R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 17 - Volume and Surface Areas of Solids

Page / Exercise

Chapter 19 - Volume and Surface Areas of Solids MCQ

Solution 1

Correct option: (a)

  

A cylindrical pencil sharpened at one edge is the combination of a cylinder and a cone. Observe the figure, the lower portion is a cylinder and the upper tapering portion is a cone.

Solution 2

Correct option: (b)

A shuttlecock used for playing badminton is the combination of a frustum of a cone and a hemisphere, the lower portion being the hemisphere and the portion above that being the frustum of the cone.

Solution 3

Correct option: (c)

A funnel is the combination of a cylinder and frustum of a cone. The lower portion is cylindrical and the upper portion is a frustum of a cone.

Solution 4

Correct option: (a)

A surahi is a combination of a sphere and a cylinder, the lower portion is the sphere and the upper portion is the cylinder.

Solution 5

Correct option: (b)

The shape of a glass (tumbler) is usually in the form of a frustum of a cone.

Solution 6

Correct option: (c)

The shape of a gill in the gilli-danda game is a combination of two cones and a cylinder. The cones at either ends with the cylinder in the middle.

Solution 7

Correct option: (a)

A plumbline (sahul) is the combination of a hemisphere and a cone, the hemisphere being on top and the lower portion being the cone.

Solution 8

Correct option: (d)

A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left over is called the frustum of a cone.

Solution 9

Correct option: (c)

During conversion of a solid from one shape to another, the volume of the new shape will remain altered.

Solution 10

Correct option: (c)

In a right circular cone, the cross section made by a plane parallel to the base is a circle.

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Solution 68

Correct option: (a)

  

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Correct option: (a)

  

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Chapter 19 - Volume and Surface Areas of Solids FA

Solution 1

Solution 2

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Solution 10

  

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Solution 20

  

Chapter 19 - Volume and Surface Areas of Solids Ex. 17A

Solution 1

Solution 2

Solution 3

Solution 4(i)

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

s  

Solution 10

Radius of the cylinder = 14 m

And its height = 3 m

Radius of cone = 14 m

And its height = 10.5 m

Let l be the slant height

Curved surface area of tent

                    = (curved area of cylinder + curved surface area of cone)

                  

Hence, the curved surface area of the tent = 1034

Cost of canvas = Rs.(1034 × 80) = Rs. 82720

Solution 11

For the cylindrical portion, we have radius = 52.5 m and height = 3 m

For the conical portion, we have radius = 52.5 m

And slant height = 53 m

Area of canvas = 2rh + rl = r(2h + l)

                                

Solution 12

Radius o f cylinder = 2.5 m

Height of cylinder = 21 m

Slant height of cone = 8 m

Radius of cone = 2.5 m

Total surface area of the rocket = (curved surface area of cone

                                                              + curved surface area of cylinder + area of base)

                                                          

                                        

Solution 13

  

  

  

Solution 14

Height of cone = h = 24 cm

Its radius = 7 cm

Total surface area of toy

                       

Solution 15

Solution 16

Solution 17

Radius of hemisphere = 10.5 cm

Height of cylinder = (14.5 10.5) cm = 4 cm

Radius of cylinder = 10.5 cm

Capacity = Volume of cylinder + Volume of hemisphere

                       

Solution 18

Solution 19

  

  

Solution 20

                    

Height of cylinder = 20 cm

And diameter = 7 cm and then radius = 3.5 cm

Total surface area of article

               = (lateral surface of cylinder with r = 3.5 cm and h = 20 cm)

                          

Solution 21

Radius of cylinder

And height of cylinder

Radius of cone r = 2.1 cm

And height of cone

Volume of water left in tub

              = (volume of cylindrical tub – volume of solid)

                        

Solution 22

(i)Radius of cylinder = 6 cm

Height of cylinder = 8 cm

 

Volume of cylinder

             

Volume of cone removed

             

 

(ii)Surface area of cylinder = 2 = 2× 6 × 8

Solution 23

Solution 24

Solution 25

Solution 26

Diameter of spherical part of vessel = 21 cm

Solution 27

 

Height of cylinder = 6.5 cm

Height of cone =

Radius of cylinder = radius of cone

                          = radius of hemisphere

                         =

Volume of solid = Volume of cylinder + Volume of cone

+ Volume of hemisphere

                      

Solution 28

Solution 29(ii)

Solution 30

Solution 31

Solution 32

Chapter 19 - Volume and Surface Areas of Solids Ex. 17B

Solution 1

Solution 2

Solution 3

Solution 4

Radius of the cone = 12 cm and its height = 24 cm

Volume of cone =

                     

