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

## Exercise Test Yourself

### Solution 14

### Solution 15

Note: Answer is incorrect in the textbook

## Volume and Surface Areas of Solids Exercise Ex. 17A

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 14

Radius = BO = r = 3.5 cm

Height of toy = AC = 15.5 cm

Height of cone = AO = h = AC - OC = 12 cm

### 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 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

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

The volume of remaining solid

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

### Solution 26

Diameter of spherical part of vessel = 21 cm

### Solution 27

Height of cylinder = 6.5 cm

Height of cone =

=

Volume of solid = Volume of cylinder + Volume of cone

+ Volume of hemisphere

### Solution 28

Total surface area of cube = 6(side)2

=6×212

=2646

Now the surface area of the remaining solid = Total surface area of the cube + curved surface area of the hemisphere - Area of the circular base of the hemisphere

=2646 + 693 - 346.5

=2992.5 cm2

### Solution 29 (i)

Note: In the question, cuboid is printed, but as the given data is insufficient, we are considering the shape as cube.

### Solution 33

Volume displaced by 1 person = 0.04m3

∴ Volume displaced by 500 persons = 0.04×500 = 20m3

Now,

Volume displaced = Area of tank × height of displacement

Also,

Area of tank = 80×50 m2

### Solution 34

Canal is shaped like a cuboid,

Hence

Volume of water = lbh

Given,

b = 6 m

h = 1.5 m

l = 10 km per hour

= 10000 m per 60 min.

= 5000 m per 30 min.

Hence, volume of water for 30 min.

= 6×1.5×5000

= 45000

Now,

Volume of water = Volume of Area irrigated

45000 = Area irrigated × Height of standing water

Height of standing water required = 8cm = 0.08m

∴ Area irrigated = 45000/0.08

= 562500 m2

### Solution 4 (ii)

Let the radius and height be 5x and 12x respectively

Hence

Volume of cone = 314cm3

Hence,

Height = 11cm

## Volume and Surface Areas of Solids Exercise Ex. 17B

### Solution 1

Solid cuboid

l = 9m

b = 8m

h = 2m

Volume = 9×8×2

Solid cube

S = 2m

Volume = s3 = 2×2×2

### Solution 4

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

Volume of cone =

### Solution 6

Hence, height of the cone = 4 cm

### 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

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 =

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

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 =

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

Volume of sphere =

Diameter of cylindrical wire = 2.8 cm

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

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 26

Canal is shaped like a cuboid,

Hence

Volume of water = lbh

Given,

b = 5.4 m

h = 1.8 m

l = 25 km per hour

= 25000 m per 60 min.

=  m per 40 min.

Hence, volume of water for 30 min.

Now,

Volume of water = Volume of Area irrigated

162000 = Area irrigated × Height of standing water

Height of standing water required = 10cm = 0.1m

∴ Area irrigated = 162000/0.1

= 1620000 m2

### 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 30

Let the number of marbles be n

n volume of marble = volume of rising water in beaker

### Solution 35

Cylindrical tank

d = 2 m

r = 1 m

h = 5 m

Volume of water for irrigation = Area of park × Standing water height

Recycling of water is a need of hour, since the reserves of fresh water are depleting very fast due to increase in population, manufacturing, production etc. We should keep the wastage of water minimum and recycle them as much as we can.

## Volume and Surface Areas of Solids Exercise Ex. 17C

(i)

(ii)

### Solution 6

= 22704 cm3

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

= 13970g

= 13.97kg

The total cost of the oil in the container = 13.97×40 = Rs. 558.80

### 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 22

We have,

cone ABC with radius 'r1' and height 'h'

cone AFG with radius 'r3' and height '3h'

Now,

∆ABC ~ ∆AFG,

## Volume and Surface Areas of Solids Exercise 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.

### Solution 24

Correct option: (c)

### Solution 68

Correct option: (a)

### Solution 69

Correct option: (a)