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

## Chapter 19 - Volume and Surface Areas of Solids MCQ

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.

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.

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.

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.

Correct option: (b)

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

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.

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.

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.

Correct option: (c)

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

Correct option: (c)

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

Correct option: (a)

Correct option: (a)

## Chapter 19 - Volume and Surface Areas of Solids FA

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

s

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

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)

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)

Height of cone = h = 24 cm

Its radius = 7 cm

Total surface area of toy

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

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)

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)

(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

Diameter of spherical part of vessel = 21 cm

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

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

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

Volume of cone =

Internal radius = 3 cm and external radius = 5 cm

Hence, height of the cone = 4 cm

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

Radius of the sphere=

Let the number of cones formed be n, then

Hence, number of cones formed = 504

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

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

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

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

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

Diameter of sphere = 21 cm

Hence, radius of sphere =

Volume of sphere = =

Volume of cube = a^{3} = (1 1 1)

Let number of cubes formed be n

Volume of sphere = n Volume of cube

Hence, number of cubes is 4851.

Volume of sphere (when r = 1 cm) = =

Volume of sphere (when r = 8 cm) = =

Let the number of balls = n

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.

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.

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

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

Let the number of marbles be n

n volume of marble = volume of rising water in beaker

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

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

Capacity of the frustum

Total surface area =

Height = 15 cm, R = and

Capacity of the bucket =

Quantity of water in bucket = 28.49 litres

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

Total surface area of container =

Cost of metal sheet used =

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 =

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)

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

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

### Other Chapters for CBSE Class 10 Maths

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 16- Areas of Circle, Sector and Segment Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- Probability### R S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects

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