R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 15 - Volume and Surface Area of Solids

Choosing the correct option for multiple choice questions in your Maths exam requires practice. For chapter revision, you can refer to TopperLearning’s RS Aggarwal and V Aggarwal Solutions for CBSE Class 9 Maths Chapter 15 Volume and Surface Area of Solids. The solutions by some of the best Maths experts will help you relearn concepts such as the total surface area of a cube, volume of a cube, volume of a cylinder etc.

With our chapter solutions, get clarity on application-type questions such as calculating the concrete mixture for building a pillar. Besides, our CBSE Class 9 Mathematics videos and sample papers will be useful during Maths exam preparation.

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Chapter 15 - Volume and Surface Area of Solids MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Correct option: (b)

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

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Solution 73

Chapter 15 - Volume and Surface Area of Solids Ex. 15C

Solution 1

Radius of a cone, r = 5.25 cm

Slant height of a cone, l = 10 cm

Solution 2

Radius of a cone, r = 12 m

Slant height of a cone, l = 21 cm

Solution 3

Radius of a conical cap, r = 7 cm

Height of a conical cap, h = 24 cm

Thus, 5500 cm² sheet will be required to make 10 caps.

Solution 4

Let r be the radius of a cone.

Slant height of a cone, l = 14 cm

Curved surface area of a cone = 308 cm²

Solution 5

Radius of a cone, r = 7 m

Slant height of a cone, l = 25 m

Cost of whitewashing = Rs. 12 per m²

⇒ Cost of whitewashing 550 m² area = Rs. (12 × 550) = Rs. 6600

Solution 6

Radius of a conical tent, r = 24 m

Height of a conical tent, h = 10 m

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Radius of a conical heap, r = 4.5 m

Height of a conical tent, h = 3.5 m

Solution 12

Radius of a conical tent, r = 7 m

Area of canvas used in making conical tent = (551 - 1) m² = 550 m²

⇒ Curved surface area of a conical tent = 550 m²

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Curved surface area of the tent = Area of the cloth = 165 m²

Chapter 15 - Volume and Surface Area of Solids Ex. 15A

Solution 1(iv)

Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.

_{Volume of cuboid= l x b x h}

_{= (24 x 0.25 x 6) m³.}

_{= 36 m³.}

_{Lateral surface area= 2(l + b) x h}

_{= [2(24 +0.25) x 6] m²}

_{= (2 x 24.25 x 6) m²}

_{= 291 m².}

_{Total surface area =2(lb+ bh + lh)}

_{=2(24 x 0.25+0.25x 6 +24 x 6) m²}

_{= 2(6+1.5+144) m²}

_{= (2 x151.5) m²=303 m².}

Solution 1(iii)

Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m

_{Volume of a cuboid = l x b x h}

_{= (15 x 6 x 0.5) m³=45 m³.}

_{Lateral surface area = 2(l + b) x h}

_{= [2(15 + 6) x 0.5] m²}

_{= (2 x 21x0.5) m²=21 m²}

_{Total surface area =2(lb+ bh + lh)}

_{= 2(15 x 6 +6 x 0.5+ 15 x 0.5) m²}

_{= 2(90+3+7.5) m²}

_{= (2 x 100.5) m²}

_{=201 m²}

Solution 1(ii)

Length 26 m, breadth =14 m and height =6.5 m

Volume of a cuboid= l x b x h

= (26 x 14 x 6.5) m³

= 2366 m³

Lateral surface area of a cuboid =2 (l + b) x h

= [2(26+14) x 6.5] m²

= (2 x 40 x 6.5) m²

= 520 m²

_{Total surface area= 2(lb+ bh + lh)}

_{= 2(26 x 14+14 x6.5 +26 x6.5)}

_{= 2 (364+91+169) m²}

_{= (2 x 624) m2= 1248 m².}

Solution 1(i)

length =12cm, breadth = 8 cm and height = 4.5 cm

Volume of cuboid = l x b x h

= (12 x 8 x 4.5) cm³= 432 cm³

Lateral surface area of a cuboid = 2(l + b) x h

= [2(12 + 8) x 4.5] cm²

= (2 x 20 x 4.5) cm² = 180 cm²

Total surface area cuboid = 2(lb +b h+ l h)

= 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm²

= 2(96 +36 +54) cm²

= (2 x186) cm²

= 372 cm²

Solution 2

For a matchbox,

Length = 4 cm

Breadth = 2.5 cm

Height = 1.5 cm

Volume of one matchbox = Volume of cuboid

= Length × Breadth × Height

= (4 × 2.5 × 1.5) cm³

= 15 cm³

Hence, volume of 12 such matchboxes = 12 × 15 = 180 cm³

Solution 3

For a cuboidal water tank,

Length = 6 m

Breadth = 5 m

Height = 4.5 m

Now,

Volume of a cuboidal water tank = Length × Breadth × Height

= (6 × 5 × 4.5) m³

= 135 m³

= 135 × 1000 litres

= 135000 litres

Thus, a tank can hold 135000 litres of water.

