RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids

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RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids

Exercise 19A

Question 1:

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

Question 2:

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)

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Question 3:

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)

Question 4:

Radius of wooden cylinder = 4.2 cm
Height of wooden cylinder = 12 cm
Lateral surface area

Radius of hemisphere = 4.2 cm
Surface area of two hemispheres

Total surface area = (100.8 + 70.56) π cm2
= 538.56 cm2
= 171.36 π
= 171.36 × [latex]frac { 22 }{ 7 }  [/latex]  cm2
= 538.56 cm2
Further, volume of cylinder = πr2h = 4.2 × 4.2 × 12 π cm2
= 211.68 π cm2
Volume of two hemispheres = 2 × [latex]frac { 2 }{ 3 }  [/latex] πr3 cu.units
= [latex]frac { 4 }{ 3 }  [/latex] π  × 4.2 × 4.2 × 4.2
= 98.784 cm3
Volume of wood left = (211.68 – 98.784) π
= 112.896 π cm3
= 112.896 × [latex]frac { 22 }{ 7 }  [/latex]  cm3
= 354.816 cm3

Question 5:
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)

Question 6:

Height of cone = h = 24 cm
Its radius = 7 cm

Total surface area of toy

Question 7:

Height of cylindrical container h1 = 15 cm
Diameter of cylindrical container = 12 cm
Volume of container =
Height of cone r2 = 12 cm
Diameter = 6 cm
Radius of r2 = 3 cm

Radius of hemisphere = 3 cm
Volume of hemisphere =
Volume of cone + volume of hemisphere
= 36π + 18π = 54π
Number of cones

Number of cones that can be filled = 10

Question 8:

Diameter of cylindrical gulabjamun = 2.8 cm
Its radius = 1.4 cm
Total height of gulabjamun = AC + CD + DB = 5 cm
1.4 + CD + 1.4 = 5
2.8 + CD = 5
CD = 2.2 cm
Height of cylindrical part h = 2.2 cm
Volume of 1 gulabjamun = Volume of cylindrical part + Volume of two hemispherical parts

Volume of 45 gulabjamuns = 45 × 25.07 cm3
Quantity of syrup = 30% of volume of gulabjamuns
= 0.3 × 45 × 25.07 = 338.46 cm3

Question 9:
Diameter = 7cm, radius = = 3.5 cm
Height of cone = 14.5 cm – 3.5 cm = 11 cm

Total surface area of toy =

Question 10:
Diameter of cylinder = 24 m
Radius of cylinder = [latex]frac { 24 }{ 2 }  [/latex] = 12 cm
Height of the cylinder = 11 m
Height of cone = (16 – 11) cm = 5 cm
Slant height of the cone l =
Area of canvas required = (curved surface area of the cylindrical part) + (curved surface area of the conical part)

Question 11:
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

Question 12:

Height of cylinder = 6.5 cm
Height of cone = h2 = (12.8-6.5) cm = 6.3 cm
Radius of cylinder = radius of cone
= radius of hemisphere
= [latex]frac { 7 }{ 2 }  [/latex] cm
Volume of solid = Volume of cylinder + Volume of cone + Volume of hemisphere

Question 13:

Radius of each hemispherical end = [latex]frac { 28 }{ 2 }  [/latex] = 14 cm
Height of each hemispherical part = Its Radius
Height of cylindrical part = (98 – 2 × 14) = 70 cm
Area of surface to be polished = 2(curved surface area of hemisphere) + (curved surface area of cylinder)

Cost of polishing the surface of the solid
= Rs. (0.15 × 8624)
= Rs. 1293. 60

Question 14:

Radius of cylinder r1 = 5 cm
And height of cylinder h1 = 9.8 cm
Radius of cone r = 2.1 cm
And height of cone h2 = 4 cm
Volume of water left in tub = (volume of cylindrical tub – volume of solid)

Question 15:
(i) Radius of cylinder = 6 cm
Height of cylinder = 8 cm

Volume of cylinder

Volume of cone removed

(ii) Surface area of cylinder = 2π = 2π × 6 × 8 cm2 = 96 π cm2

Question 16:

Diameter of spherical part of vessel = 21 cm

Question 17:

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

Question 18:

Diameter of cylinder = 5 cm
Radius = 2.5 cm
Height of cylinder = 10 cm
Volume of cylinder = πr2h  cu.units = 3.14 × 2.5 × 2.5 × 10  cm= 196.25 cm3
Apparent capacity of glass = 196.25
Radius of hemisphere = 2.5 cm
Volume of hemisphere

Actual capacity of glass = ( 196.25 – 32.608 ) cm3 = 163.54 cm3

Exercise 19B

Question 1:

Radius of the cone = 12 cm and its height = 24 cm
Volume of cone = [latex]frac { 1 }{ 3 }  [/latex] πr3 h = (frac { 1 }{ 3 } times 12times 12times 24) π cm3  = (48 × 24 )π cm3

Question 2:
Internal radius = 3 cm and external radius = 5 cm

Hence, height of the cone = 4 cm

Question 3:
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

Question 4:
Radius of the sphere = [latex]frac { 21 }{ 2 }  [/latex] cm

Let the number of cones formed be n, then

Hence, number of cones formed = 504

Question 5:
Radius of the cannon ball = 14 cm
Volume of cannon ball =
Radius of the cone = [latex]frac { 35 }{ 2 }  [/latex] cm
Let the height of cone be h cm
Volume of cone =

Hence, height of the cone = 35.84 cm

Question 6:
Let the radius of the third ball be r cm, then,
Volume of third ball = Volume of spherical ball – volume of 2 small balls

Question 7:
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 = πr2h = ( πr2 × 37 )

Hence diameter of the base of the cylinder = 12 cm

Question 8:
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

Question 9:
Diameter of sphere = 21 cm
Hence, radius of sphere = [latex]frac { 19 }{ 2 }  [/latex] cm
Volume of sphere = [latex]frac { 4 }{ 3 }  [/latex] πr3 = [latex](frac { 4 }{ 3 } times frac { 22 }{ 7 } times frac { 21 }{ 2 } times frac { 21 }{ 2 } times frac { 21 }{ 2 } )   [/latex]
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.

