# Surface Areas and Volumes

Chapters

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- Solutions
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Solutions

Find the lateral surface area and total surface area of the following right prisms.

i)L.S.A. = 4l^{2}

= 4 × 4^{2}

= 64cm^{2}

T.S.A = 6l^{2}

= 6 × 4^{2}= 96cm^{2}

ii) L.S.A. =2h(l + b)

= 2 × 5 (8 + 6)

= 10 × 14 = 140 cm^{2}

T.S.A. = 2 (lb + bh + lh)

= 2(8 × 6 + 6 × 5 + 8 × 5)

= 2 (48 + 30 + 40)

= 236 cm^{2}

The total surface area of a cube is 1350 sq.m. Find its volume.

Solution:

Given T.S.A. of a cube 6l^{2}= 1350

l^{2} = 1350/6

l^{2} = 225

∴ l = √225 = 15m

∴ Volume of the cube = l3

= 15 × 15 × 15

= 3375 m^{3}

Find the area of four walls of a room (Assume that there are no doors or windows) if its length 12 m; breadth 10 m and height 7.5 m.

Solution:

Length of the room = 12 m

Breadth of the room = 10 m

Height of the room = 7.5 m

Area of four walls of the room

A = 2h (l + b)

A = 2 × 7.5 (12 + 10)

= 15 × 22

= 330 m^{2}

The volume of a cuboid is 1200 cm

Solution:

Length of a cuboid, ‘l’ = 15 cm

Breadth of the cuboid, b = 10 cm

Volume of the cuboid, V = lbh = 1200.cm^{3}

Let the height = h

∴ 15 × 10 × h = 1200

∴ h =1200/15×10

= 8 cm

How does the total surface area of a box change if

i) Each dimension is doubled ?

Solution:

Let the original dimensions be Length - l units

Breadth - b units

Height - h units

Then T.S.A = 2 (lb + bh + lh)

If the dimensions are doubled then

Length = 2l

Breadth = 2b

Height = 2h

T.S.A. = 2 (2l. 2b + 2b . 2h + 2l . 2h)

= 2 (4lb + 4bh + 4lh)

= 4 × [2 (lb + bh + lh]

= 4 × original T.S.A.

i.e., T.S.A. increases by 4 times.

ii) Each dimension is tripled ?

Solution:

Let the original and changed dimensions are l, b, h and 31, 3b, 3h

Original T.S.A. = 2 (lb + bh + lh)

Changed T.S.A

= 2 (3l . 3b + 3b . 3h + 3l. 3h)

Changed T S.A. = 2 (9lb + 9bh + 9lh)

= 9 × [2 (lb + bh + lh)]

= 9 [original T.S.A.]

Thus original T.S.A. increased by 9 times if each dimension is tripled.

The base of a prism is triangular in shape with sides 3 cm, 4 cm and 5 cm. Find the volume of the prism if its height is 10 cm.

Solution:

Volume of triangular prism = Area of the base × height

Sides of the triangle are 3 cm, 4 cm and 5 cm.

Area = s (s - a) (s - b) (s - c)

∴ Volume of the prism = 6 × 10 = 60 cm^{3}

(OR)

3 cm, 4 cm and 5 cm are the sides of right triangle.

∴ Area of the triangle

=1/2 bh = 1/2 × 3 × 4 = 6 cm^{2}

Volume of prism = base area × height

= 6 × 10 = 60cm^{3}

A regular square pyramid is 3 m height and the perimeter of its base is16 m Find the volume of the pyramid.

Solution:

Perimeter of the base = 16 m

Height of the pyramid 3 m

Volume of the pyramid

= 1/3 × volume of prism

= 1/3 × (base area x height)

= 1/3 × 4 × 4 × 3= 16m [4 × side=16 ∴ side = 4 m Area = s^{2}= 4 × 4]

An Olympic swimming pool is in the shape of a cuboid of dimensions 50 m long and 25 m wide. If it is 3 m deep throughout, how many litres of water does it hold ?

