 m3

Now, 1 m3 = 1000 L

∴ 24.75 m3 = 24750 L

Half the capacity of tank = 12375 L

Time taken by the pipe to empty 225 litres = 1 minute

Time taken by the pipe to empty 1 litre = 1

225 minutes

Time taken by the pipe to empty 12375 litres

=1

225× 12375 = 55 minutes

Page No 14.31:

Question 53: Water is flowing at the rate of 2.52 km/h through a

cylindrical pipe into a cylindrical tank, the radius of the base is 40 cm. If

the increase in the level of water in the tank, in half an hour is 3.15 m, find

the internal diameter of the pipe. [CBSE 2015]

ANSWER:

Increase in the level of water in half an hour, h = 3.15 m = 315 cm

Radius of the water tank, r = 40 cm

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Volume of water that falls in the tank in half an hour = πr2h

= π × (40)2 × 315

= 5,04,000 π cm3

Rate of flow of water = 2.52 km/h

Length of water column in half an hour = 2.52 ÷ 2 = 1.26 km = 1,26,000

cm

Let the internal diameter of the cylindrical pipe be 𝑑.

Volume of the water that flows through the pipe in half an hour

= 𝜋 (𝑑

2)

2× 126000

We know

Volume of the water that flows through the pipe in half an hour =

Volume of water that falls in the tank in half an hour

⇒ 𝜋 (𝑑

2)

2× 126000 = 504000𝜋

⇒ (𝑑

2)

2= 4

⇒ 𝑑 = 4 cm

Thus, the internal diameter of the pipe is 4 cm.

Page No 14.31:

Question 54: Water flows at the rate of 15 km/hr 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 the pond rise by 21 cm.

ANSWER:

Let the level of water in the pond rises by 21 cm in t hours.

Speed of water = 15 km/hr = 15000 m/hr

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Diameter of the pipe = 14 cm = 14

100 m

So, the radius of the pipe, r = 14

2×100= 7100 m

Volume of water flowing out of the pipe in 1 hour

= 𝜋𝑟2ℎ

= 𝜋 × (7

100)

2× 15000 𝑚3

= 231 m3

∴ Volume of water flowing out of the pipe in t hours = 231t m3.

Volume of water in the cuboidal pond

= 50 m × 44 m ×21

100 m

= 462 m3

Volume of water flowing out of the pipe in t hours = Volume of water in

the cuboidal pond

∴ 231t = 462

∴ Required time =Desired volume

Volume of water flown in 1 hour

= (462

231) hours = 2 hours

Thus, the water in the pond rise by 21 cm in 2 hours.

Page No 14.31:

Question 55: A canal is 300 cm wide and 120 cm deep. The water in the

canal is flowing with a speed of 20 km/hr. How much area will it irrigate

in 20 minutes if 8 cm of standing water is desired?

ANSWER:

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Width of the canal = 300 cm = 3 m

Depth of the canal = 120 cm = 1.2 m

Speed of water flow = 20 km/h = 20000 m/h

Distance covered by water in 1 hour or 60 min = 20000 m

So, distance covered by the water in 20 min = 20

60× 20000 =

20000

3 m

Amount of water irrigated in 20 minutes

= 3 × 1.2 ×20000

3= 24000 m3

Area irrigated by this water if 8 cm of standing water is desired will be 224000

8

100

= 300000 m2

So, area irrigated will be 300000 m2 or 30 hectors.

Page No 14.31:

Question 56: The sum of the radius of base and height of a solid right

circular cylinder is 37 cm. If the total surface area of the solid cylinder is

1628 cm2, find the volume of cylinder. (Use π = 22 / 7)

ANSWER:

Let the radius of the base of the cylinder be r cm

Let the height be h cm.

Now given that r + h = 37 cm

Total surface area = 1628 cm2 ...(i)

Total surface area of the cylinder = 2𝜋𝑟2 + 2𝜋𝑟ℎ

= 2𝜋𝑟(𝑟 + ℎ)

= 2𝜋𝑟 × 37

= 74𝜋𝑟 ...(ii)

From equation (i) and (ii) we get

74𝜋𝑟 = 1628

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⇒ 𝑟 =1628

74𝜋= 7 cm3

Thus, the height will be 37 − 7 = 30 cm

Thus, the volume of the cylinder = 𝜋𝑟2ℎ = 𝜋(7)2 × 30 = 4620 cm3

Hence the volume is 4620 cm3

Page No 14.31:

Question 57: A tent of height 77 dm is in the form a right circular cylinder

of diameter 36 m and height 44 dm surmounted by a right circular cone.

Find the cost of the canvas at Rs 3.50 per m2. [Use π = 22/7]

ANSWER:

The height of the tent is 77dm = 7.7m. The height of the upper portion

of the tent is

44dm = 4.4m. Therefore, the height of the cylindrical part is 77 − 44 =

33 dm = 3.3m. The radius of the cylindrical part is 36

2= 18 m.

Let the slant height of the cone part is l m. Then, we have

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𝑙2 = (18)2 + (3.3)2

⇒ 𝑙2 = 324 + 10.89 = 335.89

⇒ 𝑙 = 18.3

Therefore, the slant height of the cone part is 18.3 m.

The curved surface area of the cylindrical part is

𝑆 = 2𝜋 × 18 × 4.4 m²

The curved surface area of the cone part is

𝑆1 = 𝜋 × 18 × 18.3 m²

Therefore, the total curved surface area of the tent is

𝑆 + 𝑆1 = 2𝜋 × 18 × 4.4 + 𝜋 × 18 × 18.3

= 18𝜋 × (8.8 + 18.3)

= 18𝜋 × 27.10

The cost of canvas per m2 is Rs 3.50. Hence, the total cost for canvas in

Rs is

= 18 ×22

7× 27.10 × 3.50

= 5365.80

Hence total cost is Rs. 5365.80

Page No 14.31:

Question 58: The largest sphere is to be curved out of a right circular

cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.

ANSWER:

The radius of the right circular cylinder is 7cm and the height is 14cm.

Therefore, the radius of the largest sphere curved out from the cylinder is

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the minimum of the radius and half the height of the cylinder, which is

7cm. Therefore, the volume of the sphere is

𝑉 =4

3𝜋 × (7)3

= 1437.33 cm³

Page No 14.31:

Question 59: A right angled triangle whose sides are 3 cm, 4 cm and 5

cm is revolved about the sides containing the right angle in two ways.

Find the difference in volumes of the two cones so formed. Also, find their

curved surfaces.

ANSWER:

We consider the following figure as follows

Let the angle B is right angle and the sides of the triangle are AB = 4cm,

BC = 3cm,

AC = 5cm.

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When the triangle is revolved about the side AB, then the base-radius,

height and slant height of the produced cone becomes BC, AB and AC

respectively. Therefore, the volume of the produced cone is

𝑉1 =1

3𝜋 × 𝐵𝐶2 × 𝐴𝐵

=1

3𝜋 × (3)2 × 4

= 12𝜋 cubic cm

In this case, the curved surface area of the cone is

𝑆1 = 𝜋 × 𝐵𝐶 × 𝐴𝐶

= 𝜋 × 3 × 5

= 15𝜋 square cm

When the triangle is revolved about the side BC, then the base-radius,

height and slant height of the produced cone becomes AB, BC and AC

respectively. Therefore, the volume of the produced cone is

𝑉2 =1

3𝜋 × 𝐴𝐵2 × 𝐵𝐶

=1

3𝜋 × (4)2 × 3

= 16𝜋 cubic cm

In this case, the curved surface area of the cone is

𝑆2 = 𝜋 × 𝐴𝐵 × 𝐵𝐶

= 𝜋 × 4 × 5

= 20𝜋 square cm

Therefore, the difference between the volumes of the two cones so formed

is

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𝑉2 − 𝑉1 = 16𝜋 − 12𝜋

= 4𝜋 cm³

Hence the difference between the volumes is 4𝜋 cm³

And surface areas are 15𝜋 cm² and 20𝜋 cm².

Page No 14.31:

Question 60: A 5 m wide cloth is used to make a conical tent of base

diameter 14 m and height 24 m. Find the cost of cloth used at the rate of

Rs 25 per metre. [Use π = 22/7] [CBSE 2014]

ANSWER:

Let the slant height of the cone be l.

Thus , the slant height is given by

𝑙 = √(14

2)

2+ (24)2

= √49 + 576

= √625

= 25 m

Now, the curved surface area of the tent is given by 22

7×

14

2× 25

= 550 m2

The curved surface area will be equal to the area of the cloth

⇒ 550 = Length × 5

⇒ Length = 110 m

Now, the cost of cloth is given by

110 × 25

= Rs 2750

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Page No 14.31:

Question 61: The volume of a hemi-sphere is 24251

2 cm3. Find its curved

surface area. (Use π = 22/7)

ANSWER:

Let the radius of the hemisphere be r cm.

Volume of hemisphere = 24251

2 cm3

⇒ 2

3𝜋𝑟3 =

4851

2

⇒2

3×

22

7𝑟3 =

4851

2

⇒ 𝑟3 =4851×3×7

2×2×22

⇒ 𝑟3 =441×21

2×2×2

⇒ 𝑟3 = (21

2)

3

⇒ 𝑟 =21

2 cm

Now, the curved surface area of hemisphere is given by

2𝜋𝑟2

= 2 ×22

7× (

21

2)

2

= 693 cm2

Page No 14.31:

Question 62: 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 the cylinder is 176 cm3, find the outer

and inner diameters of the cylinder. (Use π = 22/7)

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ANSWER:

The height of the hollow cylinder is 14 cm. Let the inner and outer radii

of the hollow cylinder are r cm and R cm respectively. The difference

between the outer and inner surface area of the hollow cylinder is

= 2𝜋𝑅 × 14 − 2𝜋𝑟 × 14

= 28𝜋(𝑅 − 𝑟) cm²

By the given condition, this difference is 88 square cm. Hence, we have

28𝜋(𝑅 − 𝑟) = 88

⇒ 𝑅 − 𝑟 =44×7

14×22

⇒ 𝑅 − 𝑟 =4×7

14×2

⇒ 𝑅 − 𝑟 = 1

The volume of the metal used in making the cylinder is

𝑉1 = 𝜋{(𝑅)2 − (𝑟)2} × 14 cm³

By the given condition, the volume of the metal is 176 cubic cm. Hence,

we have

𝜋{(𝑅)2 − (𝑟)2} × 14 = 176

⇒ 𝑅2 − 𝑟2 =176×7

14×22

⇒ 𝑅2 − 𝑟2 = 4

⇒ (𝑅 − 𝑟)(𝑅 + 𝑟) = 4

⇒ 1 × (𝑅 + 𝑟) = 4

⇒ 𝑅 + 𝑟 = 4

Hence, we have two equations with unknown’s R and r

𝑅 − 𝑟 = 1,

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𝑅 + 𝑟 = 4

Adding the two equations, we have

(𝑅 − 𝑟) + (𝑅 + 𝑟) = 1 + 4

⇒ 2𝑅 = 5

⇒ 𝑅 = 2.5

Then from the second equation, we have

𝑟 = 4 − 2.5 = 1.5

Therefore, the outer and inner diameters of the hollow cylinder are 5cm

and 3cm respectively.

Page No 14.31:

Question 63: The internal and external diameters of a hollow

hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of

painting 1 cm2 of the surface is 10 paise. Find the total cost to paint the

vessel all over.

ANSWER:

We are given the following hemi hollow sphere

The internal and external radii of the hollow hemispherical vessel are 21

2=

10.5 cm and 25.2

2= 12.6 cm respectively. Therefore, the total surface area

of the hollow hemispherical vessel is

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𝑆 = 2𝜋 × (12.6)2 + 2𝜋 × (10.5)2 + 𝜋{(12.6)2 − (10.5)2}

= 2 ×22

7× 158.76 + 2 ×

22

7× 110.25 +

22

7× 48.51

= 997.92 + 693 + 152.46

= 1843.38

The cost of painting 1 square cm is 10 paise. Therefore, the total cost of

painting the vessel all over is

1843.38 × 10 = 1843.38 × 10 Paise

= 184.338 Rupees

Hence total cost of painting is 184.338 rupees

Page No 14.31:

Question 64: Prove that the surface area of a sphere is equal to the curved

surface area of the circumscribed cylinder.

ANSWER:

We have the following figure to visualize the situation

Let the radius of the sphere is r. Therefore, the surface area of the sphere

is

𝑆 = 4𝜋 × 𝑟2

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= 4𝜋𝑟2

The circumscribed cylinder of the sphere must have radius r cm and

height 2r cm. Therefore, the curved surface area of the cylinder is

𝑆1 = 2𝜋𝑟 × 2𝑟2

= 4𝜋𝑟2

Hence, S and S1 are same. Thus the proof is complete.

Page No 14.31:

Question 65: If the total surface area of a solid hemisphere is 462 cm2,

find its volume (Take π = 22/7) [CBSE 2014]

ANSWER:

Let the radius of the hemisphere be r cm.

Total surface area of hemisphere = 462 cm2

⇒ 3𝜋𝑟2 = 462

⇒ 3 ×22

7× (𝑟)2 = 462

⇒ 𝑟2 = 49

⇒ 𝑟 = 7 cm

Now, the volume of hemisphere is given by

23𝜋𝑟3

=2

3×

22

7(7)3

=2156

3

= 7182

3 cm3

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Page No 14.31:

Question 66: 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?

ANSWER:

Diameter of the pipe = 5 mm = 0.5 cm

Radius of the pipe = 0.5

2cm =

1

4cm

Rate of flow of water through the pipe = 10 m/min = 1000 cm/min

Volume of water that flows out through the pipe in 1 min

= 𝜋𝑟2ℎ = 𝜋 × (1

4)

2× 1000 cm3

Volume of water flowing out through the pipe in t min

= 𝜋 (1

4)

2× 1000𝑡

Diameter of the conical vessel = 40 cm

Radius = 20 cm

Height or depth = 24 cm

Volume of the conical vessel = 1

3𝜋𝑅2𝐻 =

1

3𝜋(20)2 × 24 = 3200𝜋

Time required to fill the vessel = capacity of the vessel

volume of water flowing per min

𝑡 =3200𝜋

𝜋(14)2×1000

𝑡 = 51.2

So, the time required is 51.2 min = 51 min 12 sec

Page No 14.31:

Question 67: 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.

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ANSWER:

Height of the cone, h = 120 cm

Radius of the cone, r = 60 cm

Height of the cylinder, H = 180 cm

Radius of the cylinder, R = 60 cm

Volume of the cylinder = 𝜋𝑅2𝐻 = 𝜋(60)2 × 180 cm3

Volume of the cone = 1

3𝜋𝑟2ℎ =

1

3𝜋(60)2 × 120

Volume of water left in the cylinder = Volume of cylinder − volume of

the cone

= 𝜋(60)2 × 180 −1

3𝜋(60)2 × 120

= 𝜋(60)2[180 − 40]

= 𝜋 × 3600[140]

= 1584000 cm3

= 1.584 m3

Page No 14.32:

Question 68: A heap of rice in the form of a cone of diameter 9 m and

height 3.5 m. Find the volume of rice. How much canvas cloth is required

to cover the heap?

ANSWER:

The heap of rice is in the form of a cone.

Diameter, d = 9 m

radius, r = 92m92m

height, h = 3.5 m

Volume, 𝑉 =1

3𝜋𝑟2ℎ

=1

3𝜋 (

9

2)

2× 3.5

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= 74.25 m3

Thus, volume of rice = 74.25 m3

The canvas cloth required to cover the heap will be the curved surface

area of the cone

𝑙 = √ℎ2 + 𝑟2

𝑙 = √3.52 + (9

2)

2

𝑙 = √12.25 + 20.25

𝑙 = 5.7 m

CSA = 𝜋𝑟𝑙

= 𝜋 × (9

2) × 5.7

= 80.62 m2

Hence, the canvas cloth required to cover the heap will be 80.62 m2

Page No 14.32:

Question 69: 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.

ANSWER:

The cylinderical bucket has the height H = 32 cm

radius, R = 18 cm

Volume of the cylinderical bucket will be

𝑉 = 𝜋(𝑅)2𝐻 = 𝜋(18)2 × 32

Height of the conical heap, h = 24 cm

Volume of cylinderical bucket = Volume of the conical heap

𝜋(𝑅)2𝐻 =1

3𝜋𝑟2ℎ

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⇒ (18)2 × 32 =1

3𝑟2(24)

⇒ (18)2×32×3

24= 𝑟2

⇒ 1296 = 𝑟2

⇒ 36 = 𝑟

Hence, the radius of the conical heap = 36 cm

Slant height 𝑙 = √ℎ2 + 𝑟2

⇒ 𝑙 = √242 + 362

⇒ 𝑙 = √576 + 1296

⇒ 𝑙 = √1872

⇒ 𝑙 = 43.2 cm

Thus, the slant height l = 43.27 cm

Page No 14.32:

Question 70: 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?

ANSWER:

The radius of the hemispherical bowl, R = 9 cm

Radius of the cylinderical bottles, r = 1.5 cm

Height of the bottles, h = 4 cm

Let the number of bottles required be n.

Volume of the hemispherical bowl = n × Volume of the cylinderical

bottles

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Volume of the hemispherical Bowl

Volume of the cylinderical bottles = n

⇒2

3𝜋𝑅3

𝜋𝑟2ℎ= 𝑛

⇒2

3(9)3

(1.5)2(4)= 𝑛

⇒ 54 = 𝑛

Hence, the 54 bottles are required.