Solution 4

For a cuboidal water tank,

Length = 10 m

Breadth = 2.5 m

Volume = 50000 litres = 50 m³

Now,

Volume of a cuboidal tank = Length × Breadth × Height

⇒ 50 = 10 × 2.5 × Height

⇒ Height = 2 m = Depth

Thus, the depth of a tank is 2 m.

Solution 5

For a godown,

Length = 40 m

Breadth = 25 m

Height = 15 m

Volume of a godown = Length × Breadth × Height

= (40 × 25 × 15) m³

For each wooden crate,

Length = 1.5 m

Breadth = 1.25 m

Height = 0.5 m

Volume of each wooden crate = Length × Breadth × Height

= (1.5 × 1.25 × 0.5) m³

Solution 6

Solution 7

Solution 8

_{Length of Cistern = 8 m}

_{Breadth of Cistern = 6 m}

_{And Height (depth) of Cistern =2.5 m}

_{Capacity of the Cistern = Volume of cistern}

_{Volume of Cistern = (l x b x h)}

_{= (8 x 6 x2.5) m³}

_{=120 m³}

_{Area of the iron sheet required = Total surface area of the cistem.}

_{Total surface area = 2(lb +bh +lh)}

_{= 2(8 x 6 + 6x2.5+ 2.5x8) m²}

_{= 2(48 + 15 + 20) m²}

_{= (2 x 83) m²=166 m²}

Solution 9

Area of four walls of the room = 2(length + breadth) × Height

= [2(9 + 8) × 6.5] m²

= (34 × 6.5) m²

= 221 m²

Area of one door = Length × Breadth = (2 × 1.5) m² = 3 m²

Area of two windows = 2 × (Length × Breadth)

= [2 × (1.5 × 1)] m²

= (2 × 1.5) m²

= 3 m²

Area to be whitewashed

= Area of four walls of the room - Area of one door - Area of two windows

= (221 - 3 - 3) m²

= 215 m²

Cost of whitewashing = Rs. 25 per square metre

⇒ Cost of whitewashing 215 m² = Rs. (25 × 215) = Rs. 5375

Solution 10

Solution 11

External length of the box = 36 cm

External breadth of the box = 25 cm

External height of the box = 16.5 cm

∴ External volume of the box = (36 × 25 × 16.5) cm³ = 14850 cm³

Internal length of the box = [36 - (1.5 × 2)] cm = 33 cm

Internal breadth of the box = [25 - (1.5 × 2)] cm = 22 cm

Internal height of the box = (16.5 - 1.5) cm = 15 cm

∴ Internal volume of the box = (33 × 22 × 15) cm³ = 10890 cm³

Thus, volume of iron used in the box

= External volume of the box - Internal volume of the box

= (14850 - 10890) cm³

= 3960 cm³

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Let the edge of the cube = 'a' cm

Then, surface area of cube = 6a² cm²

Solution 26

Solution 27

Solution 28

Volume of a cuboid = (9 × 8 × 2) m³ = 144 m³

Volume of each cube of edge 2 m = (2 m)³ = 8 m³

Chapter 15 - Volume and Surface Area of Solids Ex. 15B

Solution 1

Solution 2

Radius (r) of cylindrical bowl =

Height (h) up to which the bowl is filled with soup = 4 cm

Volume of soup in 1 bowl = pr²h = 154 cm³

Hence, volume of soup in 250 bowls = (250 × 154) cm³= 38500 cm³= 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.

Solution 3

Radius (r) of pillar = 20 cm = m

Height (h) of pillar = 10 m

Solution 4

For a tin can of rectangular base,

Length = 5 cm

Breadth = 4 cm

Height = 15 cm

∴ Volume of a tin can = Length × Breadth × Height

= (5 × 4 × 15) cm³

= 300 cm³

For a cylinder with circular base,

Diameter = 7 ⇒ Radius = r = cm

Height = h = 10 cm

⇒ Volume of plastic cylinder is greater than volume of a tin can.

Difference in volume = (385 - 300) = 85 cm³

Thus, a plastic cylinder has more capacity that a tin can by 85 cm³.

