## 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.

Question 2.
Solution:
Radius of cylindrical tank (r) = 1.5 m
and height (h) = 10.5 m

= 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

Question 4.
Solution:
Volume of cylinder = 1.54 m³
= 1540000 cm³
Diameter of its base = 140 cm
Radius (r) = 70 cm

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

Question 6.
Solution:
Diameter of closed cylinder = 14 m
Radius = $$\\ \frac { 14 }{ 2 }$$
= 7 m
Height = 5

Question 7.
Solution:
Circumference of the base of cylinder = 88 cm.

Question 8.
Solution:
Lateral surface of cylinder = 220 m²
Height (h) = 14 m
Let radius of cylinder = r

Question 9.
Solution:
Volume of cylinder = 1232 cm³
height (h) = 8cm
Let r be the radius, then

Question 10.
Solution:
Ratio in radius and height of a cylinder = 7 : 2
Let radius = 7x
then height = 2x

Question 11.
Solution:
Curved surface area = 4400 cm²
circumference of base = 110 cm
Let r be the radius

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

Question 14.
Solution:
Length of rectangular vessel (l) = 22 cm.
Breadth (A) = 16 cm.
and height (A) = 14 cm.

Question 15.
Solution:
Diameter of cylindrical metal = 1 cm.
Radius (r) = $$\\ \frac { 1 }{ 2 }$$ cm.
Length. (A) = 11 cm.
Volume = πr²h

Question 16.
Solution:
Side of a solid cube = 2.2 cm
Volume = (side)³
= (2.2)³
= 10.648 cm³

Question 17.
Solution:
Diameter of a well = 7 m
Radius (r) = $$\\ \frac { 7 }{ 2 }$$ m

Question 18.
Solution:
Inner diameter of well = 14 m
Inner radius = $$\\ \frac { 14 }{ 2 }$$ = 7 m
Depth (h) = 12 m

Question 19.
Solution:
No. of revolutions = 750
Diameter of road roller = 84 cm
Length (h) = 1 m

Question 20.
Solution:
Thickness of the metal = 1.5 cm.
External diameter = 12 cm.

Question 21.
Solution:
Inner diameter of tube = 12 cm.
Inner radius (r) = $$\\ \frac { 12 }{ 2 }$$ = 6 cm.
Thickness of metal = 1 m.

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

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