Page No 14.32:

Question 71: A factory manufactures 120,000 pencils daily. The pencil

is 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 ₹0.05 per dm2.

ANSWER:

Length of the pencil, h = 25 cm

circumference of the base = 1.5 cm

Curved surface area of the pencil which needs to be painted will be

CSA = circumference × height

= 1.5 × 25 cm2

= 37.5 cm2

= 0.375 dm2

Pencils manufactured in one day = 120000

So, the total area to be painted will be 120000 × 0.375 dm2

= 45000 dm2

Cost of painting this area will be 45000 × 0.05 = Rs 2250

Page No 14.32:

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Question 72: The 3

4

th part of a conical vessel of internal radius 5 cm and

height 24 cm is full of water. The water is emptied into a cylindrical vessel

with internal radius 10 cm. Find the height of water in cylindrical vessel.

ANSWER:

Radius of conical vessel r = 5 cm

Height of conical vessel h = 24 cm

The volume of water = 3

4× volume of conical vessel.

=3

4×

1

3𝜋𝑟2ℎ

=3

4×

1

3𝜋 × 25 × 24

= 150𝜋

Let h' be the height of cylindrical vessel, which filled by the water of

conical vessel,

Radius of cylindrical vessel = 10 cm

Clearly,

Volume of cylindrical vessel = volume of water

𝜋(10)2ℎ = 150𝜋

⇒ ℎ =150𝜋

100𝜋

⇒ ℎ = 1.5 cm

Thus, the height of cylindrical vessel is 1.5 cm.

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Exercise – 14.2

Page No 14.60:

Question 1: A tent is in the form of a right circular cylinder surmounted

by a cone. The diameter of cylinder is 24 m. The height of the cylindrical

portion is 11 m while the vertex of the cone is 16 m above the ground.

Find the area of canvas required for the tent.

ANSWER:

We have a right circular cylinder surmounted by a cone.

Diameter of cylinder = 24 m, Height if cylindrical portion = 11 m and the

vertex of the cone is 16 meters above the ground. We have to find the area

of canvas required for the tent.

Suppose curved area of the cone portion is 𝑆1.

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From the above figure the slant height of the top is given by

𝑙 = √52 + 122

= 13 m

𝑟 =24

2= 12 cm

𝑆1 = 𝜋𝑟𝑙

=22

7× 12 × 13 m²

Now, Let us suppose that the curved area of cylinder is 𝑆2

𝑆2 = 2𝜋𝑟ℎ

= 2 ×22

7× 12 × 11 m²

Therefore, the area of canvas is given by

𝑆 = 𝑆1 + 𝑆2

= (22

7× 12 × 13 + 2 ×

22

7× 12 × 11) m²

=22

7(12 × 13 + 2 × 12 × 11) m²

=22

7× 420 m²

= 1320 m²

Hence, 𝑆 = 1320 m².

Page No 14.60:

Question 2: A rocket is in the form of a circular cylinder closed at the

lower end with a cone of the same radius attached to the top. The cylinder

is of radius 2.5 m and height 21 m and the cone has the slant height 8 m.

Calculate the total surface area and the volume of the rocket.

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ANSWER:

Given:

Radius of the cylinder 𝑟 = 2.5 m, height of the cylinder, ℎ = 21 m, slant

height of the cone 𝑙 = 8 m.We have to find total surface area and volume

of the rocket

Let us assume that the area of the cone is 𝑆1.

𝑆1 = 𝜋𝑟𝑙

= 3.14 × 2.5 × 8

= 62.8 m²

The area of the cylinder 𝑆2 is given by

𝑆2 = 2𝜋𝑟ℎ + 𝜋𝑟2

= 3.14 × 2.5(2 × 21 + 2.5)

= 349.33 m²

Total area 𝑆 is

𝑆 = 𝑆1 + 𝑆2

= 62.8 + 349.33

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= 412.13 m²

Now, we are going to find the volume of the rocket V.

Volume of the cone is given by

𝑉1 =1

3𝜋𝑟2ℎ

=1

3× 3.14 × 2.52√82 − 2.52

= 49.71 m²

Volume of the cylinder is

𝑉1 = 𝜋𝑟2ℎ

= 3.14 × 2.52 × 21

= 412.13 m²

Total volume of the cone is given by

𝑉 = 𝑉1 + 𝑉2

= 49.71 + 412.13

= 461.84 m³

Hence, the area and the volume of the rocket is S = 412.12 m², V = 461.84

m³.

Page No 14.60:

Question 3: A tent of height 77 dm is in the form of a right circular

cylinder of diameter 36 m and height 44 dm surmounted by a right circular

cone. Find the cost of the canvas at Rs. 3.50 per m2. (Use π = 22/7).

ANSWER:

Given:

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Height of the tent h = 77 dm = 7.7 m, diameter of cylinder 𝑑 = 36 m

Height of the cylinder h1 = 44 dm = 4.4 m, height of cone h2 = 33 dm =

3.3 m

We have the following diagram

Radius 𝑟 =𝑑

2=

36

2= 18 m

The curved area 𝑆1 of cylinder is given by

𝑆1 = 2𝜋𝑟ℎ

= 2 ×22

7× 18 × 4.4

= 497.82 m2

The slant height of the cone is

𝑙 = √ℎ2 + 𝑟2

= √3.32 + 182

= √334.89

= 18.3 m

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The curved area of the cone is given by

𝑆2 = 𝜋𝑟𝑙 =22

7× 18 × 18.3

= 1035.25 m2

The total area of the canvas required is given as

S = S1 + S2

= 497.82 + 1035.25

= 1533.07 m2

Therefore, the cost of the canvas at the rate of Rs 3.5 per square meter is

given by

1533.07 × 3.5

= Rs 5365.745

Hence the cost of the canvas is Rs 5365.745

Page No 14.60:

Question 4: A toy is in the form of a cone surmounted on a hemisphere.

The diameter of the base and the height of the cone are 6 cm and 4 cm,

respectively. Determine the surface area of the toy. (Use π = 3.14)

ANSWER:

Given that, a toy is in the form of a cone surmounted on the hemisphere.

Diameter of the base 𝑑 = 6 cm and the height of the cone ℎ = 4 cm, then

we have to find the surface area of the toy.

We have the following figure

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The radius of the base is

𝑟 =𝑑

2

=6

2= 3 cm

From the above figure, the slant height of the cone is

𝑙 = √𝑟2 + ℎ2

= √32 + 42

= 5 cm

We know that when the surface area of the cone is 𝑆1, then

𝑆1 = 𝜋𝑟𝑙

= 3.14 × 3 × 5

= 47.1 cm²

The surface area of the hemisphere is

𝑆2 = 2𝜋𝑟2

= 2 × 3.14 × 32

= 56.52 cm²

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Therefore, the surface area of the toy is

S = S1 + S2

= 47.1 + 56.52

= 103.62 cm2

Hence, 𝑆 = 103.62 cm²

Page No 14.60:

Question 5: A solid is in the form of a right circular cylinder, with a

hemisphere at one end and a cone at the other end. The radius of the

common base is 3.5 cm and the heights of the cylindrical and conical

portions are 10 cm. and 6 cm, respectively. Find the total surface area of

the solid. (Use π = 22/7)

ANSWER:

We have the following diagram

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For cone, we have

𝑟 = 3.5 cm

ℎ = 6 cm

𝑙 = √𝑟2 + ℎ2

= √3.52 + 62

= 6.95 cm

Curved surface area of the cone is given as

𝑆1 = 𝜋𝑟𝑙

=22

7× 3.5 × 6.946

= 76.408 cm²

For cylindrical part, we have

𝑟 = 3.5 cm

ℎ = 10 cm

Curved surface area of the cylinder is

𝑆2 = 2𝜋𝑟ℎ

= 2 ×22

7× 3.5 × 10

= 220 cm²

The surface area of the hemisphere is

𝑆3 = 2𝜋𝑟2

= 2 ×22

7× 3.52

= 77 cm²

Total surface area of the solid is given by

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𝑆 = 𝑆1 + 𝑆2 + 𝑆3

= 76.408 + 220 + 77

= 373.408 cm²

Hence the total surface area of the solid is 𝑆 = 373.408 cm².

Page No 14.60:

Question 6: 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 cylindrical part are 5 cm and 13 cm respectively. The radii of the

hemispherical 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.

ANSWER:

We have the following diagram

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For cylindrical part, we have

ℎ = 13 cm

𝑟 = 5 cm

Therefore, the curved surface area of the cylinder is given by

𝑆1 = 2𝜋𝑟ℎ

= 2 × 3.14 × 5 × 13

= 408.2 cm²

For conical part, we have

ℎ = 30 − 13 − 5

= 12 cm

𝑙 = √ℎ2 + 𝑟2

= √122 + 52

= 13 cm

Therefore, the curved surface area of the conical part is

𝑆2 = 𝜋𝑟𝑙

= 3.14 × 5 × 13

= 204.1 cm²

For hemisphere, we have

𝑟 = 5 cm

Therefore, the surface area of the hemisphere is

𝑆3 = 2𝜋𝑟2

= 2 × 3.14 × 52

= 157 cm²

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The total surface area of the toy is

𝑆 = 𝑆1 + 𝑆2 + 𝑆3

= 408.2 + 204.1 + 157

= 769.3 cm²

Hence, total surface area of the toy is 𝑆 = 769.3 cm²

Page No 14.60:

Question 7: A cylindrical tub of radius 5 cm and length 9.8 cm is full of

water. A solid in the form of a right circular cone mounted on a

hemisphere is immersed in the tub. If the radius of the hemisphere is

immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height

of the cone outside the hemisphere is 5 cm, find the volume of the water

left in the tub (Take π = 22/7)

ANSWER:

To find the volume of the water left in the tube, we have to subtract the

volume of the hemisphere and cone from volume of the cylinder.

For right circular cylinder, we have

𝑟 = 5 cm

ℎ = 9.8 cm

The volume of the cylinder is

𝑉1 = 𝜋𝑟2ℎ

=22

7× 52 × 9.8

= 770 cm²

For hemisphere and cone, we have

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𝑟 = 3.5 cm

ℎ = 5 cm

Therefore, the total volume of the cone and hemisphere is

𝑉2 =1

3𝜋𝑟2ℎ +

2

3𝜋𝑟3

=1

3×

22

7× 3.52 × 5 +

2

3×

22

7× 9.8 × 3.53

= 154 cm³

The volume of the water left in the tube is

𝑉 = 𝑉1 − 𝑉2

= 770 − 154

= 616 cm³

Hence, the volume of the water left in the tube is 𝑉 = 616 cm³

Page No 14.61:

Question 8: A circus tent has cylindrical shape surmounted by a conical

roof. The radius of the cylindrical base is 20 m. The heights of the

cylindrical and conical portions are 4.2 m and 2.1 m respectively. Find the

volume of the tent.

ANSWER:

Given that:

Radius of the cylindrical base 𝑟 = 20 m

Height of the cylindrical portion ℎ1 = 4.2 m

Height of the conical portion ℎ2 = 2.1 m

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The volume of the cylinder is given by the following formula

𝑉1 = 𝜋𝑟2ℎ1

=22

7× 202 × 4.2

= 5280 cm³

The volume of the conical portion is

𝑉2 =1

3𝜋𝑟2ℎ2

=1

3×

22

7× 202 × 2.1

= 880 cm³

Therefore, the total volume of the circus tent is

𝑉 = 𝑉1 + 𝑉2

= 5280 + 880

= 6160 cm³

Hence, the volume of the circus tent is 𝑉 = 6160 m³

Page No 14.61:

Question 9: A petrol tank is a cylinder of base diameter 21 cm and length

18 cm fitted with conical ends each of axis length 9 cm. Determine the

capacity of the tank.

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ANSWER:

To find the total capacity of the tank, we have to add the volume of the

cylinder and cone.

Diameter of the cylinder, 𝑑 = 21 cm

Radius of the cylinder, 𝑟 =𝑑

2=

21

2 cm

Height of the cylinder, ℎ1 = 18 cm

Also, radius of cone, 𝑟 = 21

2 cm

Height of the cone, ℎ2 = 9 cm

Now,

Total capacity of the tank

= Volume of the cylinder + Volume of 2 cones

= 𝜋𝑟2ℎ1 + 2 ×1

3𝜋𝑟2ℎ2

= 𝜋𝑟2 (ℎ1 +2

3ℎ2)

=22

7× (

21

2)

2× (18 +

2

3× 9)

=22

7× (

21

2)

2× 24

= 8316 cm3

Hence the total capacity of the tank is 8316 cm3.

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Page No 14.61:

Question 10: A conical hole is drilled in a circular cylinder of height 12

cm and base radius 5 cm. The height and the base radius of the cone are

also the same. Find the whole surface and volume of the remaining

cylinder.

ANSWER:

Given that:

𝑟 = 5 cm

ℎ = 12 cm

We have the following diagram

Slant height of cone is given by

𝑙 = √𝑟2 + ℎ2

= √52 + 122

= 13 cm

The total surface area of the remaining part is given by

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𝑆 = 2𝜋𝑟ℎ + 𝜋𝑟2 + 𝜋𝑟𝑙

= 2 × 𝜋 × 5 × 12 + 𝜋 × 52 + 𝜋 × 5 × 13

= 120𝜋 + 25𝜋 + 65𝜋

= 210𝜋 cm²

The volume of the remaining part is given by

𝑉 = 𝜋𝑟2ℎ −1

3𝜋𝑟2ℎ

=2

3𝜋𝑟2ℎ

=2

3× 𝜋 × 52 × 12

= 200𝜋 cm³

Hence, 𝑆 = 210𝜋 cm², 𝑉 = 200𝜋 cm³

Page No 14.61:

Question 11: A tent is in the form of a cylinder of diameter 20 m and

height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find

the capacity of the tent and the cost of the canvas at Rs 100 per square

metre.

ANSWER:

Given that:

Radius of the base 𝑟 =𝑑

2=

20

2= 10 m

Height of the cylinder ℎ1 = 2.5 m

Height of the cone ℎ2 = 7.5 m

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Slant height of the cone

𝑙 = √𝑟2 + ℎ2

= √102 + 7.52

= 12.5 cm

The total capacity of the tent is given by

𝑉 = 𝜋𝑟2ℎ1 −1

3𝜋𝑟2ℎ2

= 𝜋 × 102 × 2.5 +1

3× 𝜋 × 102 × 7.5

= 𝜋 × 250 + 𝜋 × 250

= 500𝜋 cm³

The total area of canvas required for the tent is

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𝑆 = 2𝜋𝑟ℎ + 𝜋𝑟𝑙

= 2 × 3.14 × 10 × 2.5 + 3.14 × 10 × 12.5

= 𝜋(2 × 10 × 2.5 + 10 × 12.5)

= 𝜋(50 + 125)

=22

7× 175

= 550𝜋 m²

Therefore, the total cost of the canvas is

= 100 × 550

= Rs. 55000

Hence, the total capacity and cost is 𝑉 = 550𝜋 m², and Rs. 55000

Page No 14.61:

Question 12: A boiler is in the form of a cylinder 2 m long with

hemispherical ends each of 2 metre diameter. Find the volume of the

boiler.

ANSWER:

Given that:

Height of the cylinder ℎ = 2 m

Radius of the cylinder and hemisphere are same and is given by

𝑟 =𝑑

2=

2

2= 1 m

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The volume of the cylinder is cylinder is

𝑉1 = 𝜋𝑟2ℎ

=22

7× 12 × 2

=22

7× 2 cm³

There are two hemispheres at each ends of the cylinder, therefore the

volume of the two hemispheres is

𝑉2 =2

3𝜋𝑟3 +

2

3𝜋𝑟3

=4

3×

22

7× 13

=22

7×

4

3 cm³

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Therefore, the total volume of the boiler is given by

𝑉 = 𝑉1 + 𝑉2

= (22

7× 2 +

22

7×

4

3) m³

=22

7×

10

3 m³

=220

21 m³

Hence the volume of the boiler is 𝑉 =220

21 m³

Page No 14.61:

Question 13: A vessel is a hollow cylinder fitted with a hemispherical

bottom of the same base. The depth of the cylinder is 14

3 m and the

diameter of hemisphere is 3.5 m. Calculate the volume and the internal

surface area of the solid.

ANSWER:

Given that:

Radius of the same base 𝑟 =3.5

2= 1.75 m

Height of the cylinder ℎ =14

3 m

The volume of the vessel is given by

𝑉 = 𝜋𝑟2ℎ +2

3𝜋𝑟3

= 3.14 × 1.752 ×14

3+

2

3× 3.14 × 1.753

= 56 m³

The internal surface area of the solid is

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𝑆 = 2𝜋𝑟2 + 2𝜋𝑟ℎ

= 2 × 3.14 × 1.752 + 2 × 3.14 × 1.75 ×14

3

= 70.51 m²

Hence, the volume of the vessel and internal surface area of the solid

is 𝑉 = 56𝜋 m³, and S = 70.51 m²

Page No 14.61:

Question 14: A solid is composed of a cylinder with hemispherical ends.

If the whole length of the solid is 104 cm and the radius of each of the

hemispherical ends is 7 cm, find the cost of polishing its surface at the rate

of Rs 10 per dm2.

ANSWER:

We have a solid composed of cylinder with hemispherical ends.

