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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 C 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 C Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids Ex 19 C Maths book pdf download. Now you will get step by step solution to each question.

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

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

Question 1:
Here h = 42 cm, R = 16 cm, and r = 11 cm
Capacity = rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19c-q1-1
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 1.1

Question 2:
Here R = 33 cm, r = 27 cm and l = 10 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 2.1
Capacity of the frustum
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 2.2
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 2.3
Total surface area
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 2.4
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 2.5

Question 3:
Height = 15 cm, R = 562 cm = 28 cm and r = 422 cm = 21 cm
Capacity of the bucket =
rs-aggarwal-class-10-solutions-volume-and-surface-areas-of-solids-19b-q3-1
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 3.1
Quantity of water in bucket = 28.49 litres

Question 4:
R = 20 cm, r = 8 cm and h = 16 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 4.1
Total surface area of container = πl (R+r) + πr2
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 4.2
Cost of metal sheet used = Rs. (1959.36×15100) = Rs. 293.90

Question 5:
R = 15 cm, r = 5 cm and h = 24 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 5.1
(i) Volume of bucket =
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 5.2
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 5.3
Cost of milk = Rs. (8.164 × 20) = Rs. 163.28
(ii) Total surface area of the bucket
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 5.4
Cost of sheet = (1711.3×10100) = Rs. 171.13

Question 6:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 6.1
R = 10cm, r = 3 m and h = 24 m
Let l be the slant height of the frustum, then
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 6.2
Quantity of canvas = (Lateral surface area of the frustum) + (lateral surface area of the cone)
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 6.3

Question 7:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 7.1
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
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 7.2
Radius of the cone = EB = 7 m
Slant height l2 of cone = 12 m
Surface area of canvas required
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 7.3

Question 8:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 8.1
In the given figure, we have
∠COD = 30°, OC = 10 cm, OE = 20 cm
Let CD = r cm and EB = R cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 8.2
Also, CE = 10 cm
Thus, ABDF is the frustum of a cone in which
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 8.3
Volume of wire of radius r and length l
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 8.4
Volume of wire = Volume of frustum
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 8.5
Length of the wire is 7964.44 m

Question 9:
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 9.1
Radii of upper and lower end of frustum are r = 8 cm, R = 32 cm
Height of frustum h = 18 cm
RS Aggarwal Solutions Class 10 Chapter 19 Volume and Surface Areas of Solids 9c 9.2
Cost of milk at Rs 20 per litre = Rs. 25.344 × 20 = Rs. 506. 88

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