Solution:

Dimensions of the swimming pool are

Length = 50 m

Breadth = 25 m

Deep = 3 m

∴ Volume of the swimming pool

V = lbh

V = 50 × 25 × 3 = 3750 m^{3}

∴ It can hold 37,50,000 litres of water.

[∵ 1 m^{3} = 1000 lit.]

A closed cylindrical tank of height 1.4 m and radius of the base is 56 cm is made up of a thick metal sheet. How much metal sheet is required ?

(Express in square metres).

Solution:

Radius of the tank r’ = 56 cm

=56/100 m = 0.56m

Height of the tank h = 1.4 m

T.S.A. of a cylinder = 2πr (r + h)

∴ Area of the metal sheet required = 2πr (r + h)

A = 2 × 22/7 × 0.56 × (0.56 + 1.4)

= 2 × 22 × 0.08 × 1.96

= 6.8992 m^{2}

= 6.90 m^{2}

The volume of a cylinder is 308 cm

Solution:

Volume of the cylinder V = πr^{2}h

= 308 cm^{3}

Height of the cylinder h = 8 cm

∴ 308 = 22/7 . r^{2} x 8

r^{2}= 308 × 7/22 x 1/8

r^{2}= 12.25

∴ r =√12.25 = 3.5cm

L.S.A. = 2πrh

= 2 × 22/7× 3.5 × 8 = 176cm^{2}

T.S.A. = 2πr (r + h)

2 × 22/7 × 3.5 (3.5 + 8)

= 2 × 22 × 0.5 × 11.5 = 253 cm^{2}

A metal cuboid of dimensions 22 cm × 15 cm × 7.5 cm was melted and cast into a cylinder of height 14 cm. What is its radius ?

Solution:

Dimensions of the metal cuboid

= 22 cm × 15 cm × 7.5 cm

Height of the cylinder, h = 14 cm

Cuboid made as cylinder

∴ Volume of cuboid = Volume of cylinder

lbh = 2πr^{2}h

⇒ 22 × 15 × 7.5 = 22/7 × r^{2}× 14

⇒ r^{2}=22×15×7.5×7/14×22

⇒ r^{2}= 7.5 × 7.5

r = 7.5 cm

An overhead water tanker is in the shape of a cylinder has capacity of 616 litres. The diameter of the tank is 5.6 m. Find the height of the tank. cagp)

Solution:

Volume of the cylinder, V = πr2h = 616

Diameter of the tank = 5.6 m

Thus its radius, r = d/2 = 5.6/2 = 2.8 m

Height = h (say)

∴ πr2 h = 616

22/7 × 2.8 × 2.8 × h = 616

h = 616×7/22×2.8×2.8 = 25

∴ Height = 25 m

A metal pipe is 77 cm long. The inner diametre of a cross section is 4 cm; the outer diameter being 4.4 cm (see figure). Find its

i) Inner curved surface area

ii) Outer curved surface area

iii) Total surface area

i) Inner curved surface area

Solution:

Height of the pipe = 77 cm

Inner diameter = 4 cm

Inner radius = d/2 = 4/2 = 2 cm

∴ Inner C.S.A. = 2πrh

= 2 × 22/7 × 2 × 77

= 88 × 11 = 968cm^{2}

ii) Outer curved surface area

Solution:

Outer diameter = 4.4 cm

∴ Outer radius, r = d/2 = 4.4/2 = 2.2 cm

Height of the pipe, h = 77 cm

∴ Outer C.S.A. = 2πrh

= 2 × 22/7 × 2.2 × 77

= 96.8 × 11

= 1064.8 cm^{2}

iii) Total surface area Sol. Total surface area .