Radius of the two curved surfaces (𝑟) = 7 cm

Height of cylinder is h.

Total height of the body (ℎ + 2𝑟) = 104 cm

So, total surface area is given by,

Total surface area

= Curved surface area of cylinder + 2(Curves surface area of hemisphere)

= 2𝜋𝑟ℎ + 2(2𝜋𝑟2)

= 2𝜋𝑟(ℎ + 2𝑟)

= 2(3.14)(7)(104) cm²

= 4571.84 cm²

Change the units of curved surface area as,

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Total surface area = 4571.84

100 dm²

= 45.7184 dm²

Cost of polishing the surface is Rs 10 per dm².

So total cost,

= Rs. (45.7184)(10)

= Rs. 457.18

Page No 14.61:

Question 15: A cylindrical vessel of diameter 14 cm and height 42 cm is

fixed symmetrically inside a similar vessel of diameter 16 cm and height

42 cm. The total space between the two vessels is filled with cork dust for

heat insulation purposes. How many cubic centimeters of cork dust will

be required?

ANSWER:

We have to find the volume of cork dust filled between the two vessels.

Radius of outer vessel (𝑟2) = 8 cm

Radius of inner vessel (𝑟1) = 7 cm

Height of the cylinder (ℎ) = 42 cm

So, volume of cork dust filled between the two vessels,

= 𝜋ℎ(𝑟22 − 𝑟1

2)

= (3.14)(42)(64 − 49)

= 1978.2 cm²

= 1980 cm³

Volume of cork dust filled between the two vessels is 1980 cm³.

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Page No 14.61:

Question 16: A cylindrical road roller made of iron is 1 m long, Its

internal diameter is 54 cm and the thickness of the iron sheet used in

making the roller is 9 cm. Find the mass of the roller, if 1 cm3 of iron has

7.8 gm mass. (Use π = 3.14)

ANSWER:

We have to find the mass of the roller.

Radius of inner cylinder (𝑟1) = 27 cm

Radius of outer cylinder

(𝑟2) = (27 + 9) cm

= 36 cm

Length of the cylinder (ℎ) = 100 cm

So, volume of iron,

= 𝜋ℎ(𝑟22 − 𝑟1

2)

= (3.14)(100)(1296 − 729)

= 178038 cm³

It is given that, 1 cm³ of iron has a mass of 7.8 gm.

So the mass of iron used,

= (178038)(7.8) gm

= 1388696.4 gm

= 1388.7 kg

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Page No 14.61:

Question 17: A vessel 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.

ANSWER:

We have to find the inner surface area of a vessel which is in the form of

a hemisphere mounted by a hollow cylinder.

Radius of hemisphere and cylinder (𝑟) = 7 cm

Total height of vessel (𝑟 + ℎ) = 13 cm

So, the inner surface area of a vessel,

= 2𝜋𝑟(𝑟 + ℎ)

= 2 (22

7) (7)(13) cm²

= 572 cm²

Page No 14.61:

Question 18: 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.

ANSWER:

We have to find the total surface area of a toy which is a cone surmounted

on a hemisphere.

Radius of hemisphere and the base of the cone (𝑟) = 3.5 cm

Height of the cone,

h = 15.5 − 3.5 = 12 cm

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slant height (𝑙) = √ℎ2 + 𝑟2

= √122 + 3.52

= √156.25

= 12.5 cm

So, total surface area of toy,

𝑆 = 𝜋𝑟𝑙 + 2𝜋𝑟2

= 𝜋𝑟(𝑙 + 2𝑟)

=22

7× 3.5(12.5 + 2 × 3.5)

= 214.5 cm2

Page No 14.61:

Question 19: The difference between outside and inside surface areas of

cylindrical metallic pipe 14 cm long is 44 m2. If the pipe is made of 99

cm3 of metal, find the outer and inner radii of the pipe.

ANSWER:

We have to find the outer and inner radius of a hollow pipe.

Radius of inner pipe be (𝑟1)

Radius of outer cylinder be (𝑟2)

Length of the cylinder (ℎ) = 14 cm

Difference between the outer and the inner surface area is 44 cm²

So,

= 2𝜋ℎ(𝑟2 − 𝑟1) = 44

= 2 (22

7) (14)(𝑟2 − 𝑟1) = 44

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So,

(𝑟2 − 𝑟1) =1

2 …… (1)

So, volume of metal used is 99 cm³, so,

𝜋ℎ(𝑟22 − 𝑟1

2) = 99

(22

7) (14)(𝑟2 − 𝑟1)(𝑟2 + 𝑟1) = 99

Use equation (1) in the above to get,

(22

7) (14) (

1

2) (𝑟2 + 𝑟1) = 99

Therefore,

(𝑟2 + 𝑟1) =9

2 …… (2)

Solve equation (1) and (2) to get,

𝑟2 =5

2 cm

𝑟1 = 2 cm

Page No 14.61:

Question 20: A right circular cylinder having diameter 12 cm and height

15 cm is full ice-cream. The ice-cream is to be filled in 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.

ANSWER:

We have to find the number of cones which can be filled using the ice

cream in the cylindrical vessel.

Radius of the cylinder (𝑟1) = 6 cm

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Height of cylinder (ℎ) = 15 cm

Radius of cone and the hemisphere on it (𝑟2) = 3 cm

Height of cone (𝑙) = 12 cm

Let ‘n’ number of cones filled. So we can write it as,

𝑛(Volume of each cone) = Volume of cylinder

So,

(𝑛) (1

3𝜋𝑟2

2𝑙 +2

3𝜋𝑟2

3) = 𝜋𝑟12ℎ

(𝑛) (𝑟2

2(𝑙+2𝑟2)

3) = 𝑟1

2ℎ

Now put the values to get,

(𝑛) (9(12+6)

3) = 36(15)

54𝑛 = 540

Therefore, 𝑛 = 10

Page No 14.61:

Question 21: A solid iron pole having cylindrical portion 110 cm high

and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the

mass of the pole, given that the mass of 1 cm3 of iron is 8 gm.

ANSWER:

We have to find the mass of a pole having a cylindrical base surmounted

by a cone.

Radius of cone and cylinder (𝑟) = 6 cm

Height of cylinder (ℎ) = 110 cm

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Height of cone (𝑙) = 9 cm

So volume of the pole is,

= 𝜋𝑟2ℎ +1

3𝜋𝑟2𝑙

= 𝜋𝑟2 (ℎ +1

3𝑙)

Put the values to get,

= (22

7) (36)(110 + 3) cm³

= 12785.14 cm³

Mass of 1 cm³ of iron is 8 gm.

Therefore, mass of the iron,

= (12785.14)(8) gm

= 102.2 kg

Page No 14.61:

Question 22: 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.

ANSWER:

We have to find the remaining volume of the cylinder when the toy is

inserted into it. The toy is a hemisphere surmounted by a cone.

Radius of cone, cylinder and hemisphere (𝑟) = 2 cm

Height of cone (𝑙) = 2 cm

Height of the cylinder (ℎ) = 4 cm

So the remaining volume of the cylinder when the toy is inserted into it,

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= 𝜋𝑟2ℎ − (1

3𝜋𝑟2𝑙 +

2

3𝜋𝑟3)

Put the values to get,

= 16𝜋 − (8𝜋

3+

16𝜋

3)

= 16𝜋 − 8𝜋

= 8𝜋

Page No 14.61:

Question 23: 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

bottoms. Find the volume of water left in the cylinder, if the radius of the

cylinder is 60 cm and its height is 180 cm.

ANSWER:

We have to find the remaining volume of water left in the cylinder when

the solid is inserted into it. The solid is a hemisphere surmounted by a

cone.

Radius of cone, cylinder and hemisphere (𝑟) = 60 cm

Height of cone (𝑙) = 120 cm

Height of the cylinder (ℎ) = 180 cm

So the remaining volume of water left in the cylinder when the solid is

inserted into it,

= 𝜋𝑟2ℎ − (1

3𝜋𝑟2𝑙 +

2

3𝜋𝑟3)

= 𝜋𝑟2 (ℎ −1

3𝑙 +

2

3𝑟)

Put the values to get,

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= (22

7) (3600)(180 − 40 − 40) m³

= 1.131 m³.

Page No 14.62:

Question 24:

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is

full of water. A solid cone of base diameter 7 cm and height 6 cm is

completely immersed in water. Find the value of water (i) displaced out

of the cylinder (ii) left in the cylinder. (Take π 22/7)

ANSWER:

We have a cylindrical vessel in which a cone is inserted. We have,

Radius of the cylinder (𝑟1) = 5 cm

Radius of cone (𝑟2) = 3.5 cm

Height of cylinder (ℎ) = 10.5 cm

Height of cone (𝑙) = 6 cm

(i) We have to find the volume of water displaced from the cylinder when

cone is inserted.

So,

Volume of water displaced = Volume of cone

So volume of water displaced,

=1

3𝜋𝑟2

2𝑙

=1

3(

22

7) (12.25)(6) cm²

= 77 cm³

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(ii) We have to find the volume of water remaining in the cylinder.

Volume of water left = Volume of cylinder – Volume of cone

So volume of the water left in the cylinder,

= [(22

7(25)(10.5)) − (77)] cm³

= (825 − 77) cm³

= 748 cm³

Page No 14.62:

Question 25: A hemispherical depression is cut out from one face of a

cubical wooden block of edge 21 cm, such that the diameter of the

hemisphere is equal to the edge of the cube. Determine the volume and

total surface area of the remaining block.

ANSWER:

We have to find the remaining volume and surface area of a cubical box

when a hemisphere is cut out from it.

Edge length of cube (𝑎) = 21 cm

Radius of hemisphere (𝑟) = 10.5 cm

Therefore, volume of the remaining block,

= Volume of box – Volume of hemisphere

So,

= (𝑎)3 −2

3𝜋𝑟3

= (21)3 −2

3(

22

7) (

21

2)

3

= (9261 − 2425.5) cm³

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= 6835.5 cm³

So, remaining surface area of the box,

= Surface area of box – Area of base of hemisphere + Curved surface area

of hemisphere

Therefore,

= 6(𝑎)2 − 𝜋𝑟2 + 2𝜋𝑟2

= 6𝑎2 + 𝜋𝑟2

Put the values to get the remaining surface area of the box,

= [6(441) +22

7(

21

2)

2] cm²

= 2992.5 cm²

Page No 14.62:

Question 26: A toy is in the form of a hemisphere surmounted by a right

circular cone of the same base radius as that of the hemisphere. If the

radius of the base of the cone is 21 cm and its volume is 2/3 of the volume

of hemisphere, calculate the height of the cone and the surface area of the

toy. (Use 𝜋 =22

7).

ANSWER:

Solution:

Let the height of the conical part be h.

Radius of the cone = Radius of the hemisphere = r = 21 cm

The toy can be diagrammatically represented as

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Volume of the cone = 1

3𝜋𝑟2ℎ

Volume of the hemisphere = 2

3𝜋𝑟3

According to given information:

Volume of the cone =2

3× Volume of the hemisphere

∴1

3𝜋𝑟2ℎ =

2

3×

2

3𝜋𝑟3

⇒ ℎ =2

3×

2

3𝜋𝑟3

1

3𝜋𝑟2

⇒ ℎ =4

3𝑟

∴ ℎ =4

3× 21 cm = 28 cm

Thus, surface area of the toy = Curved surface area of cone + Curved

surface area of hemisphere

= 𝜋𝑟𝑙 + 2𝜋𝑟2

= 𝜋𝑟√ℎ2 + 𝑟2 + 2𝜋𝑟2

= 𝜋𝑟(√ℎ2 + 𝑟2 + 2𝑟)

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=22

7× 21 cm (√(28 cm)2 + (21cm)2 + 2 × 21cm)

= 66(√784 + 441 + 42) cm²

= 66(√1225 + 42) cm²

= 66(35 + 42) cm²

= 66 × 77 cm²

= 5082 cm²

Page No 14.62:

Question 27: A solid is in the shape of a cone surmounted on a

hemisphere, the radius of each of them is being 3.5 cm and the total height

of solid is 9.5 cm. Find the volume of the solid. (Use π = 22/7).

ANSWER:

Height of cone = 9.5 − 3.5 = 6 cm

Volume of the solid = Volume of cone + Volume of hemisphere

= 1

3𝜋𝑟2ℎ +

2

3𝜋𝑟3

=1

3×

22

7× (3.5)2 × 6 +

2

3×

22

7× (3.5)3

= 77 + 89.83

= 166.83 cm3

Page No 14.62:

Question 28: An wooden toy is made by scooping out a hemisphere of

same radius from each end of a solid cylinder. If the height of the cylinder

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is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the

toy. (Use π = 22/7).

ANSWER:

Volume of wood in the toy = Volume of cylinder − 2(Volume of

hemisphere)

= 𝜋𝑟2ℎ − 2 ×2

3𝜋𝑟3

=22

7× (3.5)2 × 10 − 2 ×

2

3×

22

7× (3.5)3

= 385 − 179.67

= 205.33 cm3

Page No 14.62:

Question 29: The largest possible sphere is carved out of a wooden solid

cube of side 7 cm. Find the volume of the wood left. (Use π =22

7).

[CBSE 2014]

ANSWER:

The radius of the largest possible sphere is carved out of a wooden solid

cube is equal to the half of the side of the cube.

Radius of the sphere = 7

2= 3.5

Volume of the wood left = Volume of cube − Volume of sphere

= (Side)3 −4

3𝜋𝑟3

= 73 −4

3×

22

7× (3.5)3

= 343 − 179.67

= 163.33 cm3

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Page No 14.62:

Question 30: From a solid cylinder of height 2.8 cm and diameter 4.2 cm,

a conical cavity of the same height and same diameter is hollowed out.

Find the total surface area of the remaining solid. (Take π=227π=227).

[CBSE 2014]

ANSWER:

Total surface area of the remaining solid

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

= 2𝜋𝑟ℎ + 𝜋𝑟𝑙 + 𝜋𝑟2

= 2 ×22

7×

4.2

2× 2.8 +

22

7×

4.2

2× √(

4.2

2)

2+ (2.8)2 +

22

7× (

4.2

2)

2

= 2 ×22

7× 2.1 × 2.8 +

22

7× 2.1 × √(2.1)2 + (2.8)2 +

22

7× (2.1)2

= 2 ×22

7× 2.1 × 2.8 +

22

7× 2.1 × 3.5 +

22

7× (2.1)2

= 36.96 + 23.1 + 13.86

= 73.92 cm2

The total surface area of the remaining solid is 73.92 cm2.

Page No 14.62:

Question 31: The largest cone is curved out from one face of solid cube

of side 21 cm. Find the volume of the remaining solid. [CBSE 2015]

ANSWER:

The radius of the largest possible cone is carved out of a solid cube is

equal to the half of the side of the cube.

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Also, the height of the cone is equal to the side of the cube.

Radius of the cone = 21

2= 10.5 cm

Volume of the remaining solid = Volume of cube − Volume of cone

= (Side)3 −1

3𝜋𝑟2ℎ

= (21)3 −1

3×

22

7× (10.5)2 × 21

= 9261 − 2425.5

= 6835.5 cm3

Disclaimer: The answer given in the book is not correct.

Page No 14.62:

Question 32: A solid wooden toy is in the form of a hemisphere

surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm

and the total wood used in the making of toy is 1665

6 cm3. Find the height

of the toy. Also, find the cost of painting the hemispherical part of the toy

at the rate of Rs 10 per cm2. (Take 𝜋 =22

7). [CBSE 2015]

ANSWER:

Volume of solid wooden toy = 1665

6 cm3

⇒ Volume of cone + Volume of hemisphere = 1665

6

⇒1

3𝜋𝑟2ℎ +

2

3𝜋𝑟3 =

1001

6

⇒1

3×

22

7× (3.5)2 × ℎ +

2

3×

22

7× (3.5)3 =

1001

6

⇒1

3×

22

7× (3.5)2(ℎ + 7) =

1001

6

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⇒ 38.53(ℎ + 7) =1001

6

⇒ ℎ + 7 = 13

⇒ ℎ = 6 cm

Height of the solid wooden toy

= Height of cone + Radius of hemisphere

= 6 + 3.5

= 9.5 cm

Now, curved surface area of hemisphere = 2𝜋𝑟2

= 2 ×22

7× (3.5)2 = 77 cm2

Cost of painting the hemispherical part of the toy = 10 × 77 = Rs 770

Page No 14.62:

Question 33: In the given figure, from a cuboidal solid metalic block, of

dimensions 15 cm ⨯ 10 cm ⨯ 5 cm, a cylindrical hole of diameter 7 cm is

drilled out. Find the surface area of the remaining block. (Take 𝜋 =22

7).

[CBSE 2015]

ANSWER:

Surface area of the remaining block

= Total Surface area of cubic block + Curved Surface area of cylinder −

2(Area of circular base)

= 2(𝑙𝑏 + 𝑏ℎ + 𝑙ℎ) + 2𝜋𝑟ℎ − 2𝜋𝑟2

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= 2(15 × 10 + 10 × 5 + 15 × 5) + 2 ×22

7×

7

2× 5 − 2 ×

22

7× (

7

2)

2

= 2 × 275 + 110 − 77

= 583 cm2

Page No 14.62:

Question 34: A building is in the form of a cylinder surmounted by a

hemi-spherical vaulted dome and contains 4119

21 m3 of air. If the internal

diameter of dome is equal to its total height above the floor, find the height

of the building?