Solution 5

Radius (r) of 1 pillar =

Height (h) of 1 pillar = 4 m

Solution 6

Curved surface area of a cylinder = 4.4 m²

Radius (r) of a cylinder = 0.7 m

Solution 7

Lateral surface area of a cylinder = 94.2 cm²

Height (h) of a cylinder = 5 cm

Solution 8

Volume of a cylinder = 15.4 litres = 15400 cm³

Height (h) of a cylinder = 1 m = 100 cm

Solution 9

Internal diameter of a cylinder = 24 cm

⇒ Internal radius of a cylinder, r = 12 cm

External diameter of a cylinder = 28 cm

⇒ External radius of a cylinder, R = 14 cm

Length of the pipe, i.e height, h = 35 cm

Solution 10

Diameter of a cylindrical pipe = 5 cm

⇒ Radius (r) of a cylindrical pipe = 2.5 cm

Height (h) of a cylindrical pipe = 28 m = 2800 cm

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Diameter of a cylinder = 140 cm

⇒ Radius, r = 70 cm

Height (h) of a cylinder = 1 m = 100 cm

Solution 24

Radius (r) of cylindrical vessel = 15 cm

Height (h) of cylindrical vessel = 32 m

Radius of small cylindrical glass = 3 cm

Height of a small cylindrical glass = 8 cm

Solution 25

Radius of the well = 5 m

Depth of the well = 8.4 m

Width of the embankment = 7.5 m

External radius of the embankment, R = (5 + 7.5) m = 12.5 m

Internal radius of the embankment, r = 5 m

Area of the embankment = π (R² - r²)

Volume of the embankment = Volume of the earth dug out = 660 m²

Solution 26

Speed of water = 30 cm/sec

∴ Volume of water that flows out of the pipe in one second

= Area of cross-section × Length of water flown in one second

= (5 × 30) cm³

= 150 cm³

Hence, volume of water that flows out of the pipe in 1 minute

= (150 × 60) cm³

= 9000 cm³

= 9 litres

Solution 27

Suppose the tank is filled in x minutes. Then,

Volume of the water that flows out through the pipe in x minutes

= Volume of the tank

Hence, the tank will be filled in 28 minutes.

Solution 28

Let the rise in the level of water = h cm

Then,

Volume of the cylinder of height h and base radius 28 cm

= Volume of rectangular iron solid

Thus, the rise in the level of water is 4 cm.

Solution 29

Radius, r = 1.5 m

Height, h = 280 m

Solution 30

Let the length of the wire = 'h' metres

Then,

Volume of the wire × 8.4 g = (13.2 × 1000) g

Thus, the length of the wire is 125 m.

Solution 31

Solution 32

Let R cm and r cm be the outer and inner radii of the cylindrical tube.

We have, length of tube = h = 14 cm

Now,

Outside surface area - Inner surface area = 88 cm²

⇒ 2πRh - 2πrh = 88

⇒ 2π(R - r)h = 88

It is given that the volume of the tube = 176 cm³

⇒ External volume - Internal volume = 176 cm³

⇒ πR²h - πr²h = 176

⇒ π (R² - r²)h = 176

Adding (i) and (ii), we get

2R = 5

⇒ R = 2.5 cm

⇒ 2.5 - r = 1

⇒ r = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm respectively.

Solution 33

When the sheet is folded along its length, it forms a cylinder of height, h₁ = 18 cm and perimeter of base equal to 30 cm.

Let r₁ be the radius of the base and V₁ be is volume.

Then,

Again, when the sheet is folded along its breadth, it forms a cylinder of height, h₂ = 30 cm and perimeter of base equal to 18 cm.

Let r₂ be the radius of the base and V₂ be is volume.

Then,

Chapter 15 - Volume and Surface Area of Solids Ex. 15D

Solution 1(iii)

Solution 1(ii)

Solution 1(i)

Solution 2

Solution 3

Solution 4

Surface area of sphere = 154 cm²

⇒ 4πr² = 154

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

$We space have comma straight r equals Inner space radius space of space the space bowl equals 4 space cm straight R equals outer space radius space of space the space bowl equals left parenthesis 4 plus 0.5 right parenthesis space cm equals 4.5 space cm Volume space of space the space inner space hemisphere equals 2 over 3 πr cubed equals 2 over 3 cross times 22 over 7 cross times 4 cross times 4 space cm cubed Volume space of space the space outer space hemisphere equals 2 over 3 πR cubed equals 2 over 3 cross times 22 over 7 cross times 4.5 cross times 4.5 space cm cubed therefore space Volume space of space steel space used space space space space space equals Volume space of space the space outer space hemisphere minus Volume space of space the space inner space hemisphere space space space space space equals 2 over 3 cross times 22 over 7 cross times 4.5 cross times 4.5 minus 2 over 3 cross times 22 over 7 cross times 4 cross times 4 space space space space space equals 2 over 3 cross times 22 over 7 cross times open square brackets open parentheses 4.5 close parentheses squared minus open parentheses 4 close parentheses squared close square brackets space space space space space equals 44 over 21 cross times open parentheses 91.125 minus 64 close parentheses space space space space space equals 44 over 21 cross times 27.125 space space space space space equals 56.83 space space cm cubed$