RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B.

Other Exercises

Question 1.
Solution:
(i) Radius of the base of the cylinder (r) = 7 cm.
Height (h) = 50 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 1.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 1.2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 1.3

Question 2.
Solution:
Radius of cylindrical tank (r) = 1.5 m
and height (h) = 10.5 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 2.1
= 74.25 x 1000l
= 74250 l

Question 3.
Solution:
Radius of the base of pole (r)
= 10 dm
= \(\\ \frac { 10 }{ 100 } \) m
= \(\\ \frac { 1 }{ 10 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 3.1

Question 4.
Solution:
Volume of cylinder = 1.54 m³
= 1540000 cm³
Diameter of its base = 140 cm
Radius (r) = 70 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 4.1

Question 5.
Solution:
Volume of cylindrical rod = 3850 cm³
Length of rod (h) = 1 m = 100 cm
Let radius of the base of the rod = r
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 5.1

Question 6.
Solution:
Diameter of closed cylinder = 14 m
Radius = \(\\ \frac { 14 }{ 2 } \)
= 7 m
Height = 5
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 6.1

Question 7.
Solution:
Circumference of the base of cylinder = 88 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 7.1

Question 8.
Solution:
Lateral surface of cylinder = 220 m²
Height (h) = 14 m
Let radius of cylinder = r
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 8.1

Question 9.
Solution:
Volume of cylinder = 1232 cm³
height (h) = 8cm
Let r be the radius, then
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 9.1

Question 10.
Solution:
Ratio in radius and height of a cylinder = 7 : 2
Let radius = 7x
then height = 2x
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 10.1

Question 11.
Solution:
Curved surface area = 4400 cm²
circumference of base = 110 cm
Let r be the radius
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 11.1

Question 12.
Solution:
In first case,
Side of square base (a) = 5 cm.
and height (h) = 14 cm.
Volume = 5 x 5 x 14 = 350 cm³
In second case,
Radius of the circular base (r) = 3.5 cm.
Height (h) = 12 cm.
Volume = πr²h
= \(\\ \frac { 22 }{ 7 } \) x 3.5 x 3.5 x 12 cm³
= 462 cm²
Hence second type of circular plastic can has greater capacity.
Difference = 462 – 350
= 112 cm³

Question 13.
Solution:
Diameter of a cylindrical pillar = 48 cm.
Radius (r) = \(\\ \frac { 48 }{ 2 } \) = 24 cm.
\(\\ \frac { 24 }{ 100 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 13.1

Question 14.
Solution:
Length of rectangular vessel (l) = 22 cm.
Breadth (A) = 16 cm.
and height (A) = 14 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 14.1

Question 15.
Solution:
Diameter of cylindrical metal = 1 cm.
Radius (r) = \(\\ \frac { 1 }{ 2 } \) cm.
Length. (A) = 11 cm.
Volume = πr²h
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 15.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 15.2

Question 16.
Solution:
Side of a solid cube = 2.2 cm
Volume = (side)³
= (2.2)³
= 10.648 cm³
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 16.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 16.2

Question 17.
Solution:
Diameter of a well = 7 m
Radius (r) = \(\\ \frac { 7 }{ 2 } \) m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 17.1

Question 18.
Solution:
Inner diameter of well = 14 m
Inner radius = \(\\ \frac { 14 }{ 2 } \) = 7 m
Depth (h) = 12 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 18.1
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 18.2

Question 19.
Solution:
No. of revolutions = 750
Diameter of road roller = 84 cm
Length (h) = 1 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 19.1

Question 20.
Solution:
Thickness of the metal = 1.5 cm.
External diameter = 12 cm.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 20.1

Question 21.
Solution:
Inner diameter of tube = 12 cm.
Inner radius (r) = \(\\ \frac { 12 }{ 2 } \) = 6 cm.
Thickness of metal = 1 m.
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20B 21.1

Hope given RS Aggarwal Solutions Class 8 Chapter 20 Volume and Surface Area of Solids Ex 20B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.