ANSWER:

let the total height of the building be H m.

let the radius of the base be r m. Therefore, the radius of the

hemispherical dome is r m.

Now given that internal diameter = total height

⇒ 2𝑟 = 𝐻

Total height of the building = height of the cylinder +radius of the dome

⇒ H = h + r

⇒ 2r = h + r

⇒ r = h

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Volume of the air inside the building = volume of the cylinder+ volume

of the hemisphere

⇒ 4119

21= 𝜋𝑟2ℎ +

2

3𝜋𝑟3

⇒ 880

21= 𝜋ℎ2ℎ +

2

3𝜋ℎ3

⇒ 880

21= 𝜋ℎ3 (1 +

2

3)

⇒ 880

21= 𝜋ℎ3 (

5

3)

⇒ ℎ = 2 m

Hence, height of the building H = 2 × 2 = 4m

Page No 14.62:

Question 35: 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 is 10 cm × 5 cm ×

4 cm. The radius of each of the conical depression 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.

ANSWER:

The dimensions of the cuboid = 10 cm × 5 cm × 4 cm

Volume of the total cuboid = 10 cm × 5 cm × 4 cm = 200 cm3

Radius of the conical depressions, r = 0.5 cm

Depth, h = 2.1 cm

Volume of the conical depression

= 1

3𝜋𝑟2ℎ =

1

3𝜋(0.5)2(2.1) = 0.5495 cm3

Edge of cubical depression, a = 3 cm

Volume of the cubical depression = 𝑎3 = 33 = 27 cm3

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Volume of wood used to make the entire stand = Volume of the total

cuboid − volume of conical depression − volume of cubical depression

= 200 − 4 × 0.5495 − 27

= 170.802 cm3

Page No 14.63:

Question 36: A building is in the form of a cylinder surrounded by a

hemispherical dome. The base diameter of the dome is equal to 2

3 of the

total height of the building. Find the height of the building, if it

contains 671

21 m3 of air.

ANSWER:

Let the radius of the dome be r.

Diameter be 𝑑.

Let the height of the building be H.

Given 𝑑 =2

3𝐻

⇒ 2𝑟 =2

3𝐻

⇒ 𝑟 =𝐻

3

⇒ 3𝑟 = 𝐻

Also, h + r = H

⇒ 3𝑟 = ℎ + 𝑟

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⇒ 2𝑟 = ℎ

⇒ 𝑟 =ℎ

2

Volume of air = Volume of air in the cylinder + Volume of air in the

hemispherical dome

⇒ 𝜋𝑟2ℎ +2

3𝜋𝑟3 = 67

1

21

⇒ 𝜋 (ℎ

2)

2ℎ +

2

3𝜋 (

ℎ

2)

3=

1408

21

⇒ ℎ3 =11264

176= 64

⇒ ℎ = 4 m

Hence, the radius will be 𝑟 =ℎ

2=

4

2= 2 m

Height of the building, H = 3r = 3 × 2 = 6 m

Page No 14.63:

Question 37: A solid toy s in the form of a hemisphere surrounded by a

right circular cone. The height of 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.

ANSWER:

The height of the cone, h = 4 cm

Diameter of the base, d = 8 cm

Radius of the cone, r = 4 cm

lateral side will be

𝑙 = √ℎ2 + 𝑟2

𝑙 = √(4)2 + (4)2 = 4√2

Volume of the toy = volume of hemisphere + volume of cone

⇒ 𝑉 =2

3𝜋𝑟3 +

1

3𝜋𝑟2ℎ

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=𝜋𝑟2

3[2𝑟 + ℎ]

=𝜋(4)2

3[2 × 4 + 4]

= 201.14 cm3

When the cube circumscribes the toy, then

Volume of the cube = 𝑎3 = (8)3 = 512 cm3

Volume of cube−volume of toy = 512 − 201.14 = 310.86 cm3

Total surface area of the toy = curved surface area of the hemisphere +

curved surface area of the cone

= 2𝜋𝑟2 + 𝜋𝑟𝑙

= 2𝜋(4)2 + 𝜋 × 4 × 4√2

= 𝜋(4)2[2 + √2]

= 171.68 cm2

Page No 14.63:

Question 38: A circus tent is in the shape of cylinder surmounted by a

conical top of same diameter. If their common diameter is 56 m, the height

of the cylindrical part is 6 m and the total height of the tent above the

ground is 27 m, find the area of the canvas used in making the tent.

ANSWER:

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Total height of the tent above the ground = 27 m

Height of the cylinderical part, ℎ1 = 6 m

Height of the conical part, ℎ2 = 21 m

Diameter = 56 m

Radius = 28 m

Curved surface area of the cylinder, CSA1 = 2𝜋𝑟ℎ1 = 2𝜋 × 28 × 6 =

336𝜋

Curved surface area of the cylinder, CSA2 will be

𝜋𝑟𝑙 = 𝜋𝑟(√ℎ2 + 𝑟2) = 𝜋 × 28 × (√212 + 282) = 28𝜋(√441 + 784)

= 28𝜋 × 35

= 980𝜋

Total curved surface area = CSA of cylinder + CSA of cone

= CSA1 + CSA2

= 336𝜋 + 980𝜋

= 1316𝜋 = 4136 m2

Thus, the area of the canvas used in making the tent is 4136 m2 .

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Exercise – 14.3

Page No 14.78:

Question 1: A bucket has top and bottom diameter of 40 cm and 20 cm

respectively. Find the volume of the bucket if its depth is 12 cm. Also,

find the cost of tin sheet used for making the bucket at the rate of Rs. 1.20

per dm2. (Use π = 3.14)

ANSWER:

The radii of the top and bottom circles are r1 = 20 cm and r2 = 10 cm

respectively. The height of the bucket is h = 12 cm. Therefore, the volume

of the bucket is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟1 + 𝑟12) × ℎ

=1

3𝜋(202 + 20 × 10 + 102) × 12

=1

3×

22

7× 700 × 12

= 8800 cm³

The slant height of the bucket is

𝑙 = √(𝑟1 − 𝑟2)2 + ℎ2

= √(20 − 10)2 + 122

= √244

= 2√61 cm

The total surface area of the bucket is

= 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟22

=22

7× (20 + 10) × 2√61 +

22

7× 102

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=1320√61+2200

7 cm²

=1320√61+2200

7×100 dm²

The total cost of tin sheet used for making the bucket is

= 1.20 × (1320√61+2200

7×100)

= 21.40

Page No 14.78:

Question 2: A frustum of a right circular cone has a diameter of base 20

cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and

volume.

ANSWER:

The radii of the bottom and top circles are r1 = 10 cm and r2 = 6 cm

respectively. The height of the frustum cone is h = 3 cm. Therefore, the

volume of the bucket is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟1 + 𝑟12) × ℎ

=1

3𝜋(102 + 10 × 6 + 62) × 3

=1

3×

22

7× 196 × 3

= 616 cm³

Hence volume = 616 cm³

The slant height of the bucket is

𝑙 = √(𝑟1 − 𝑟2)2 + ℎ2

= √(10 − 6)2 + 32

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= √25

= 5 cm

The total surface area of the frustum cone is

= 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟12 + 𝜋𝑟2

2

=22

7× (10 + 6) × 5 +

22

7× 102 +

22

7× 62

=4752

7 Square cm

= 678.85 Square cm

Hence Total surface area = 678.85

Page No 14.78:

Question 3: The slant height of the frustum of a cone is 4 cm and the

perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface

of the frustum.

ANSWER:

The slant height of the frustum of the cone is l = 4 cm. The perimeters of

the circular ends are 18 cm and 6 cm. Let the radii of the bottom and top

circles are r1 cm and r2 cm respectively. Then, we have

2𝜋𝑟1 = 18

⇒ 𝜋𝑟1 = 9

2𝜋𝑟2 = 6

⇒ 𝜋𝑟2 = 3

The curved surface area of the frustum cone is

= 𝜋(𝑟1 + 𝑟2) × 𝑙

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= (𝜋𝑟1 + 𝜋𝑟2) × 𝑙

= (9 + 3) × 4

= 48 Square cm

Page No 14.78:

Question 4: The perimeters of the ends of a frustum of a right circular

cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its

volume, the slant surface and the total surface.

ANSWER:

The height of the frustum of the cone is h = 16 cm. The perimeters of the

circular ends are 44 cm and 33 cm. Let the radii of the bottom and top

circles are r1 cm and r2 cm respectively. Then, we have

2𝜋𝑟1 = 44

⇒ 𝜋𝑟1 = 22

⇒ 𝑟1 =22×7

22

⇒ 𝑟1 = 7

2𝜋𝑟2 = 33

⇒ 𝜋𝑟2 =33

2

⇒ 𝑟2 =33

2×

7

22

⇒ 𝑟2 =21

4

The slant height of the bucket is

𝑙 = √(𝑟1 − 𝑟2)2 + ℎ2

= √(7 −21

4)

2+ 162

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= 16.1 cm

The curved/slant surface area of the frustum cone is

= 𝜋(𝑟1 + 𝑟2) × 𝑙

= (𝜋𝑟1 + 𝜋𝑟2) × 𝑙

= (22 + 16.5) × 16.1

= 6.19.85 cm²

Hence Curved surface area = 619.85 cm²

The volume of the frustum of the cone is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(72 + 7 × 5.25 + 5.252) × 16

= 1900 cm³

Hence Volume of frustum = 1900 cm³

The total surface area of the frustum cone is

= 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟12 + 𝜋𝑟2

2

= 619.85 +22

7× 72 +

22

7× 5.252

= 860.25 Square cm

Hence Total surface area = 860.25 cm²

Page No 14.78:

Question 5: If the radii of the circular ends of a conical bucket which is

45 cm high be 28 cm and 7 cm, find the capacity of the bucket. (Use π =

22/7).

ANSWER:

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The height of the conical bucket is h = 45 cm. The radii of the bottom and

top circles are r1 = 28cm and r2 =7cm respectively.

The volume/capacity of the conical bucket is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(282 + 28 × 7 + 72) × 45

=1

3×

22

7× 1029 × 45

= 22 × 147 × 15

= 48510 cm³

Hence volume = 48510 cm³

Page No 14.78:

Question 6: The height of a cone is 20 cm. A small cone is cut off from

the top by a plane parallel to the base. If its volume be 1/125 of the volume

of the original cone, determine at what height above the base the section

is made.

ANSWER:

We have the following situation as shown in the figure

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Let VAB be a cone of height h1 = VO1 =20cm. Then from the symmetric

triangles VO 1A and VOA1, we have

VO1

VO=

O1A

OA1

⇒20

VO=

O1A

OA1

It is given that, volume of the cone VA1O is 1

125 times the volume of the

cone VAB. Hence, we have

1

3πO1A2 × VO =

1

125×

1

3πO1A2 × 20

⇒ (O1A

OA1)

2× VO =

4

25

⇒ (𝑉𝑂

20)

2× VO =

4

25

⇒ VO3 =400×4

25

⇒ VO3 = 16 × 4

⇒ VO = 4

Hence, the height at which the section is made is 20 − 4 = 16 cm.

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Page No 14.78:

Question 7: If the radii of the circular ends of a bucket 24 cm high are 5

cm and 15 cm respectively, find the surface area of the bucket.

ANSWER:

The height of the conical bucket is h = 24 cm. The radii of the bottom and

top circles are r1 = 15cm and r2 = 5cm respectively.

The slant height of the bucket is

𝑙 = √(𝑟1 − 𝑟2)2 + ℎ2

= √(15 − 5)2 + 242

= √676

= 26 cm

The curved surface area of the bucket is

= 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟22

=22

7× (15 + 5) × 26 + 𝜋 × 5

= 𝜋 × 20 × 26 + 25𝜋

= 545𝜋 cm2

Hence the curved surface area of the bucket is 545𝜋 cm2

Page No 14.79:

Question 8: The radii of the circular bases of a frustum of a right circular

cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface

area and the volume of the frustum.

ANSWER:

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The height of the frustum cone is h = 12 cm. The radii of the bottom and

top circles are r1 = 12cm and r2 = 3cm respectively.

The slant height of the frustum cone is

𝑙 = √(𝑟1 − 𝑟2)2 + ℎ2

= √(12 − 3)2 + 122

= √225

= 15 cm

The total surface area of the frustum cone is

= 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟22 + 𝜋𝑟2

2

= 𝜋 × (12 + 3) × 15 + 𝜋 × 122 + 𝜋 × 32

= 𝜋 × 225 × 26 + 144𝜋 + 9𝜋

= 378𝜋 cm2

Hence Total surface area = 378𝜋 cm2

The volume of the frustum cone is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(122 + 12 × 3 + 32) × 12

=1

3× 𝜋 × 189 × 12

= 756𝜋 cm³

Hence Volume of frustum = 756𝜋 cm³

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Page No 14.79:

Question 9: A tent consists of a frustum of a cone capped by a cone. If

the radii of the ends of the frustum be 13 m and 7 m, the height of the

frustum be 8 m and the slant height of the conical cap be 12 m, find the

canvas required for the tent. (Take: π = 22/7)

ANSWER:

The height of the frustum cone is h = 8 m. The radii of the end circles of

the frustum are r1 = 13m and r2 =7m.

The slant height of the frustum cone is

𝑙 = √(𝑟1 − 𝑟2)2 + ℎ2

= √(13 − 7)2 + 82

= √100

= 10 meter

The curved surface area of the frustum is

𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙

= 𝜋(13 + 7) × 10

= 𝜋 × 20 × 10

= 200𝜋 m²

The base-radius of the upper cap cone is 7m and the slant height is 12m.

Therefore, the curved surface area of the upper cap cone is

𝑆2 = 𝜋 × 7 × 12

=22

7× 7 × 12

= 22 × 12

= 264 m²

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Hence, the total canvas required for the tent is

𝑆1 + 𝑆2 = 200𝜋 + 264

= 892.57 m²

Hence total canvas is 892.57 m²

Page No 14.79:

Question 10: 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 ₹44 per

litre which the container can hold.

ANSWER:

Radius, r1 = 8 cm and r2 = 20 cm

Height, h = 16 cm

Volume of milk that the container can hold

𝑉 =1

3𝜋ℎ(𝑟1

2 + 𝑟1𝑟2 + 𝑟22)

𝑉 =1

3𝜋 × 16(82 + 8 × 20 + 202)

𝑉 = 10459.42 cm3

= 10.45942 liters

Cost of milk will be 10.45942 × 44 = Rs 460.21

Page No 14.79:

Question 11: 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 surface area of the bucket. Also, find the cost of

milk which can completely fill the container, at the rate of ₹25 per litre.

(Use π = 3.14).

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ANSWER:

Height of the bucket, h = 30 cm

Radii r1 = 10 cm and r2 = 20 cm

Capacity of the bucket,

𝑉 =1

3𝜋ℎ(𝑟1

2 + 𝑟1𝑟2 + 𝑟22)

=1

3𝜋 × 30(102 + 10 × 20 + 202)

= 21980 cm3

= 21.980 liters

𝑙 = √ℎ2 + (𝑟2 − 𝑟1)2

𝑙 = √302 + (20 − 10)2

= 1010

Surface area of the bucket

𝑆 = CSA + area of the base

𝑆 = 𝜋(𝑟1 + 𝑟2)𝑙 + 𝜋𝑟12

𝑆 = 𝜋(10 + 20)10√10 + 𝜋(10)2

𝑆 = 2978.86 + 314 = 3292.86 cm2

Cost of milk which can completely fill the container at Rs 25/litre

= 21.980 × 25

= Rs 549.50

Page No 14.79:

Question 12: A bucket is in the form of a frustum of a cone with a

capacity of 12308.8 cm3 of water. The radii of the top and bottom circular

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ends are 20 cm and 12 cm respectively. Find the height of the bucket and

the area of the metal sheet used in its making. (Use π = 3.14).

ANSWER:

Let the depth of the bucket is h cm. The radii of the top and bottom circles

of the frustum bucket are r1 =20cm and r2 =12cm respectively.

The volume/capacity of the bucket is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(202 + 20 × 12 + 122) × ℎ

=1

3×

22

7× 784 × ℎ

=1

3× 22 × 112 × ℎ cm3

Given that the capacity of the bucket is 12308.8 Cubic cm. Thus, we have

1

3× 22 × 112 × ℎ = 12308.8

⇒ ℎ =12308.8×3

22×12

⇒ ℎ = 15

Hence, the height of the bucket is 15 cm

The slant height of the bucket is

𝑙 = √(𝑟2 − 𝑟1)2 + ℎ2

= √(20 − 12)2 + 152

= √289

= 17 cm

The surface area of the used metal sheet to make the bucket is

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𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟22

= 𝜋 × (20 + 12) × 17 + 𝜋 × 122

= 𝜋 × 32 × 17 + 144𝜋

= 2160.32 cm2

Hence Surface area of the metal = 2160.32 cm²

Page No 14.79:

Question 13: A bucket made of 𝑎 aluminium sheet is of height 20 cm and

its upper and lower ends are of radius 25 cm and 10 cm respectively. Find

the cost of making the bucket if the aluminium sheet costs Rs 70 per 100

cm2. (Use π = 3.14).

ANSWER:

The height of the bucket is 20cm. The radii of the upper and lower circles

of the bucket are r1 =25cm and r2 =10cm respectively.