Solution 5

Solution 6

Internal radius = 3 cm and external radius = 5 cm

Hence, height of the cone = 4 cm

Solution 7

Solution 8

Inner radius of the bowl = 15 cm

Volume of liquid in it =

Radius of each cylindrical bottle = 2.5 cm and its height = 6 cm

Volume of each cylindrical bottle

                                      

Required number of bottles =

                                      

   Hence, bottles required = 60

Solution 9

Radius of the sphere=

Let the number of cones formed be n, then

              

            Hence, number of cones formed = 504

Solution 10

Radius of the cannon ball = 14 cm

Volume of cannon ball =

Radius of the cone =

Let the height of cone be h cm

Volume of cone =

           

            Hence, height of the cone = 35.84 cm

Solution 11

Let the radius of the third ball be r cm, then,

           Volume of third ball = Volume of spherical ball volume of 2 small balls

          

Solution 12

External radius of shell = 12 cm and internal radius = 9 cm

Volume of lead in the shell =

Let the radius of the cylinder be r cm

Its height = 37 cm

Volume of cylinder =

Hence diameter of the base of the cylinder = 12 cm

Solution 13

Volume of hemisphere of radius 9 cm

                                       

Volume of circular cone (height = 72 cm)

                                      

Volume of cone = Volume of hemisphere

Hence radius of the base of the cone = 4.5 cm

Solution 14

Diameter of sphere = 21 cm

Hence, radius of sphere =

Volume of sphere =  =

Volume of cube = a3 = (1 1 1)

Let number of cubes formed be n

Volume of sphere = n Volume of cube

Hence, number of cubes is 4851.

Solution 15

Volume of sphere (when r = 1 cm) =  =

Volume of sphere (when r = 8 cm) =  =

Let the number of balls = n

Solution 16

Radius of sphere = 3 cm

Volume of sphere =

Radius of small sphere =

Volume of small sphere =

                                

Let number of small balls be n

Hence, the number of small balls = 1000.

Solution 17

Diameter of sphere = 42 cm

Radius of sphere =

Volume of sphere =

Diameter of cylindrical wire = 2.8 cm

Radius of cylindrical wire =

Volume of cylindrical wire =

                                  

Volume of cylindrical wire = volume of sphere

Hence length of the wire 63 m.

Solution 18

Diameter of sphere = 18 cm

Radius of copper sphere =

Length of wire = 108 m = 10800 cm

Let the radius of wire be r cm

                         

But the volume of wire = Volume of sphere

        

Hence the diameter = 2r = (0.3 2) cm = 0.6 cm

Solution 19

Solution 20

Solution 21

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Solution 26

Solution 27

Height of cylindrical tank = 2.5 m

Its diameter = 12 m, Radius = 6 m

Volume of tank =

Water is flowing at the rate of 3.6 km/ hr = 3600 m/hr

Diameter of pipe = 25 cm, radius = 0.125 m

Volume of water flowing per hour

                         


Solution 28

Solution 29

Solution 30

Let the number of marbles be n

n volume of marble = volume of rising water in beaker

Solution 31

Solution 32

Solution 33

Solution 34

  

Chapter 19 - Volume and Surface Areas of Solids Ex. 17C

Solution 1

Solution 2

Solution 3

  

Solution 4

Solution 5(i)

Solution 6

Here R = 33 cm, r = 27 cm and l = 10 cm

Capacity of the frustum

                                

Total surface area =

                       

Solution 7

Height = 15 cm, R = and 

Capacity of the bucket =

                              

Quantity of water in bucket = 28.49 litres

Solution 8

R = 20 cm, r = 8 cm and h = 16 cm

Total surface area of container =

                                         

Cost of metal sheet used =

Solution 9

R = 15 cm, r = 5 cm and h = 24 cm

 

(i)Volume of bucket =

          

Cost of milk = Rs. (8.164 20) = Rs. 163.28

 

(ii)Total surface area of the bucket

         

Cost of sheet =

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

           

R = 10cm, r = 3 m and h = 24 m

Let l be the slant height of the frustum, then

    

Quantity of canvas = (Lateral surface area of the frustum)

                                + (lateral surface area of the cone)

                          

Solution 15

ABCD is the frustum in which upper and lower radii are EB = 7 m and FD = 13 m

Height of frustum= 8 m

Slant height of frustum

            

Radius of the cone = EB = 7 m

Slant height of cone = 12 m

Surface area of canvas required

             

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Chapter 19 - Volume and Surface Areas of Solids Ex. 17D

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