Question 10:
Volume of sphere (when r = 1 cm) = [latex]frac { 4 }{ 3 }  [/latex] πr3 = (frac { 4 }{ 3 } times 1times 1times 1) π cm3
Volume of sphere (when r = 8 cm) = [latex]frac { 4 }{ 3 }  [/latex] πr3 = (frac { 4 }{ 3 } times 8times 8times 8) π cm3
Let the number of balls = n

Question 11:
Radius of marbles = [latex]frac { Diameter }{ 2 } =frac { 1.4 }{ 2 } cm  [/latex]

Let the number of marbles be n
∴ n × volume of marble = volume of rising water in beaker

Question 12:
Radius of sphere = 3 cm
Volume of sphere = [latex]frac { 4 }{ 3 }  [/latex] πr3 = (frac { 4 }{ 3 } times 3times 3times 3) π cm3  = 36π cm3
Radius of small sphere = [latex]frac { 0.6 }{ 2 }  [/latex] cm = 0.3 cm
Volume of small sphere = (frac { 4 }{ 3 } times 0.3times 0.3times 0.3) π cm3

Let number of small balls be n

Hence, the number of small balls = 1000.

Question 13:
Diameter of sphere = 42 cm
Radius of sphere = [latex]frac { 42 }{ 2 }  [/latex] cm = 21 cm
Volume of sphere = [latex]frac { 4 }{ 3 }  [/latex] πr3 = (frac { 4 }{ 3 } times 21times 21times 21) π cm3
Diameter of cylindrical wire = 2.8 cm
Radius of cylindrical wire = [latex]frac { 2.8 }{ 2 }  [/latex] cm = 1.4 cm
Volume of cylindrical wire =  πr2h = ( π × 1.4 × 1.4 × h ) cm3 = ( 1.96πh ) cm3
Volume of cylindrical wire = volume of sphere

Hence length of the wire 63 m.

Question 14:
Diameter of sphere = 6 cm
Radius of sphere = [latex]frac { 6 }{ 2 }  [/latex] cm = 3 cm
Volume of sphere = [latex]frac { 4 }{ 3 }  [/latex] πr3 = (frac { 4 }{ 3 } times 3times 3times 3) π cm3  = 36π cm3
Radius of wire = [latex]frac { 2 }{ 2 }  [/latex] mm = 1 mm = 0.1 cm
Volume of wire = πr2l = ( π × 0.1 × 0.1 × l ) cm2 = ( 0.01 πl ) cm2
36π = 0.01 π l
∴ [latex]l=frac { 36 }{ 0.01 } =3600   [/latex] cm
Length of wire = [latex]frac { 3600 }{ 100 }  [/latex] m = 36 m

Question 15:
Diameter of sphere = 18 cm
Radius of copper sphere = [latex]frac { 3600 }{ 100 }  [/latex] m = 36 m

Length of wire = 108 m = 10800 cm
Let the radius of wire be r cm
= πr2l cm3 = ( πr2 × 10800 ) cm3
But the volume of wire = Volume of sphere

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

Question 16:
The radii of three metallic spheres are 3 cm, 4 cm and 5 cm respectively.
Sum of their volumes

Let r be the radius of sphere whose volume is equal to the total volume of three spheres.

Exercise 19C

Question 1:
Here h = 42 cm, R = 16 cm, and r = 11 cm
Capacity =

Question 2:
Here R = 33 cm, r = 27 cm and l = 10 cm

Capacity of the frustum

Total surface area
=

Question 3:
Height = 15 cm, R = [latex]frac { 56 }{ 2 }  [/latex] cm = 28 cm and r = [latex]frac { 42 }{ 2 }  [/latex] cm = 21 cm
Capacity of the bucket =

Quantity of water in bucket = 28.49 litres

Question 4:
R = 20 cm, r = 8 cm and h = 16 cm

Total surface area of container = πl (R+r) + πr2

Cost of metal sheet used = Rs. [latex](1959.36times frac { 15 }{ 100 } )   [/latex] = Rs. 293.90

Question 5:
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 = [latex]( 1711.3times frac { 10 }{ 100 } )   [/latex] = Rs. 171.13

Question 6:

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)

Question 7:

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 l1 of frustum

Radius of the cone = EB = 7 m
Slant height l2 of cone = 12 m
Surface area of canvas required

Question 8:

In the given figure, we have
∠COD = 30°, OC = 10 cm, OE = 20 cm
Let CD = r cm and EB = R cm

Also, CE = 10 cm
Thus, ABDF is the frustum of a cone in which

Volume of wire of radius r and length l

Volume of wire = Volume of frustum

Length of the wire is 7964.44 m

Question 9:

Radii of upper and lower end of frustum are r = 8 cm, R = 32 cm
Height of frustum h = 18 cm

Cost of milk at Rs 20 per litre = Rs. 25.344 × 20 = Rs. 506. 88

Complete RS Aggarwal Solutions Class 10

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