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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 A Maths book pdf download. Now you will get step by step solution to each question.

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

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) 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 × 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) Question 6: Height of cone = h = 24 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
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 = 242 = 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
= 72 cm
Volume of solid = Volume of cylinder + Volume of cone + Volume of hemisphere Question 13: 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) 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
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

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