The slant height of the bucket is

𝑙 = √(𝑟2 − 𝑟1)2 + ℎ2

= √(25 − 10)2 + 202

= √625

= 25 cm

The surface area of the used aluminium sheet to make the bucket is

𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟22

= 𝜋 × (25 + 10) × 25 + 𝜋 × 102

= 𝜋 × 35 × 25 + 100𝜋

= 3061.5 cm2

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Therefore, the total cost of making the bucket is

=3061.5

100× 70

= 2143.05

Hence the total cost is Rs. 2143.05

Page No 14.79:

Question 14: The radii of the circular ends of a solid frustum of a cone

are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface

area.

ANSWER:

The slant height of the frustum of a cone is l=10cm. The radii of the upper

and lower circles of the bucket are r1 =33cm and r2 =27cm respectively.

The total surface area of the frustum of the cone is

𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟12 + 𝜋𝑟2

2

= 𝜋 × (33 + 27) × 10 + 𝜋 × 332 + 𝜋 × 272

= 600𝜋 + 1089𝜋 + 729𝜋

= 7599.42 cm2

Hence total surface area is 7599.42 cm2

Page No 14.79:

Question 15: A bucket made up of a metal sheet is in the form of a

frustum of a cone of height 16 cm with diameters of its lower and upper

ends as 16 cm and 40 cm respectively. Find the volume of the bucket.

Also, find the cost of the bucket if the cost of metal sheet used in Rs 20

per 100 cm2. (Use π = 3.14)

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ANSWER:

The height of the bucket is h=16cm. The radii of the upper and lower

circles of the bucket are r1 =20 cm and r2 = 8 cm respectively.

The slant height of the bucket is

𝑙 = √(𝑟2 − 𝑟1)2 + ℎ2

= √(20 − 8)2 + 162

= √400

= 20 cm

The volume of the bucket is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(202 + 20 × 8 + 82) × 16

=1

3× 3.14 × 624 × 16

= 3.14 × 208 × 16

= 10449.92 cm3

Hence the volume of the bucket is 10449.92 cm3

The surface area of the used metal sheet to make the bucket is

𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟22

= 𝜋 × (20 + 8) × 20 + 𝜋 × 82

= 𝜋 + 28 × 20 + 64𝜋

= 624π cm2

Therefore, the total cost of making the bucket is

=624π

100× 20

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= RS. 391.9

Page No 14.79:

Question 16: A solid is in the shape of a frustum of a cone, the diameters

of the two circular ends are 60 cm and 36 cm and the height is 9 cm. Find

the area of its whole surface and the volume.

ANSWER:

The height of the frustum of a cone is h=9cm. The radii of the upper and

lower circles of the frustum of the cone are r1 =30cm and r2 =18cm

respectively.

The slant height of the frustum of the cone is

𝑙 = √(𝑟2 − 𝑟1)2 + ℎ2

= √(30 − 18)2 + 92

= √225

= 15 cm

The volume of the frustum of the cone is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(302 + 30 × 18 + 182) × 9

=1

3× 1764 × 9 × 𝜋

= 5292π cm3

The total surface area of the frustum of the cone is

𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟12 + 𝜋𝑟2

2

= 𝜋 × (30 + 18) × 15 + 𝜋 × 302 + 𝜋 × 182

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= 𝜋 × 48 × 15 + 900𝜋 + 324𝜋

= 1944π cm2

Page No 14.79:

Question 17: A milk container is made of metal sheet in the shape of

frustum of cone whose volume is 104593737cm3. The radii of its lower

and upper circular ends are 8 cm and 20 cm respectively. Find the cost of

metal sheet used in making the container at the rate of Rs. 1.40 per cm2 .

(Use π = 22/7)

ANSWER:

Let the depth of the container is h cm. The radii of the top and bottom

circles of the container are r1 =20cm and r2 =8cm respectively.

The volume/capacity of the container is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(202 + 20 × 8 + 82) × ℎ

=1

3×

22

7× 624 × ℎ

=22

7× 208 × ℎ cm3

Given that the capacity of the bucket is 104593

7 cm³. Thus, we have

22

7× 208 × ℎ = 10459

3

7

⇒ ℎ =73216

22×208

⇒ ℎ = 16 cm

Hence, the height of the container is 16 cm.

The slant height of the container is

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𝑙 = √(𝑟2 − 𝑟1)2 + ℎ2

= √(20 − 8)2 + 162

= √400

= 20 cm

The surface area of the used metal sheet to make the container is

𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙 + 𝜋𝑟22

= 𝜋 × (20 + 8) × 20 + 𝜋 × 82

= 𝜋 + 28 × 20 + 64𝜋

= 624π cm2

The cost to make the container is = 624π × 1.4 = 624 ×22

7× 1.4 =

Rs. 2745.6.

Page No 14.79:

Question 18: A solid cone of base radius 10 cm is cut into two parts

through the mid-point of its height, by a plane parallel to its base. Find the

ratio in the volumes of two parts of the cone.

ANSWER:

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Let the height of the cone be H.

Now, the cone is divided into two parts by the parallel plane

∴ OC = CA

Now, In ∆OCD and OAB

∠OCD = OAB (Corresponding angles)

∠ODC = OBA (Corresponding angles)

By AA-similarity criterion ∆OCD ∼ ∆OAB

∴𝐶𝐷

𝐴𝐵=

𝑂𝐶

𝑂𝐴

⇒𝐶𝐷

10=

𝐻

2×𝐻

⇒ 𝐶𝐷 = 5 cm

Volume of first part

Volume of second part=

1

3𝜋(𝐶𝐷)2(𝑂𝐶)

1

3𝜋𝐶𝐴[(𝐴𝐵)2+(𝐴𝐵)(𝐶𝐷)+𝐶𝐷2]

=(5)2

[(10)2+(10)(5)+52]

=25

100+50+25

=25

175

=1

7

Page No 14.79:

Question 19: A bucket opens at the top, and made up of a metal sheet is

in the form of a frustum of a cone. The depth of the bucket is 24 cm and

the diameters of its upper and lower circular ends are 30 cm and 10 cm

respectively. Find the cost of metal sheet used in it at the rate of Rs. 10

per 100 cm2. (Use π = 3.14).

ANSWER:

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The slant height of the bucket is given by

𝑙 = √ℎ2 + (𝑅 − 𝑟)2

= √(24)2 + (15 − 5)2

= √576 + 100

= √676

= 26 cm

Surface area of bucket

= Curved surface area of bucket + Area of the smaller circular base

= 𝜋𝑙(𝑅 + 𝑟) + 𝜋𝑟2

= 3.14 × 26 × (15 + 5) + 3.14 × 5 × 5

= 1632.8 + 78.5

= 1711.3 cm2

Cost of metal sheet used = 10

100× 1711.3 = Rs 171.13

Page No 14.79:

Question 20: In the given figure, from the top of a solid cone of height

12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane

parallel to the base. Find the total surface area of the remaining solid.

(Use π =22

7 and √5 = 2.236). [CBSE 2015]

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ANSWER:

Now, In ∆OCD and OAB

∠OCD = OAB (Corresponding angles)

∠ODC = OBA (Corresponding angles)

By AA-similarity criterion ∆OCD ∼ ∆OAB

∴𝐶𝐷

𝐴𝐵=

𝑂𝐶

𝑂𝐴

⇒𝐶𝐷

6=

4

12

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⇒ 𝐶𝐷 = 2 cm

The slant height of the bucket is given by

𝑙 = √ℎ2 + (𝑅 − 𝑟)2

= √(8)2 + (6 − 2)2

= √64 + 16

= √80

= 4√5 cm

Surface area of the remaining solid

= Curved surface area of figure + Area of the smaller circle + Area of

the larger circle

= 𝜋𝑙(𝑅 + 𝑟) + 𝜋𝑅2 + 𝜋𝑟2

=22

7× 4√5 × (6 + 2) +

22

7× 6 × 6 +

22

7× 2 × 2

=22

7× 4√5 × 8 +

22

7× 6 × 6 +

22

7× 2 × 2

= 224.88 + 113.14 + 12.57

= 350.59 cm2

Page No 14.80:

Question 21: The height of a cone is 10 cm. The cone is divided into two

parts using a plane parallel to its base at the middle of its height. Find the

ratio of the volumes of the two parts.

ANSWER:

We have,

Radius of the solid cone, R = CP

Height of the solid cone, AP = H

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Radius of the smaller cone, QD = r

Height of the smaller cone, AQ = h

Also, 𝐴𝑄 =𝐴𝑃

2 i.e. ℎ =

𝐻

2 or 𝐻 = 2ℎ .....(i)

Now, in ΔAQD and ΔAPC,

∠𝑄𝐴𝐷 = ∠𝑃𝐴𝐶 (Common angle)

∠𝐴𝑄𝐷 = ∠𝐴𝑃𝐶 = 90°

So, by AA criteria

𝛥𝐴𝑄𝐷~𝐴𝑃𝐶

⇒𝐴𝑄

𝐴𝑃=

𝑄𝐷

𝑃𝐶

⇒ℎ

𝐻=

𝑟

𝑅

⇒ℎ

2ℎ=

𝑟

𝑅 [Using (i)]

⇒1

2=

𝑟

𝑅

⇒ 𝑅 = 2𝑟 .... (ii)

As, Volume of smaller cone =1

3𝜋𝑟2ℎ

And, Volume of solid cone =1

3𝜋𝑅2𝐻

=1

3𝜋(2𝑟)2(2ℎ) [Using (i) and (ii)

=8

3𝜋𝑟2ℎ

So,

Volume of frustum = Volume of solid cone − Volume of smaller cone

=8

3𝜋𝑟2ℎ −

1

3𝜋𝑟2ℎ

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=7

3𝜋𝑟2ℎ

Now, the ratio of the volumes of the two parts=Volume of the smaller cone

Volume of the frustum

= (1

3𝜋𝑟2ℎ) (

7

3𝜋𝑟2ℎ)

=1

7

= 1: 7

So, the ratio of the volume of the two parts of the cone is 1: 7.

Page No 14.80:

Question 22: A bucket, made of metal sheet, is in the form of a cone

whose height is 35 cm and radii of circular ends are 30 cm and 12 cm.

How many litres of milk it contains if it is full to the brim? If the milk is

sold at Rs 40 per litre, find the amount received by the person.

ANSWER:

The given bucket is in the form of the frustum of a cone.

Height, h = 35 cm

r1 = 30 cm

r2 = 12 cm

Volume =𝜋

3ℎ(𝑟1

2 + 𝑟22 + 𝑟1𝑟2)

=𝜋

3× 35(302 + 122 + 30 × 12)

=𝜋

3× 35(900 + 144 + 360)

=𝜋

3× 35 × (1404)

= 51433.2 cm3

= 51.4 litres

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Selling price of the milk = Rs 40/litre

So, selling price of 51.4 litres of milk will be 51.4 × 40 = Rs 20565

Page No 14.80:

Question 23: The diameters of the lower and upper ends of a bucket in

the form of a frustum of a cone are 10 cm and 30 cm respectively. If its

height is 24 cm,

(i) Find the area of the metal sheet used to make the bucket.

(ii) Why we should avoid the bucket made by ordinary plastic? (use π =

3.14)

ANSWER:

We have:

Radius of upper end of frustum, R = 15 cm; Radius of lower end of

frustum, r = 5 cm; Height of frustum, h = 24 cm

We know,

Slant height, l2 = h2 + (R – r)2

⇒ l2 = ((24)2 + (15 – 5)2} = (576 + 100) = 676

⇒ l = 26 cm

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(i)

Required area of the metal sheet = 𝜋[𝑟2 + 𝑙(𝑅 + 𝑟)] sq. cm

= 3.14[52 + 26(15 + 5)] cm2

= 3.14 × (25 + 520) cm2

= 3.14 × 545cm2

= 1711.3 cm2

(ii)Plastic is harmful to the environment and to protect the environment

its use should be avoided.

Page No 14.80:

Question 24: A reservoir in the form of the frustum of a right circular

cone contains 44 × 107 litres of water which fills it completely. The radii

of the bottom and top of the reservoir are 50 metres and 100 metres

respectively. Find the depth of water and the lateral surface area of the

reservoir. (Take: π = 22/7)

ANSWER:

Let the depth of the frustum cone like reservoir is h m. The radii of the

top and bottom circles of the frustum cone like reservoir are r1 =100m

and r2 =50m respectively.

The volume of the reservoir is

𝑉 =1

3𝜋(𝑟1

2 + 𝑟1𝑟2 + 𝑟22) × ℎ

=1

3𝜋(1002 + 100 × 50 + 502) × ℎ

=1

3×

22

7× 17500 × ℎ

=1

3× 22 × 2500 × ℎ cm3

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=1

3× 22 × 2500 × ℎ × 106 m3

=1

3× 22 × 2500 × ℎ × 103 litres

Given that the volume of the reservoir is 44 × 107 litres. Thus, we have

1

3× 22 × 2500 × ℎ × 103 = 44 × 107

⇒ ℎ =3×44×107

22×2500×103

⇒ ℎ = 24

Hence, the depth of water in the reservoir is 24 m

The slant height of the reservoir is

𝑙 = √(𝑟1 − 𝑟2)2 + ℎ2

= √(100 − 50)2 + 242

= √3076

= 55.46169 meter

The lateral surface area of the reservoir is

𝑆1 = 𝜋(𝑟1 + 𝑟2) × 𝑙

= 𝜋 × (100 + 50) × 55.46169

= 𝜋 × 150 × 55.46169

= 26145.225 m2

Hence, the lateral surface area is 26145.225 m2

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VERY SHORT ANSWER TYPE QUESTIONS (VSAQs)

Page No 14.86:

Question 1: The radii of the base of a cylinder and a cone are in the ratio

3: 4 and their heights are in the ratio 2: 3. What is the ratio of their

volumes?

ANSWER:

Let r1 and r2 be the radii of the base of a cylinder and a cone.

The volume of cylinder 𝑉1 = 𝜋𝑟12ℎ1 …… (i)

The volume of cone 𝑉2 =1

3𝜋𝑟2

2ℎ2 …… (ii)

Dividing (i) by (ii), the, we get

𝑉1

𝑉2=

𝜋𝑟12ℎ1

1

3𝜋𝑟2

2ℎ2

= 3 × (𝑟1

𝑟2)

2× (

ℎ1

ℎ2)

(𝑟1

𝑟2=

3

4,

ℎ1

ℎ2=

2

3, given)

𝑉1

𝑉2= 3 × (

3

4)

2×

2

3

𝑉1

𝑉2=

9

8

𝑉1: 𝑉2 = 9: 8

Page No 14.86:

Question 2: If the heights of two right circular cones are in the ratio 1: 2

and the perimeters of their bases are in the ratio 3: 4, what is the ratio of

their volumes?

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ANSWER:

Given that,

ℎ1: ℎ2 = 1: 2 and 2𝜋𝑟1: 2𝜋𝑟2 = 3: 4

i.e. 𝑟1: 𝑟2 = 3: 4

Therefore,

The ratios of volume of their cones will be

𝑉1: 𝑉2 =1

3𝜋𝑟1

2ℎ1:1

3𝜋𝑟2

2ℎ2

𝑉1

𝑉2=

1

3𝜋𝑟1

2ℎ1

1

3𝜋𝑟2

2ℎ2

= (𝑟1

𝑟2)

2× (

ℎ1

ℎ2)

𝑉1

𝑉2= (

3

4)

2× (

1

2)

𝑉1

𝑉2=

9

32

𝑉1: 𝑉2 = 9: 32

Page No 14.86:

Question 3: If a cone and a sphere have equal radii and equal volumes.

What is the ratio of the diameter of the sphere to the height of the cone?

ANSWER:

Given that,

A cone and a sphere have equal radii and equal volume

i.e., volume of cone = volume of sphere

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1

3𝜋𝑟1

2ℎ =4

3𝜋𝑟3

𝑟2ℎ = 4𝑟3

ℎ = 4𝑟

ℎ = (2𝑟) × 2

ℎ

2𝑟=

2

1

ℎ

2=

2

1 (diameter (d) = 2𝑟)

ℎ: 𝑑 = 2: 1 or 𝑑: ℎ = 1: 2

Page No 14.86:

Question 4: A cone, a hemisphere and a cylinder stand on equal bases

and have the same height. What is the ratio of their volumes?

ANSWER:

Let r be the radius of the base.

And h is the height.

Here, h = r.

Now,

The ratio of their volumes will be

Volume of cone: volume of hemisphere: volume of a cylinder

1

3𝜋𝑟2ℎ:

2

3𝜋𝑟3: 𝜋𝑟2ℎ

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𝑉1: 𝑉2: 𝑉3 = 1: 2: 3

Hence, 𝑉1: 𝑉2: 𝑉3 = 1: 2: 3

Page No 14.86:

Question 5: The radii of two cylinders are in the ratio 3: 5 and their

heights are in the ratio 2: 3. What is the ratio of their curved surface areas?

ANSWER:

Given that, 𝑟1: 𝑟2 = 3: 5 and ℎ1: ℎ2 = 2: 3

Now, the ratio of their curved surface area 𝑠1: 𝑠2 = 2𝜋𝑟1ℎ1: 2𝜋𝑟2ℎ2

𝑠1: 𝑠2 = 2𝜋𝑟1ℎ1: 2𝜋𝑟2ℎ2

𝑠1

𝑠2=

2𝜋𝑟1ℎ1

2𝜋𝑟2ℎ2

= (𝑟1

𝑟2) (

ℎ1

ℎ2)

𝑠1

𝑠2=

3

5×

2

3=

2

5

Hence 𝑠1: 𝑠2 = 2: 5

Page No 14.86:

Question 6: Two cubes have their volumes in the ratio 1: 27. What is the

ratio of their surface areas?