= Inner C.S.A + Outer C.S.A

= 968 + 1064.8

= 2032.8 cm^{2}

A cylindrical pillar has a diameter of 56 cm and is of 35 m high. There are 16 pillars around the building. Find the cost of painting the curved surface area of all the pillars at the rate of ? 5.50 per 1 m

Solution:

Diametre of the cylindrical pillar = 56 cm

Thus its radius, r = d/2

=56/2 = 28cm = 28/100 m = 0.28m

Height of the pillar, h = 35 m

Total number of pillars =16

Cost of painting = ₹ 5.50 per sq. m.

C.S.A. of each pillar = 2πrh

= 2 × 22/7 × 0.28 × 35

= 2 × 22 × 0.04 × 35 = 61.6 m^{2}

∴ C.S.A. of 16 pillars = 16 × 61.6 = 985.6 m^{2}

Cost of painting 16 pillars at the rate of ₹ 5.5 per sq.m. = 985.6 × 5.5

= ₹ 5420.8

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to roll once over the play ground to level. Find the area of the play ground in m

Soi. Diameter of the roller = 84 cm

Thus radius = 84/2 = 42 cm

= 42/100 m = 0.42m

Length of the roller =120 cm

= 120/100 = 1.2m

It takes 500 complete revolutions to roll over the play ground.

Thus 500 × L.S.A. of the roller

= Area of the play ground

∴ Area of the play ground = 500 × 2πrh

= 500 × 2 × 22/7 × 0.42 × 1.2 = 1584 m^{2}

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find (i) its inner curved surface area (ii) the cost of plastering this curved surface at the rate of Rs. 40 per m

Solution:

Inner diameter of the circular well, d = 3.5 m

Thus its radius, r = d/2 = 3.52 = 1. 75 m

Depth of the well (height) = 10 m

i) Inner C.S.A. = 2πrh

= 2 × 227 × 1.75 × 10

= 110 m^{2}

ii) Cost of plastering at the rate of

₹ 40 / m^{2} = 110 × 40 = ₹ 4400

Find (i) the total surface area of a closed cylindrical petrol storage tank whose diameter 4.2 m and height 4.5 m.

Solution:

Diameter of the cylindrical tank ‘d’ = 4.2m

Thus its radius, r = d/2 = 4.2/2 = 2.1 m

Height of the tank, h = 4.5 m

T.S.A. of the tank = 2πr (r + h)

= 2 × 22/7 × 2.1 (2.1 + 4.5)

= 2 × 22 × 0.3 × 6.6 = 87.12m^{2}

ii) How much steel sheet was actually used, if 1/12 of the steel was wasted in making the tank ?

Solution:

1/12 of the sheet was wasted.

= > 1 - 1/12 = 11/12of the sheet was used

in making the tank.

Let the metal sheet originally brought was = x m^{2}

11/12 x = 87.12m^{2}

∴ x = 87.12 x 12/11 = 95.04m^{2}

A one side open cylindrical drum has inner radius 28 cm and height 2.1 m. How much water you can store in the drum? Express in litres.

(1 litre = 1000 c.c)

Solution:

Inner radius of the cylindrical drum ‘r’ = 28 cm

. Its height, h = 2.1 m = 2.1 x 100 = 210 cm

Volume of the drum = πr^{2}h

= 22/7 x 28 x 28 x 210

= 22 x 4 x 28 x 210

= 517440 cc

=517440/1000

= 517.44 lit.

The curved surface area of the cylinder is 1760 cm

Solution:

C.S.A of the cylinder = 2πrh = 1760 cm^{2}

Volume of the cylinder = πr^{2}h

= 12320 cm^{3}

Height = h (say)

Volume/C.S.A.= πr^{2}h/2πrh= 12320/1760

⇒r/2 = 7

∴ r = 7 x 2 = 14cm

Now 2πrh = 1760cm^{2}

2 x22/7 x 14h = 1760

h =1760x7/2x22x14= 20cm

∴Height of the cylinder = 20cm

The base area of a cone is 38.5 cm Its volume is 77 cm

Solution:

Base area of the cone, πr^{2}= 38.5 cm^{2}

Volume of the cone, V =1/3πr^{2}h = 77

πr^{2}= 38.5

22/7r^{2}= 38.5

r^{2}= 38.5 x 7/22

r^{2}= 12.25

r =√12.25 = 3.5

V=1/22 x 22/7 x 3.5 x 3.5 x h = 77

∴ h =77x3x7/22x12.25= 6

∴ Height of the cone = 6 cm

The volume of a cone is 462 m

Solution:

The volume of a cone”V’=13πr^{2}h = 462

Radius ‘r’ = 7 m

Height = h (say)

1/3 x 22/7 x 7 x h = 462

h =462x3/22x7= 9

∴ Height = 9m

Curved surface area of a cone is 308 cm

Solution:

C.S.A. of the cone, πrl = 308

Slant height, l = 14 cm

i) πrl = 308; l = 14 cm

22/7 x r x 14 = 308

r =308/44= 7cm

ii) T.S.A. = πrl + πr^{2}

= πr (r + l) =22/7x 7 x (7 + 14)

= 22 x 21 = 462 cm^{3}

The cost of painting the total surface area of a cone at 25 paise per cm2 is ₹176. Find the volume of the cone, if its slant height is 25 cm.

Solution:

Slant height of the cone, l = 25 cm

Total cost at the rate of 25 p/cm^{2}

= ₹176

∴ Total surface area of the cone

= 176/25 x 100 = 176 x 4 = 704cm^{2}

But T.S.A. of the cone = πr (r + l) = 704

Thus 22/7 r(r + 25) = 704

r(r + 25) = 704x7/22 = 224

r^{2}+ 25r = 224

⇒ r^{2}+ 32r - 7r - 224 = 0

⇒ r (r + 32) - 7 (r + 32) = 0

⇒ (r + 32) (r - 7) = 0

⇒ r = 7 (∵ ’r’ can’t be negative)

∴ Volume of the cone = 1/3 πr^{2}h

= 1/3 x 22/7 x 7 x 7 24

= 22 x 7 x 8 = 1232cm^{3}

From a circle of radius 15 cm, a sector with angle 216° is cut out and its bounding radii are bent so as to form a cone. Find its volume.

Solution:

Radius of the sector, ‘r’ = 15 cm

Angle of the sector, ‘x’ = 216°

∴ Length of the arc, l = x/360 x 2πr

= 216/360 2πr= 3/5 (2πr)

Perimeter of the base of the cone = Length of the arc

2πr of cone == 6/5 πr of the circle

Radius of the cone ‘r’ =35x 15 = 9

Radius ‘r’ of the circle = slant height l of the cone = 9 cm

= 1018.3 cm^{3}

The height of a tent is 9 m. Its base diameter is 24 m. What is its slant height ? Find the cost of canvas cloth required if it costs ₹14 per sq.m.

Solution:

Height of a conical tent ‘h’ = 9 m

Base diameter = 24 m

Thus base radius ’r’= d/2 = 24/2 = 12m

Cost of canvas = ₹ 14 per sq.m.

C.S.A. of the cone = πrl

The curved surface area of a cone is 1159 5/7 cm

Solution:

C.S.A. of the cone = πrl

A tent is cylindrical to a height of 4.8 m and conical above it. The ra¬dius of the base is 4.5 m and total height of the tent is 10.8 m. Find the canvas required for the tent in square

meters.

