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

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 19
Chapter NameVolume and Surface Areas of Solids
Exercise19 A
CategoryRS Aggarwal Solutions

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

Question 1:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 1.1
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
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 1.2
Curved surface area of tent
= (curved area of cylinder + curved surface area of cone)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 1.3
Hence, the curved surface area of the tent = 1034
Cost of canvas = Rs.(1034 × 80) = Rs. 82720

Question 2:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 2.1
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)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 2.2

Question 3:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 3.1
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)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 3.2

Question 4:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 4.1
Radius of wooden cylinder = 4.2 cm
Height of wooden cylinder = 12 cm
Lateral surface area
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 4.2
Radius of hemisphere = 4.2 cm
Surface area of two hemispheres
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 4.3
Total surface area = (100.8 + 70.56) π cm2
= 538.56 cm2
= 171.36 π
= 171.36 × 227  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 × 23 πr3 cu.units
= 43 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 × 227  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)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 5.1

Credit: G.S. Classes

Question 6:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 6.1
Height of cone = h = 24 cm
Its radius = 7 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 6.2
Total surface area of toy
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 6.3

Question 7:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 7.1
Height of cylindrical container h1 = 15 cm
Diameter of cylindrical container = 12 cm
Volume of container = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Q7
Height of cone r2 = 12 cm
Diameter = 6 cm
Radius of r2 = 3 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 7.2
Radius of hemisphere = 3 cm
Volume of hemisphere = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Q7.1
Volume of cone + volume of hemisphere
= 36π + 18π = 54π
Number of cones
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 7.3
Number of cones that can be filled = 10

Question 8:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 8.1
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
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 8.2
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
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 9.1
Total surface area of toy = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Q9
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 9.2

Question 10:
Diameter of cylinder = 24 m
Radius of cylinder = 242 = 12 cm
Height of the cylinder = 11 m
Height of cone = (16 – 11) cm = 5 cm
Slant height of the cone l = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 10.1
Area of canvas required = (curved surface area of the cylindrical part) + (curved surface area of the conical part)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 10.2

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
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 11.1

Question 12:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 12.1
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
= 72 cm
Volume of solid = Volume of cylinder + Volume of cone + Volume of hemisphere
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 12.2

Question 13:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 13.1
Radius of each hemispherical end = 282 = 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)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 13.2
Cost of polishing the surface of the solid
= Rs. (0.15 × 8624)
= Rs. 1293. 60

Question 14:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 14.1
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)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 14.2

Question 15:
(i) Radius of cylinder = 6 cm
Height of cylinder = 8 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 15.1
Volume of cylinder
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 15.2
Volume of cone removed
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 15.3
(ii) Surface area of cylinder = 2π = 2π × 6 × 8 cm2 = 96 π cm2
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 15.4

Question 16:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 16.1
Diameter of spherical part of vessel = 21 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 16.2

Question 17:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 17.1
Height of cylindrical tank = 2.5 m
Its diameter = 12 m, Radius = 6 m
Volume of tank = RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 17.2
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
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 17.3

Question 18:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 18.1
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
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9a 18.2
Actual capacity of glass = ( 196.25 – 32.608 ) cm3 = 163.54 cm3

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