ANSWER:

The rate of the value of cubes = 1:27

𝑎13

𝑎23 =

1

27

𝑎1

𝑎2=

1

3 …… (i)

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Now,

The ratio of their surface area

𝑠1: 𝑠2 = 6𝑎12: 6𝑎2

2

𝑠1

𝑠2=

6𝑎12

6𝑎22

= (𝑎1

𝑎2)

2

𝑠1

𝑠2=

1

9

Hence, 𝑠1: 𝑠2 = 1: 9

Page No 14.86:

Question 7: Two right circular cylinders of equal volumes have their

heights in the ratio 1: 2. What is the ratio of their radii?

ANSWER:

Let r1 and r2 be the radii of two right circular cylinders and h1 and h2 be

the heights.

Since,

Both the cylinder has the same volume.

Therefore,

𝜋𝑟12ℎ1 = 𝜋𝑟2

2ℎ2

(𝑟1

𝑟2)

2=

ℎ2

ℎ1

(ℎ1: ℎ2 = 1: 2, given)

(𝑟1

𝑟2)

2= (

2

1)

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𝑟1: 𝑟2 = √2: 1

Page No 14.87:

Question 8: If the volumes of two cones are in the ratio 1: 4 and their

diameters are in the ratio 4: 5, then write the ratio of their weights.

ANSWER:

The ratio of the volume of cones

𝑣1: 𝑣2 = 1: 4

And 2𝑟1: 2𝑟2 = 4: 5,

i.e., 𝑟1

𝑟2=

4

5

Now, 𝑣1

𝑣2=

4

5

𝑉1

𝑉2=

1

3𝜋𝑟1

2ℎ1

1

3𝜋𝑟2

2ℎ2

1

4= (

𝑟1

𝑟2)

2× (

ℎ1

ℎ2)

ℎ1

ℎ2=

1

4×

25

16

=25

64

Hence, ℎ1: ℎ2 = 25: 64

Page No 14.87:

Question 9: A sphere and a cube have equal surface areas. What is the

ratio of the volume of the sphere to that of the cube?

ANSWER:

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Surface area of sphere = sphere area of cube

i.e., 4πr2 = 6a2

𝑟2

𝑎2 =6

4𝜋

𝑟

𝑎= (

6

4𝜋)

1

2 …… (i)

Now, volume of sphere

volume of cube=

4

3𝜋𝑟3

𝑎3

𝑉1

𝑉2=

4𝜋𝑟3

3𝑎3

=4

3𝜋 (

𝑟

𝑎) (

𝑟

𝑎)

3

=4

3𝜋√

6

𝜋

1

2

6

4𝜋

𝑉1: 𝑉2 = √6

𝜋

Page No 14.87:

Question 10: What is the ratio of the volume of a cube to that of a sphere

which will fit inside it?

ANSWER:

Ratio of sphere

=1

2× side of cube

𝑟 =𝑎

2

Now,

Volume of cube 𝑣1 = 𝑎3

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Volume of sphere

𝑣2 =4

3𝜋𝑟3

=4

3𝜋 (

𝑎

2)

3

=4

3𝜋

𝑎3

8

𝑣2 =1

6𝜋𝑎3

The ratio of their volumes

𝑉1: 𝑉2 = 𝑎3:1

6𝜋𝑎3

𝑉1

𝑉2=

𝑎3

1

6𝜋𝑎3

=6

𝜋

Hence, 𝑉1: 𝑉2 = 6: 𝜋

Page No 14.87:

Question 11: What is the ratio of the volumes of a cylinder, a cone and a

sphere, if each has the same diameter and same height?

ANSWER:

Given that the diameter and the height of the cylinder, cone and sphere

are the same.

The volume of cylinder, 𝑣1 = 𝜋𝑟12ℎ1 = 𝜋 (

𝑑

2)

2𝑑

The volume of cone, 𝑣2 =1

3𝜋𝑟2

2ℎ2 =1

3𝜋 (

𝑑

2)

2𝑑

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And the volume of sphere, 𝑣3 =4

3𝜋𝑟3

3ℎ3 =4

3𝜋 (

𝑑

2)

3

Therefore,

The ratio of their volumes,

𝑣1 = 𝑣2 = 𝑣3

⇒ 𝜋 (𝑑

2)

2𝑑 =

1

3𝜋 (

𝑑

2)

2𝑑 =

4

3𝜋 (

𝑑

2)

3

⇒ 3: 1: 2

Hence, the ratio is 3: 1: 2

Page No 14.87:

Question 12: A sphere of maximum volume is cut-out from a solid

hemisphere of radius r, what is the ratio of the volume of the hemisphere

to that of the cut-out sphere?

ANSWER:

Since, a sphere of maximum volume is cut out from a solid hemisphere of

radius.

i.e., radius of sphere

Therefore,

The volume of sphere

=4

3𝜋 (

𝑟

2)

3

𝑣1 =1

6𝜋𝑟3 …… (i)

The volume of hemisphere 𝑣2 =2

3𝜋𝑟3 …… (ii)

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Divide (i) by (ii).

𝑣1

𝑣2=

1

6𝜋𝑟3

2

3𝜋𝑟3

=1

6×

3

2

𝑣1

𝑣2=

1

4

Hence, 𝑣2: 𝑣1 = 4: 1

Page No 14.87:

Question 13: A metallic hemisphere is melted and recast in the shape of

a cone with the same base radius R as that of the hemisphere. If H is the

height of the cone, then write the values of 𝐻

𝑅.

ANSWER:

Given,

Radius of the hemisphere = Radius of the cone.

Now,

Volume of the hemisphere =2

3𝜋𝑅3

and

Volume of the cone =1

3𝜋𝑅2𝐻

Volume of the hemisphere = volume of the cone

2

3𝜋𝑅3 =

1

3𝜋𝑅2𝐻

2𝑅 = 𝐻

or 𝐻

𝑅= 2

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Page No 14.87:

Question 14: A right circular cone and a right circular cylinder have equal

base and equal height. If the radius of the base and height are in the ratio

5: 12, write the ratio of the total surface area of the cylinder to that of the

cone.

ANSWER:

Given that

𝑟: ℎ = 5: 12

i.e. 𝑟 = 5𝑥, ℎ = 12

Since,

Right, circular cone and right circular cylinder have equal base and equal

right.

Therefore,

The total surface area of cylinder 𝑆1 = 2𝜋𝑟(ℎ + 𝑟)

The total surface area of cone 𝑆2 = 𝜋𝑟(𝑙 + 𝑟)

𝑙 = √𝑟2 + ℎ2

= √25𝑥2 + 144𝑥2

= √169𝑥2

𝑙 = 13𝑥

Now,

𝑆1

𝑆2=

2𝜋𝑟(ℎ+𝑟)

𝜋𝑟(𝑙+𝑟)

=2(ℎ+𝑟)

(𝑙+𝑟)

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𝑆1

𝑆2=

2(12𝑥+5𝑥)

13𝑥+5𝑥

=2×17𝑥

18𝑥

𝑆1

𝑆2=

17

9

Hence, 𝑆1: 𝑆2 = 17: 9

Page No 14.87:

Question 15: A cylinder, a cone and a hemisphere are of equal base and

have the same height. What is the ratio of their volumes?

ANSWER:

Let the diameter of the base for all three be x cm and height be y cm.

For hemisphere radius =𝑥

2 cm

Height = 𝑦 =𝑥

2 cm

(As height of the hemisphere is equal to the radius of hemisphere)

For cone

Radius =𝑥

2 cm

Height =𝑥

2 cm

(As height is same for all)

For cylinder

Radius =𝑥

2 cm

Height =𝑥

2 cm

The ratio of their volume is

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= cylinder volume: conic volume: hemispherical volume

= 𝜋 (𝑥

2)

2 𝑥

2:

1

3𝜋 (

𝑥

2)

2(

𝑥

2) :

2

3𝜋 (

𝑥

2)

3

= 1:1

3:

2

3

= 3: 1: 2

Page No 14.87:

Question 16: The radii of two cones are in the ratio 2: 1 and their volumes

are equal. What is the ratio of their heights?

ANSWER:

Let the radius of the first cone = 2x

And height of the first cone = h1

Then,

Volume of the first cone = 1

3𝜋𝑟2ℎ

= 1

3𝜋(2𝑥)2ℎ1 ….. (i)

The radius of the second cone = x

Height of the second cone = h2

Then,

Volume of the second cone = 1

3𝜋𝑟2ℎ

= 1

3𝜋(𝑥)2ℎ2

Since,

The volumes of the two cones are equal.

Or ℎ1: ℎ2 = 1: 4

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Page No 14.87:

Question 17: Two cones have their heights in the ratio 1: 3 and radii 3: 1.

What is the ratio of their volumes?

ANSWER:

Let the radius of the cone is 3x and x,

And the height of the cone is y and 3y.

Then,

Volume of the first cone

𝑣1 =1

3𝜋𝑟2ℎ

=1

3𝜋(3𝑥)2𝑦

=1

3𝜋9𝑥2𝑦

= 3𝜋𝑥2𝑦 ……. (i)

Volume of the second cone

𝑣2 =1

3𝜋(𝑥)2 × 3𝑦

= 𝜋𝑥2𝑦 ……. (ii)

Then the radius of their volume

𝑣1

𝑣2=

3𝜋𝑥2𝑦

𝜋𝑥2𝑦

Or

𝑣1

𝑣2=

3

1

𝑣2: 𝑣1 = 3: 1

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Page No 14.87:

Question 18: A hemisphere and a cone have equal bases. If their heights

are also equal, then what is the ratio of their curved surfaces?

ANSWER:

The base of the cone and hemisphere are equal. So radius of the two is

also equal.

and

Height of the hemisphere = height of the cone

Then the slant height of the cone

𝑙 = √𝑟2 + ℎ2

= √𝑟2 + 𝑟2

= √2𝑟2

= 𝑟√2 ……. (i)

Now, the curved surface area of

Hemisphere = 2𝜋𝑟2

and

The curved surface area of cone = 𝜋𝑟𝑙

Putting the value of l from eq. (i)

We get

= 𝜋𝑟√2𝑟

= 𝜋𝑟2√2

Now,

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C.S.A.of hemisphare

C.S.A.of cone=

2𝜋𝑟2

𝜋𝑟2√2

=2

√2×

√2

√2

=2√2

2

= √2: 1

Page No 14.87:

Question 19: If r1 and r2 denote the radii of the circular bases of the

frustum of a cone such that r1 > r2, then write the ratio of the height of the

cone of which the frustum is a part to the height of the frustum.

ANSWER:

Since, ∆𝑉𝑂′𝐵~∆𝑉𝑂𝐴

Therefore,

In ∆𝑉𝑂′𝐵~∆𝑉𝑂𝐴

𝑂′𝐵

𝑂𝐴=

𝑂′𝑉

𝑂𝑉

𝑟2

𝑟1=

ℎ−ℎ1

ℎ

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𝑟2

𝑟1= 1 −

ℎ1

ℎ

ℎ1

ℎ= 1 −

𝑟2

𝑟1

= 𝑟1−𝑟2

𝑟1

Hence,

The ratio of the height of cone of which the frustum is a part to the height

of the frustum.

𝑂𝑉

𝑂𝑂′ =ℎ

ℎ1=

𝑟1

𝑟1−𝑟2

Hence, ℎ: ℎ1 = 𝑟1: (𝑟1 − 𝑟2)

Page No 14.87:

Question 20: If the slant height of the frustum of a cone is 6 cm and the

perimeters of its circular bases are 24 cm and 12 cm respectively. What is

the curved surface area of the frustum?

ANSWER:

The parameter of upper base

= 2𝜋𝑟1

2𝜋𝑟1 = 12

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𝑟1 =6

𝜋 cm

The parameter of lower base

= 2𝜋𝑟2

2𝜋𝑟2 = 24

𝑟2 =12

𝜋 cm

The surface area of frustum

= 𝜋 (6

𝜋+

12

𝜋) × 6

= 𝜋 ×18

𝜋× 6

= 108 cm²

Page No 14.87:

Question 21: If the areas of circular bases of a frustum of a cone are 4

cm2 and 9 cm2 respectively and the height of the frustum is 12 cm. What

is the volume of the frustum?

ANSWER:

Area of circular bases of frustum is

𝐴1 = 4 cm²

𝐴2 = 9 cm²

The height of frustum h = 12 cm

Now, the volume of frustum

𝑉 =ℎ

3{𝐴1 + 𝐴2 + √𝐴1𝐴2}

=12

3{4 + 9 + √4 × 9}

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= 4{13 + 6}

𝑉 = 76 cm³

Page No 14.87:

Question 22: The surface area of a sphere is 616 cm2. Find its radius.

ANSWER:

The surface area of sphere = 616k cm2

We know that

4𝜋𝑟2 = 616

𝑟2 =616

4𝜋

Taking squire root both the side

√𝑟2 = √616

4𝜋

𝑟 = 7 cm

Page No 14.87:

Question 23: A cylinder and a cone are of the same base radius and of

same height. Find the ratio of the value of the cylinder to that of the cone

ANSWER:

Since, cylinder and a cone both are have same radius and height.

Therefore,

𝑉1

𝑉2=

𝜋𝑟2ℎ1

3𝜋𝑟2ℎ

© PRAADIS

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𝑉1

𝑉2=

11

3

𝑉1

𝑉2=

3

1

𝑉1: 𝑉2 = 3: 1

Page No 14.87:

Question 24: The slant height of the frustum 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.

ANSWER:

Slant height of the Frustum = 5 cm

i.e. l = 5 cm.

𝑟1 − 𝑟2 = 4 cm.

𝑙 = √ℎ2 + (𝑟1 − 𝑟2)2

5 = √ℎ2 + (4)2

Squaring both sides, we get

25 = ℎ2 + 42

25 = ℎ2 + 16

25 − 16 = 42

Or ℎ2 = 9 cm

4 = 3 cm

Height of the Frustum = 3 cm

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Page No 14.87:

Question 25: Volume and surface area of a solid hemisphere are

numerically equal. What is the diameter of hemisphere?

ANSWER:

Let the radius of the hemisphere be r units.

Volume of a hemisphere = Surface area of the hemisphere

⇒2

3𝜋𝑟3 = 3𝜋𝑟2

⇒2

3𝑟 = 3

⇒ 𝑟 =9

2

⇒ 𝑑 = 9 units

Hence, diameter of the hemisphere is equal to 9 units.

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MULTIPLE CHOICE QUESTION

Page No 14.88:

Question 1: The diameter of a sphere is 6 cm. It is melted and drawn in

to a wire of diameter 2 mm. The length of the wire is

(a) 12 m

(b) 18 m

(c) 36 m

(d) 66 m

ANSWER:

The diameter of a sphere = 6 cm

Then radius of a sphere =6

2 cm = 3 cm

The diameter of a wire = 2 mm

Then radius of wire = 1 mm = 0.1 cm

Now,

Volume of sphere = volume of wire

4

3𝜋𝑟3 = 𝜋𝑟2ℎ

Here,

r = radius

l = length of wire

4

3× 3 × 3 × 2 = 0.1 × 0.1 × 𝑙

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36 = 0.01𝑙

𝑙 =36

0.01

To remove the decimal from base we should multiply both numerator and

denominator by 100,

We get,

𝑙 = 3600 = 36 m

Hence, the correct answer is choice (c).

Page No 14.88:

Question 2: A metallic sphere of radius 10.5 cm is melted and then recast

into small cones, each of radius 3.5 cm and height 3 cm. The number of

such cones is

(a) 63

(b) 126

(c) 21

(d) 130

ANSWER:

Radius of metallic sphere = 10.5 cm

Therefore,

Volume of the sphere

=4

3𝜋𝑟3

=4

3× 𝜋 × 10.5 × 10.5 × 10.5

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=4630.5𝜋

3 …… (i)

Now,

Radius of the cone = 3.5 cm

and Height of the cone = 3 cm

Therefore,

Volume of the cone

=1

3𝜋𝑟2ℎ

=1

3× 𝜋 × 3.5 × 3.5 × 3

=36.75𝜋

3 …… (i)

Number of cone =volume of sphere

volume of cone

Dividing eq. (i) and (ii) we get

Number of cone =4630.5𝜋

336.75𝜋

3

= 126

Number of cone = 126

Hence, the correct answer is choice (b).

Page No 14.88:

Question 3: A solid is hemispherical at the bottom and conical above. If

the surface areas of the two parts are equal, then the ratio of its radius and

the height of its conical part is

(a) 1 : 3

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(b) 1 : √3

(c) 1 : 1

(d) √3 : 1

ANSWER:

Let r be the radius of the base and h be the height of conical part.

Since,

Surface area of both part of solid is equal.

i.e.,

𝜋𝑟𝑙 = 2𝜋𝑟2

𝑟𝑙 = 2𝑟𝑙

𝑙 = 2𝑟 ….. (i)

But,

𝑙 = √ℎ2 + 𝑟2

Squaring on both side,

Then we get,

𝑙2 = ℎ2 + 𝑟2

From equation (i) putting the value of l in above equation

4𝑟2 = ℎ2 + 𝑟2

3𝑟2 = ℎ2

𝑟

ℎ=

1

√3

𝑟: ℎ = 1: √3

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Hence, the correct answer is choice (b).