Solution:

Radius of cylinder, l = 4.5 m

Height of the cylinder = h = 4.8 m

∴ C.S.A. of the cylinder = 2πrh

= 2 x 22/7 x 4.5 x 4.8

= 135.771 m^{2}

Radius of the cone ‘r’ =

Radius of the cylinder = 4.5 m

Height of the cone ’h’ = 10.8 - 4.8 = 6 m

∴ Slant height of the cone

∴ C.S.A. of the cone = πrl

= 227 x 4.5 x 7.5

= 742.57 = 106.071m^{2}

∴ Total canvas required

= C.S.A of cylinder + C.S.A. of cone

= 135.771 + 106.071

= 241.842 m^{2}

What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? Assume that extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (use π = 3.14) [Note : Take 20 cm as 0.6 m

Solution:

Radius of the cone, r = 6 m

Height of the cone, h = 8 m

∴ C.S.A. = πrl = 3.14 x 6 x 10 = 188.4 m^{2}

Let the length of the tarpaulin = l

∴ Area of the tarpaulin, lb = 188.4 + 0.6

= 189 m^{2}

⇒ 3l = 189

⇒ l = 189/3 = 63m

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 27 cm. Find the area of the sheet required to make 10 such caps.

Solution:

Radius of the cone, r = 7 cm

Height of the cone, h = 27 cm

Slant height of the cone (l)

∴ Total area of the sheet required for

10 caps = 10 x 22 √778

= 6136.383 cm2

Water is pouring into a conical vessel (as shown in the given figure), at the rate of 1.8 m

Solution:

From the figure, diameter of the cone 5.2 m

Thus its radius ’r’ = 5.2/2 = 2.6 m

∴ Height of the cone = h = 6.8 m

Volume of the cone = 1/3 πr^{2}h

= 1/3 x 22/7 x 2.6 x 2.6 x 6.8

= 1011.296/21

= 48.156 m^{3}

Quantity of water that flows per minute

= 1.8 m^{3}

∴ Total time required = Total volume/1.8

= 48.156/1.8 = 26.753

27 minutes.

Two similar cones have volumes 12π CU. units and 96π CU. units. If the curved surface area of smaller cone is 15π sq.units, what is the curved surface area of the larger one?

Solution:

The radius of a sphere is 3.5 cm. Find its surface area and volume.

Solution:

Radius of the sphere, r = 3.5 cm

The surface area of a sphere is 1018 2/7 cm

Solution:

Surface area of sphere = 4πr^{2}

= 1018 2/7 cm^{2}

= 3054.857cm^{3}

≅ 3054.86cm^{3}

The length of equator of the globe is 44 cm. Find its surface area.

Solution:

Length of the equator of the globe 2πr = 44 cm.

2 x 22/7 x r = 44

∴ r = 44x7/2x22= 7cm

∴ surface area = 4πr^{2}

= 4 x 22/7 x 7 x 7

= 4 x 22 x 7

= 616cm^{2}

The diameter of a spherical ball is 21 cm. How much leather is required to prepare 5 such balls?

Solution:

Diameter of the spherical ball d’ = 21 cm

Thus, its radius r = d/2 = 21/2 = 10.5 cm

Surface area of one ball = 4πr^{2}

= 4 x 22/7 x 10.5 x 10.5

= 88 x 1.5 x 10.5 = 1386 cm^{2}

∴ Leather required for 5 such balls

= 5 x 1386 = 6930 cm^{2}

The ratio of radii of two spheres is 2 : 3. Find the ratio of their surface areas and volumes.

Solution:

Ratio of radii r_{1}: r_{2}= 2 : 3

Ratio of surface area

= 4πr_{1}^{2}: 4πr_{2}^{2}

= 2^{2}: 3^{2}= 4 : 9

Ratio of volumes

= 4/3 πr_{1}^{3}: 4/3 πr_{2}^{3}

= 2^{3}: 3^{3}= 8 : 27

Find the total surface area of hemisphere of radius 10 cm. (Use π = 3.14)

Solution:

Radius of the hemisphere = 10 cm

Total surface area of the hemisphere = 3πr^{2}

= 3 x 3.14 x 10 x 10

= 9.42 x 100

= 942 cm^{2}

The diameter of a spherical balloon increases from 14 cm. to 28 cm. as air is being pumped into it. Find the ratio of surface areas of the balloons in the

two cases.