Page No 14.88:

Question 4: A solid sphere of radius r is melted and cast into the shape

of a solid cone of height r, the radius of the base of the cone is

(a) 2r

(b) 3r

(c) r

(d) 4r

ANSWER:

Volume of sphere = volume of the cone

4

3𝜋𝑟32

=1

3𝜋𝑅2 × 𝑟

𝑅2 = 4𝑟2

𝑅 = 2𝑟

Hence, the correct answer is choice (a).

Page No 14.88:

Question 5: The material of a cone is converted into the shape of a

cylinder of equal radius. If height of the cylinder is 5 cm, then height of

the cone is

(a) 10 cm

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(b) 15 cm

(c) 18 cm

(d) 24 cm

ANSWER:

A cone is converted into a cone.

So,

Volume of cone = Volume of cylinder

1

3𝜋𝑅2ℎ = 𝜋𝑅2 × 5

ℎ = 15 cm

Hence, the correct answer is choice (b).

Page No 14.88:

Question 6: A circus tent is cylindrical to a height of 4 m and conical

above it. If its diameter is 105 m and its slant height is 40 m, the total area

of the canvas required in m2 is

(a) 1760

(b) 2640

(c) 3960

(d) 7920

ANSWER:

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For conical portion

𝑟 = 52.5 m and 𝑙 = 40 m

Curved surface area of the conical portion

= 𝜋𝑟𝑙

= 𝜋 × 52.5 × 40

= 2100𝜋 m²

For cylindrical portion we have

𝑟 = 52.5 m and ℎ = 4 m

Then,

Curved surface area of cylindrical portion

= 2𝜋𝑟𝑙

= 2 × 𝜋 × 52.5 × 4

= 420𝜋 m²

Area of canvas used for making the tent

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= (2100 m + 420)𝜋

= 2520 ×22

7

= 7920 m²

Hence, the correct answer is choice (d).

Page No 14.88:

Question 7: The number of solid spheres, each of diameter 6 cm that

could be moulded to form a solid metal cylinder of height 45 cm and

diameter 4 cm, is

(a) 3

(b) 4

(c) 5

(d) 6

ANSWER:

Here,

Diameter of sphere = 6 cm

Radius of sphere

=6

2 cm

= 3 cm

Volume of the sphere

=4

3𝜋𝑟3

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=4

3× 𝜋 × 3 × 3 × 3

= 36𝜋 ….. (i)

Now,

Diameter of cylinder = 4 cm

=4

2 cm

Radius of cylinder = 2 cm

Height of the cylinder = 45 cm

Then,

Volume of the cylinder

= 𝜋𝑟2ℎ

= 𝜋 × 2 × 2 × 45

= 180𝜋

The number of solid sphere

= Volume of cylinder

Volume of sphere

= 5

The number of solid sphere is 5.

Hence, the correct answer is choice (c).

Page No 14.88:

Question 8: A sphere of radius 6 cm is dropped into a cylindrical vessel

partly filled with water. The radius of the vessel is 8 cm. If the sphere is

submerged completely, then the surface of the water rises by

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(a) 4.5 cm

(b) 3

(c) 4 cm

(d) 2 cm

ANSWER:

Radius of the sphere = 6 cm.

Volume of the sphere

=4

3𝜋𝑟3

=4

3𝜋 × 6 × 6 × 6

and

Radius of the cylinder = 8 cm

Volume of the cylinder

= 𝜋𝑟2ℎ

= 𝜋 × 8 × 8 × ℎ

Therefore,

Volume of the sphere = volume of the cylinder

4

3𝜋(6)3 = 𝜋(8)2ℎ

Or

ℎ =4×72

64= 4.5 cm

Hence, the correct answer is choice (a).

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Page No 14.88:

Question 9: If the radii of the circular ends of a bucket of height 40 cm

are of lengths 35 cm and 14 cm, then the volume of the bucket in cubic

centimeters, is

(a) 60060

(b) 80080

(c) 70040

(d) 80160

ANSWER:

Height of the bucket = 40 cm

Radius of the upper part of bucket = 35 cm

R1 = 35 cm and

R2 = 14 cm

The volume of the bucket

=1

3𝜋ℎ(𝑟1

2 + 𝑟22 + 𝑟1𝑟2)

=1

3×

22

7× 40((35)2 + (14)2 + (35 × 14))

=1

3×

22

7× 40[1225 + 196 + 490]

=1

3×

22

7× 40 × 1911

=1681680

21

= 80080 cm³

Hence, the correct answer is choice (b).

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Page No 14.88:

Question 10: If a cone is cut into two parts by a horizontal plane passing

through the mid-point of its axis, the ratio of the volumes of the upper part

and the cone is

(a) 1 : 2

(b) 1: 4

(c) 1 : 6

(d) 1 : 8

ANSWER:

Since,

∆𝑉𝑂𝐴~∆𝑉𝑂′𝐶

Therefore,

In ∆𝑉𝑂𝐴 and ∆𝑉𝑂′𝐶

𝑂′𝑉

𝑂𝑉=

𝑂′𝐶

𝑂𝐴

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ℎ

2

ℎ=

𝑂′𝐶

𝑂𝐴

1

2=

𝑂′𝐶

𝑂𝐴

𝑂′𝐶

𝑂𝐴=

1

2

The ratio of the volume of upper part and the cone,

𝑉1

𝑉2=

1

3𝜋(𝑂′𝐶)×

ℎ

21

3𝜋(𝑂𝐴)2×ℎ

𝑉1

𝑉2= (

𝑂′𝐶

𝑂𝐴)

2

×1

2 ….. (ii)

From eq. (i) and (ii),

We get,

𝑉1

𝑉2= (

1

2)

2×

1

2

𝑉1

𝑉2=

1

4×

1

2

𝑉1: 𝑉2 = 1: 8

Hence, the correct answer is choice (d).

Page No 14.88:

Question 11: The height of a cone is 30 cm. A small cone is cut off at the

top by a plane parallel to the base. If its volume be 127127 of the volume

of the given cone, then the height above the base at which the section has

been made, is

(a) 10 cm

(b) 15 cm

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(c) 20 cm

(d) 25 cm

ANSWER:

Let VAB be cone of height 30 cm and base radius r1 cm.

Suppose it is cut off by a plane parallel to the base at a height h2 from the

base of the cone.

Clearly 𝛥𝑉𝑂𝐷~𝛥𝑉𝑂′𝐵

Therefore,

𝑂𝑉

𝑂′𝑉=

𝑂𝐷

𝑂′𝐵

⇒ℎ1

30=

𝑟2

𝑟1

But,

Volume of cone 𝑉𝐶𝐷 =1

27Volume of cone 𝑉𝐴𝐵

⇒1

3𝜋(𝑟2)2ℎ1 =

1

27(

1

3𝜋(𝑟1)230)

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⇒ (𝑟2

𝑟1) 2ℎ1 =

10

9

⇒ (ℎ1

30)

2ℎ1 =

10

9

⇒ ℎ1 = 10

Hence,

Required height

= 30 – 10

= 20 cm

Hence, the correct answer is choice (c).

Page No 14.88:

Question 12: A solid consists of a circular cylinder with an exact fitting

right circular cone placed at the top. The height of the cone is h. If the

total volume of the solid is 3 times the volume of the cone, then the height

of the circular is

(a) 2h

(b) 2ℎ

3

(c) 3ℎ

2

(d) 4h

ANSWER:

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Let r be the radius of the base of solid.

Clearly,

The volume of solid = 3 × volume of cone

Vol. of cone + Vol. of cylinder = 3 Volume of cone

Vol. of cylinder = 2 Vol. of cone

𝜋𝑟2𝑥 = 2 ×1

3𝜋𝑟2ℎ

𝑥 =2

3ℎ

Thus,

The height of cylinder =2ℎ

3

Hence, the correct answer is choice (b).

Page No 14.89:

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Question 13: A reservoir is in the shape of a frustum of a right circular

cone. It is 8 m across at the top and 4 m across at the bottom. If it is 6 m

deep, then its capacity is

(a) 176 m3

(b) 196 m3

(c) 200 m3

(d) 110 m3

ANSWER:

𝑟1 =8

2

= 4 m

𝑟2 =4 m

𝛼

= 2 m and

ℎ = 6 m

The volume of reservoir

=ℎ

3𝜋{𝑟1

2 + 𝑟22 + 𝑟1𝑟2}

=62

3𝜋(16 + 4 + 8)

= 56𝜋

= 568 ×22

7

= 176 m²

The volume of reservoir = 176 m2

Hence, the correct answer is choice (a).

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Page No 14.89:

Question 14: Water flows at the rate of 10 metre per minute from a

cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical

vessel whose diameter at the base is 40 cm and depth 24 cm?

(a) 48 minutes 15 sec

(b) 51 minutes 12 sec

(c) 52 minutes 1 sec

(d) 55 minutes

ANSWER:

The radius of cylindrical pipe

𝑟 =5

2 mm = 0.25 cm

The volume per minute of water flow from the pipe

= 𝜋 × (0.25)2 × 1000

= 62.5𝜋 cm³/minute

The radius of cone

=40

2

= 20 cm

Depth of cone = 24 cm

The volume of cone

=1

3𝜋(20)2 × 248

= 3200𝜋 cm³

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The time it will take to fill up a conical vessel

=3200𝜋

62.5𝜋

= 51125

625 min

= 51 min + 125

625× 60 sec

= 51 min + 12 sec

Hence, the correct answer is choice (b).

Page No 14.89:

Question 15: A cylindrical vessel 32 cm high and 18 cm as the radius of

the base, 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,

the radius of its base is

(a) 12 cm

(b) 24 cm

(c) 36 cm

(c) 48 cm

ANSWER:

Volume of sand filled in cylindrical vessel

= 𝜋(18)2 × 32

= 32 × 324𝜋

Clearly,

The volume of conical heap = volume of sand

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1

3𝜋𝑟2 × 248 = 32 × 324𝜋

𝑟2 =324×324

8

𝑟 = √4 × 324

𝑟 = 2 × 18

= 36

𝑟 = 36 cm

Hence, the correct answer is choice (c).

Page No 14.89:

Question 16: The curved surface area of a right circular cone of height 15

cm and base diameter 16 cm is

(a) 60π cm2

(b) 68π cm2

(c) 120π cm2

(d) 136π cm2

ANSWER:

Height,

ℎ = 15 cm,

𝑟 =16

2

= 8 cm

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𝑙 = √𝑟2 + ℎ2

= √82 + 152

= √64 + 225

= √289

𝑙 = 17 cm

The C.S.A. of cone

= 𝜋𝑟𝑙

= 𝜋 × 8 × 17

= 136𝜋 cm³

Hence, the correct answer is choice (d).

Page No 14.89:

Question 17: A right triangle with sides 3 cm, 4 cm and 5 cm is rotated

about the side of 3 cm to form a cone. The volume of the cone so formed

is

(a) 12π cm3

(b) 15π cm3

(c) 16π cm3

(d) 20π cm3

ANSWER:

Radius of cone VAOB

r = 4 cm

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Height of cone VAOB

h = 3 cm

The volume of cone VAOB

=1

3𝜋𝑟2ℎ

=1

3𝜋(3)2 × 4

= 12𝜋 cm³

Hence, the correct answer is choice (a).

Page No 14.89:

Question 18: The curved surface area of a cylinder is 264 m2 and its

volume is 924 m3. The ratio of its diameter to its height is

(a) 3 : 7

(b) 7 : 3

(c) 6 : 7

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(d) 7 : 6

ANSWER:

The C.S.A. of cylinder

S = 264 m2

The volume of cylinder

V = 924 m3

2𝜋𝑟ℎ = 264

2𝑟ℎ =26412×7

22 …… (i)

2𝑟ℎ = 84

𝑟ℎ = 42

𝜋𝑟2ℎ = 924

𝑟(𝑟ℎ) =92442×7

22 …… (ii)

𝑟(𝑟ℎ) = 42 × 7

From eq. (i) and (ii),

We get 𝑟 = 7

Putting the value in (i) ℎ = 6

Hence, 𝑑

ℎ=

14

6=

7

3

𝑑: ℎ = 7: 3

Hence, the correct answer is choice (b).

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Page No 14.89:

Question 19: A cylinder with base radius of 8 cm and height of 2 cm is

melted to form a cone of height 6 cm. The radius of the cone is

(a) 4 cm

(b) 5 cm

(c) 6 cm

(d) 8 cm

ANSWER:

Volume of cylinder = 𝜋𝑟2ℎ = 𝜋 × (8)2 × 2 = 128𝜋 cm

Let r be the radius of cone

But,

The volume of cone = volume of cylinder

1

3𝜋𝑟2ℎ = 128𝜋

𝑟2 =128×31

62= 64

𝑟 = 8 cm

Hence, Radius of cone = 8 cm.

Hence, the correct answer is choice (d).

Page No 14.89:

Question 20: The volumes of two spheres are in the ratio 64: 27. The ratio

of their surface areas is

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(a) 1 : 2

(b) 2 : 3

(c) 9 : 16

(d) 16 : 9

ANSWER:

Ist sphere

𝑉1 =4

3𝜋𝑟1

3 …… (i)

IInd sphere

𝑉2 =4

3𝜋𝑟2

3 …… (ii)

Divide (i) by (ii) we get,

𝑉1

𝑉2=

4

3𝜋𝑟1

3

4

3𝜋𝑟2

3

64

27= (

𝑟1

𝑟2)

3

𝑟1

𝑟2= √

64

27

𝑟1

𝑟2=

4

3

Now, the ratio of their C.S.A

𝑆1

𝑆2=

4𝜋𝑟12

4𝜋𝑟22 = (

𝑟1

𝑟2)

2

𝑆1

𝑆2= (

4

3)

2=

16

9

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Hence, 𝑆1: 𝑆2 = 16: 9

Hence, the correct answer is choice (d).

Page No 14.89:

Question 21: If three metallic spheres of radii 6 cm, 8 cm and 10 cm are

melted to form a single sphere, the diameter of the sphere is

(a) 12 cm

(b) 24 cm

(c) 30 cm

(d) 36 cm

ANSWER:

Let r be the radius of single sphere.

Now,

The volume of single sphere = sum of volume of three spheres

4

3𝜋𝑟3 =

4

3𝜋(61)3 +

4

3𝜋(8)3 +

4

3𝜋(10)3

4

3𝜋𝑟3 =

4

3𝜋(216 + 512 + 1000)

𝑟3 = 1728

𝑟 = 12 cm

Hence, the diameter = 20 × r = 24 cm

Hence, the correct answer is choice (b).

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Page No 14.89:

Question 22: The surface area of a sphere is same as the curved surface

area of a right circular cylinder whose height and diameter are 12 cm each.

The radius of the sphere is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 12 cm

ANSWER:

Let r be the radius of sphere

But,

Surface area of sphere = C.S.A. of cylinder

4𝜋𝑟2 = 2𝜋𝑟ℎ

4𝑟2 = 2 × 6 × 12

𝑟2 =2×6×12

4

𝑟 = 6 cm

Hence, the correct answer is choice (c).

Page No 14.89:

Question 23: The volume of the greatest sphere that can be cut off from

a cylindrical log of wood of base radius 1 cm and height 5 cm is

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(a) 4

3π

(b) 10

3π

(c) 5π

(d) 20

3π

ANSWER:

The radius of greatest sphere cut off from cylindrical log of wood should

be radius of cylindrical log.

i.e., r = 1 cm

The volume of sphere

=4

3𝜋(1)3

=4

3𝜋 cm³

Hence, the correct answer is choice (a).

Page No 14.89:

Question 24: A cylindrical vessel of radius 4 cm contains water. A solid

sphere of radius 3 cm is lowered into the water until it is completely

immersed. The water level in the vessel will rise by

(a) 2

9 cm

(b) 4

9 cm

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(c) 9

4 cm

(d) 9

2 cm

ANSWER:

The radius of sphere, r = 3 cm

The volume of sphere

=4

3𝜋𝑟3

=4

3𝜋(3)3

= 36𝜋 cm3

Since,

The sphere fully immersed into the vessel, the level of water be raised

by x cm.

Then,

The volume of raised water = volume of sphere

𝜋(4)2 × 𝑥 = 36𝜋

𝑥 =36

16

𝑥 =9

4 cm

Hence, the correct answer is choice (c).

Page No 14.90:

Question 25: 12 spheres of the same size are made from melting a solid

cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere

is

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(a) √3 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

ANSWER:

The volume of solid cylinder = 12 × volume of one sphere

𝜋 × (8)2 × 2 = 12 ×4

3𝜋𝑟3

128 = 16𝑟3

𝑟3 =12864

168

𝑟 = √64

8

3

=4

2

𝑟 = 2 cm

The required diameter d = 2 × 2 = 4 cm

Hence, the correct answer is choice (d).

Page No 14.90:

Question 26: A solid metallic spherical ball of diameter 6 cm is melted

and recast into a cone with diameter of the base as 12 cm. The height of

the cone is

(a) 2 cm

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(b) 3 cm

(c) 4 cm

(d) 6 cm

ANSWER:

Clearly,

The volume of recasted cone = volume of sphere

1

3𝜋 (

12

2)

2× ℎ =

4

3𝜋 (

6

3)

3

1

3× 36 × ℎ =

4

3× 27

ℎ =4×27×3

3×36

ℎ = 3 cm

Hence, the correct answer is choice (b).

Page No 14.90:

Question 27: A hollow sphere of internal and external diameters 4 cm

and 8 cm respectively is melted into 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

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ANSWER:

External radius 𝑟1 =8

2= 4 cm

Internal radius 𝑟2 =4

2= 2 cm

The volume of hollow sphere

𝑉 =4

3𝜋(𝑅3 − 𝑟3)

=4

3𝜋(43 − 23)

Let h be the height of cone.

Clearly,

The volume of recasted cone = volume of hollow sphere

1

3𝜋𝑟2ℎ =

4

3𝜋(43 − 23)

⇒ 42ℎ = 4(43 − 23)

⇒ ℎ = 14 cm

Hence, the height of cone = 14 cm

Hence, the correct answer is choice (b).

Page No 14.90:

Question 28: A solid piece of iron of dimensions 49 × 33 × 24 cm is

moulded into a sphere. The radius of the sphere is

(a) 21 cm

(b) 28 cm

(c) 35 cm

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(d) none of these

ANSWER:

The volume of iron piece = 49 × 33 × 24 cm3

Let, r is the radius sphere.

Clearly,

The volume of sphere = volume of iron piece

4

3𝜋𝑟3 = 49 × 33 × 24

4

3×

22

7× 𝑟3 = 49 × 33 × 24

𝑟3 =49×333×246×3×7

4×222

𝑟3 = 49 × 3 × 3 × 3 × 7

𝑟 = 7 × 3

𝑟 = 21 cm

Hence, the correct answer is choice (a).

Page No 14.90:

Question 29: The ratio of lateral surface area to the total surface area of

a cylinder with base diameter 1.6 m and height 20 cm is

(a) 1 : 7

(b) 1 : 5

(c) 7 : 1

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(d) 8 : 1

ANSWER:

𝑟 =1.6m

2

=160

2

= 80 cm

ℎ = 20 cm

The ratio of lateral surface to the total surface area of cylinder

𝑠1

𝑠2=

2𝜋𝑟ℎ

2𝜋𝑟(ℎ+𝑟)

𝑠1

𝑠2=

ℎ

(ℎ+𝑟)

=20

(20+80)

𝑠1

𝑠2=

20

100

=1

5

𝑠1: 𝑠2 = 1: 5

Hence, the correct answer is choice (b).

Page No 14.90:

Question 30: A solid consists of a circular cylinder surmounted by a right

circular cone. The height of the cone is h. If the total height of the solid is

3 times the volume of the cone, then the height of the cylinder is

(a) 2h

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(b) 3ℎ

2

(c) ℎ

2

(d) 2ℎ

2

ANSWER:

Disclaimer: In the question, the statement given is incorrect. Instead of

total height of solid being equal to 3 times the volume

of cone, the volume of the total solid should be equal to 3 times the volume

of the cone.

Let x be the height of cylinder.

Since, volume of the total solid should be equal to 3 times the volume of

the cone,

So,

1

3𝜋𝑟2ℎ + 𝜋𝑟2𝑥 = 3 (

1

3𝜋𝑟2ℎ)

⇒1

3𝜋𝑟2ℎ − 𝜋𝑟2ℎ + 𝜋𝑟2𝑥 = 0

⇒ 𝜋𝑟2𝑥 =2

3𝜋𝑟2ℎ

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⇒ 𝑥 =2

3ℎ

Hence, the height of cylindrical part =2ℎ

3

Hence, the correct answer is choice (d).

Page No 14.90:

Question 31: The maximum volume of a cone that can be carved out of a

solid hemisphere of radius r is

(a) 3πr2

(b) 𝜋𝑟3

3

(c) 𝜋𝑟2

3

(d) 3πr3

ANSWER:

Radius of hemisphere = r

Therefore,

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The radius of cone = r

and height h = r

Then,

Volume of cone

=1

3𝜋𝑟2ℎ

=1

3𝜋𝑟2 × 𝑟

=1

3𝜋𝑟3 (unit)3

Hence, the correct answer is choice (b).

Page No 14.90:

Question 32: The radii of two cylinders are in the ratio 3 : 5. If their

heights are in the ratio 2 : 3, then the ratio of their curved surface areas is

(a) 2 : 5

(b) 5 : 2

(c) 2 : 3

(d) 3 : 5

ANSWER:

Given that

𝑟1: 𝑟2 = 3: 5 and ℎ1: ℎ2 = 2: 3

Then,

The ratio of C.S.A. of cylinders

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𝑠1

𝑠2=

2𝜋𝑟1ℎ1

2𝜋𝑟2ℎ2

𝑠1

𝑠2= (

𝑟1

𝑟2) × (

ℎ1

ℎ2)

=3

5×

2

3

𝑠1

𝑠2=

2

3

𝑠1: 𝑠2 = 2: 5

Hence, the correct answer is choice (a).

Page No 14.90:

Question 33: A right circular cylinder of radius r and height h (h = 2r)

just encloses a sphere of diameter

(a) h

(b) r

(c) 2r

(d) 2h

ANSWER:

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Radius of cylinder = r

Height = h

= 2r

Since, the sphere fitted the cylinder.

i.e., diameter of sphere = height of cylinder.

𝑑 = ℎ = 2𝑟

𝑑 = 2𝑟

Hence, the correct answer is choice (c).

Page No 14.90:

Question 34: The radii of the circular ends of a frustum are 6 cm and 14

cm. If its slant height is 10 cm, then its vertical height is

(a) 6 cm

(b) 8 cm

(c) 4 cm

(d) 7 cm

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ANSWER:

Radii of circular ends of frustum

𝑟1 = 14 cm, 𝑟2 = 6 cm

Slant height

𝑙 = 10 cm, ℎ =?

𝑙 = √ℎ2 + (𝑟1 − 𝑟2)2

𝑙2 = ℎ2 + (𝑟1 − 𝑟2)2

{squaring on both sides}

ℎ2 = 𝑙2 + (𝑟1 − 𝑟2)2

ℎ2 = (10)2 + (14 − 6)2

ℎ2 = 100 − 64

ℎ2 = 36

ℎ = 6 cm

Hence, the correct answer is choice (a).

Page No 14.90:

Question 35: The height and radius of the cone of which the frustum is a

part are h1 and r1 respectively. If h2 and r2 are the heights and radius of the

smaller base of the frustum respectively and h2: h1 = 1: 2, then r2: r1 is

equal to

(a) 1 : 3

(b) 1 : 2

(c) 2 : 1

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(d) 3 : 1

ANSWER:

Since,

∆𝐴𝑂𝑉 and 𝐿𝑂′𝑉 are similar triangles,

i.e., In ∆𝐴𝑂𝑉 and 𝐿𝑂′𝑉

𝑂𝐴

𝑂′𝐿=

𝑂𝑉

𝑂′𝑉

⇒𝑟1

𝑟2=

ℎ1

ℎ1−ℎ2

⇒ (ℎ1 − ℎ2)𝑟1 = ℎ1𝑟2

⇒ 𝑟1ℎ1 − 𝑟1ℎ2 = ℎ1𝑟2

⇒ 𝑟1ℎ1 − ℎ1𝑟2 = 𝑟1ℎ2

⇒ ℎ1(𝑟1 − 𝑟2) = 𝑟1ℎ2

⇒(𝑟1−𝑟2)

𝑟1=

ℎ2

ℎ1

⇒(𝑟1−𝑟2)

𝑟1=

1

2

⇒ 1 −𝑟2

𝑟1=

1

2

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⇒𝑟2

𝑟1= 1 −

1

2=

1

2

Thus, 𝑟2: 𝑟1 = 1: 2

Hence, the correct answer is choice (b).

Page No 14.90:

Question 36: The diameters of the ends of a frustum of a cone are 32 cm

and 20 cm. If its slant height is 10 cm, then its lateral surface area is

(a) 321 π cm2

(b) 300 π cm2

(c) 260 π cm2

(d) 250 π cm2

ANSWER:

𝑟1 =32

2

= 16 cm

𝑟2 =20

2

= 10 cm

Slant height = 10 cm

Total lateral surface area

= 𝜋(𝑟1 + 𝑟2)𝑙

= 𝜋(16 + 10)10

= 260 π cm2

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Hence, the correct answer is choice (c).

Page No 14.90:

Question 37: A solid frustum is of height 8 cm. If the radii of its lower

and upper ends are 3 cm and 9 cm respectively, then its slant height is

(a) 15 cm

(b) 12 cm

(c) 10 cm

(d) 17 cm

ANSWER:

𝑟1 = 9 cm

𝑟2 = 3 cm

ℎ = 8 cm

slant height of frustum, 𝑙 = √ℎ2 + (𝑟1 − 𝑟2)2

= √82 + (9 − 3)2

= √64 + 36

= √100

= 10 cm

Hence, the correct answer is choice (c).

Page No 14.91:

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Question 38: The radii of the ends of a bucket 16 cm height are 20 cm

and 8 cm. The curved surface area of the bucket is

(a) 1760 cm2

(b) 2240 cm2

(c) 880 cm2

(d) 3120 cm2

ANSWER:

Radius of top of bucket r1 = 20 cm

Radius of bottom of bucket r2 = 8 cm

Height of bucket = 16 cm

The curved surface area of bucket = 𝜋(𝑟1 + 𝑟2)𝑙

𝑙 = √ℎ2 + (𝑟1 − 𝑟2)2

= √162 + (20 − 8)2

= √256 + 144

= √400

= 20 cm

C.S.A. of bucket

= 𝜋(20 + 8) × 20

=22

7× 28 4 × 20

= 22 × 80

= 1760 cm2

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Hence, the correct answer is choice (a).

Page No 14.91:

Question 39: The diameters of the top and the bottom portions of a bucket

are 42 cm and 28 cm respectively. If the height of the bucket is 24 cm,

then the cost of painting its outer surface at the rate of 50 paise / cm2 is

(a) Rs. 1582.50

(b) Rs. 1724.50

(c) Rs. 1683

(d) Rs. 1642

ANSWER:

Radius of top of bucket 𝑟1 =42

2= 21 cm

Radius of bottom of bucket 𝑟2 =28

2= 14 cm

Height of bucket, h = 24 cm.

𝑙 = √ℎ2 + (𝑟1 − 𝑟2)2

= √576 + (21 − 14)2

= √576 + 49

= √625

= 25 cm

C.S.A. of the bucket

C.S.A. of bucket

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= 𝜋(𝑟1 + 𝑟2)𝑙

= 𝜋(21 + 14) × 25

=22

7× 35 5 × 25

= 2750 cm2

Area of bottom

= 𝜋𝑟2

=22

7× 196

= 616 cm²

The cost of painting its C.S.,

= (2750 + 616) ×1

2

= 3366 ×1

2

= Rs. 1683

Hence, the correct answer is choice (c).

Page No 14.91:

Question 40: If four times the sum of the areas of two circular faces of a

cylinder of height 8 cm is equal to twice the curve surface area, then

diameter of the cylinder is

(a) 4 cm

(b) 8 cm

(c) 2 cm

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(d) 6 cm

ANSWER:

Let r be the radius of cylinder.

Area of circular base of cylinder = 𝜋𝑟2

The height of cylinder h = 8 cm

The C.S.A. of cylinder = 2𝜋𝑟 × 8 = 16𝜋𝑟

Clearly,

4 × (𝜋𝑟2 + 𝜋𝑟2) = 2 × (𝜋𝑟)

8𝜋𝑟2 = 32𝜋𝑟

8𝑟2 = 32𝑟

𝑟 = 4 cm

The diameter of cylinder

𝑑 = 4 × 2 = 8 cm

Hence, the correct answer is choice (b).

Page No 14.91:

Question 41: If the radius of the base of a right circular cylinder is halved,

keeping the height the same, then the ratio of the volume of the cylinder

thus obtained to the volume of original cylinder is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1 [CBSE 2012]

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ANSWER:

Let the radius and height of the original cylinder be R and h, respectively.

Now, the radius of the new cylinder = 𝑅

2

Then, the ratio of the volume of the cylinder thus obtained to the volume

of original cylinder is given by

𝜋 (𝑅

2)

2ℎ : 𝜋 𝑅2ℎ

=1

4 : 1

= 1 : 4

Hence, the correct answer is option C.

Page No 14.91:

Question 42: A metalic solid cone is melted to form a solid cylinder of

equal radius. If the height of the cylinder is 6 cm, then the height of the

cone was

(a) 10 cm

(b) 12 cm

(c) 18 cm

(d) 24 cm [CBSE 2014]

ANSWER:

Let the height of the cone be h.

Volume of cylinder = Volume of cone

⇒ 𝜋𝑟2(6) =1

3𝜋𝑟2ℎ

⇒ ℎ = 18 cm

Hence, the correct answer is option C.

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Page No 14.91:

Question 43:

A rectangular sheet of paper 40 cm ⨯ 22 cm, is rolled to form a hollow

cylinder of height 40 cm. The radius of the cylinder (in cm) is

[CBSE 2014]

(a) 3.5

(b) 7

(c) 80

7

(d) 5

ANSWER:

Let the radius of the cylinder be r cm.

Curved surface area of cylinder = Area of rectangular sheet

⇒ 2𝜋𝑟(40) = 40 × 22

⇒ 2 ×22

7× 𝑟 × 40 = 40 × 22

⇒ 𝑟 = 3.5 cm

Hence, the correct answer is option A.

Page No 14.91:

Question 44: The number of solid spheres, each of diameter 6 cm that can

be made by melting a solid metal cylinder of height 45 cm and diameter

4 cm is [CBSE 2014]

(a) 3

(b) 5

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(c) 4

(d) 6

ANSWER:

Let the number of solid spheres be n.

Now, Volume of n solid sphere = Volume of cylinder

⇒ 𝑛 ×4

3×

22

7× (

6

2)

3=

22

7× (

4

2)

2× 45

⇒ 𝑛 ×4

3× 27 = 4 × 45

⇒ 𝑛 = 5

Hence, the correct answer is option B.

Page No 14.91:

Question 45: Volumes of two spheres 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

ANSWER:

Volume of the spheres are in the ratio 64: 27.

⇒ 𝑉1: 𝑉2 = 64: 27

⇒4

3𝜋𝑟1

3:4

3𝜋𝑟2

3 = 64: 27

⇒ 𝑟13: 𝑟2

3 = 64: 27

⇒ (𝑟1

𝑟1)

3=

64

27

⇒ (𝑟1

𝑟1)

3= (

4

3)

3

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Thus, ratio of the radii = 4 : 3

Ratio of the surface area of the spheres will be 4𝜋𝑟1

2

4𝜋𝑟22

= (𝑟1

𝑟1)

2= (

4

3)

2=

16

9

So, the ratio is 16 : 9.

Hence the correct answer is option (d).

Page No 14.91:

Question 46: A right circular cylinder of radius r and height h (h > 2r)

just encloses a sphere of diameter

(a) r (b) 2r (c) h (d) 2h

ANSWER:

Since h > 2r where h is the height of the cylinder and r is the radius

So, when a sphere is enclosed in it, the radius of the sphere will be r.

Thus, the diameter of the sphere will be 2r.

Hence, the correct answer is option (b)

Page No 14.91:

Question 47: 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

ANSWER:

When a plane parallel to the base of a cone cuts it, then a frustum and a

smaller cone is formed.

The cross-section thus formed will be a circle.

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Hence, the correct answer is option (a).

Page No 14.91:

Question 48: If two solid-hemisphere s of same base radius 𝑟 are joined

together along their bases , then curved surface area of this new solid is

(a) 4πr2 (b) 6πr2 (c) 3πr2 (d) 8πr2

ANSWER:

Base radius of the hemisphere = 𝑟

Since the two hemispheres are joined end to end, it becomes a complete

sphere.

Curved surface area of the new solid = total surface area of the sphere.

Curved surface area of the new solid = 4𝜋𝑟2

Hence, the correct answer is option (a).

Page No 14.91:

Question 49: The diameters of 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

ANSWER:

The bucket is in the form of a frustum.

The diameters are respectively, 𝑑1 = 44 cm and 𝑑2 = 24 cm

Radii of the circular ends = 𝑟1 = 22 cm and 𝑟2 = 12 cm

Volume, 𝑉 =1

3𝜋ℎ(𝑟1

2 + 𝑟1𝑟2 + 𝑟22)

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𝑉 =1

3𝜋 × 35(222 + 22 × 12 + 122)

𝑉 = 32706.6 cm3 = 32.7 liters

Hence, the correct answer is option (a)

Page No 14.91:

Question 50:

A spherical ball of radius 𝑟 is melted to make 8 new identical balls each

of radius 𝑟1. Then 𝑟: 𝑟1 =

(a) 2: 1 (b) 1: 2 (c) 4: 1 (d) 1: 4

ANSWER:

Radius of the bigger sphere = r cm

Radius of smaller spheres = r1 cm

Volume of bigger sphere

Volume of small spheres=

4

3𝜋𝑟3

4

3𝜋𝑟1

3=

𝑟3

𝑟13 = 8

⇒ (𝑟

𝑟1)

3= (

2

1)

3

⇒ (𝑟

𝑟1) = (

2

1)

Hence, 𝑟 ∶ 𝑟1 = 2 ∶ 1. Hence, the correct answer is option (a).

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