Solution:

The diameter of the balloon, d = 14 cm

Thus, its radius, r = d/2 = 14/2 = 7 cm

∴ Surface area = 4πr^{2}= 4 x 22/7 x 7 x 7

= 88 x 7 = 616cm^{2}

When air is pumped, the diameter = 28 cm

thus its radius = d/2 = 28/2 = 14 cm

Its surface area = 4πr^{2}

= 4 x 22/7 x 14 x 14

= 88 x 28 = 2464 cm^{2}

Ratio of areas = 616 : 2464

= 1 : 4

(OR)

Original radius = 14/2 = 7 cm

Increased radius = 28/2 = 14cm

Ratio of areas = r_{1}^{2}: r_{2}^{2}

= 7^{2}: 14^{2}

= 7 x 7 : 14 x 14

= 1:4

A hemispherical bowl is made of brass, 0.25 cm thickness. The inner radius of the bowl is 5 cm. Find the ratio of outer surface area to inner surface area.

Solution:

Inner radius of the hemisphere ‘r’ = 5 cm

Outer radius of the hemisphere ‘R’

= inner radius + thickness

= (5 + 0.25) cm = 5.25 cm

Ratio of areas = 3πR^{2}: 3πr^{2}

= R^{2}: r^{2}

= (5.25)^{2}: 5^{2}

= 27.5625 : 25

= 1.1025:1

= 11025 : 10000

= 441 : 400

[Note : If we read “radius as diameter” then we get the T.B. answer]

The diameter of a lead ball is 2.1 cm. The density of the lead used is 11.34 g/c

Solution:

The diameter of the ball = 2.1 cm

Thus, its radius, r = d/2 = 2.1/2 = 1.05 cm

Volume of the ball V’ = 4/3 πr^{3}

=4/3 x 22/7 x 1.05^{3}=101.87/21

∴Weight of the ball = Volume x density

= 4.851 x 1.34

= 55.010

A metallic cylinder of diameter 5 cm 1 and height 3 1/3 cm is melted and cast into a sphere. What is its diameter ?

Solution:

Diameter of the cylinder’d’ = 5 cm

Thus, its radius, r = d/2 = 5/2 = 2.5 cm

Height of the cylinder,

Volume of the cylinder

Given that cylinder melted to form sphere

∴ Volume of the sphere = Volume of the cylinder

(Where r is the radius of the sphere)

r^{3} = 3/4 x 2.5 x 2.5 x 10/3

r^{3} = 2.5^{3}

∴ r = 2.5 cm

Hence its diameter, d = 2r

= 2 x 2.5 = 5 cm

How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold ?

Solution:

Diameter of the hemispherical bowl = 10.5 cm

Thus its radius = d/2 = 10.5/2 = 5.25cm

Quantity of milk, the bowl can hold = Volume of the bowl = 2/3πr^{3}

= 2/3 x 22/7 x 5.25 x 5.25 x 5.25

= 303.1875 cm^{3}

=303.1875/1000 lit = 0.303 lit.

A hemispherical bowl has diameter 9 cm. The liquid is poured into cylindrical bottles of diameter 3 cm and height 3 cm. If a full bowl of liquid is Riled in the bottles, find how many

bottles are required ?

Solution:

Diameter of the hemispherical bowl ‘d’ = 9 cm

Its radius, r = d/2 = 9/2 = 4.5cm

Volume of its liquid = Volume of the bowl = 2/3 πr^{3}

= 2/3 x 22/7 x 4.5 x 45 x 4.5

Diameter of the cylindrical bottle, d = 3 cm

Its radius, r = d/2

=3.02

= 1.5cm

Height of the bottle, h = 3 cm

Let the number of bottles required = n

Then total volumes of these n bottles = n πr2h

But this is equal to volume of the bowl

Hence n. 22/7 × 1.5 × 1.5 × 3

= 2/3 × 22/7 × 4.5 × 4.5 × 4.5

∴ n = 23×20.25/1.5 = 9

∴ Number of bottles required = 9

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