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

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 20C.

Other Exercises

Tick the correct answer in each of the following:

Question 1.
Solution:
Answer = (b)
Length (l) = 12 cm
Breadth (b) = 9cm
height (h) = 8 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 1.1

Question 2.
Solution:
Total surface area of cube = 150 cm2
Side = \( \sqrt { \frac { 150 }{ 6 } } \)
= √25
= 5 cm
Volume = (side)3
= (5)3
= 125 cm3 (b)

Question 3.
Solution:
Volume of cube = 343 cm2
Side = \( \sqrt [ 3 ]{ 343 } =\sqrt [ 3 ]{ 7\times 7\times 7 } \)
= 7 cm
Total surface area = 6 (side)2
= 6 x (7)2
= 6 x 49 cm2
= 294 cm2 (c)

Question 4.
Solution:
Rate of painting = 10 paise per cm2
Total cost = Rs. 264.60
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 4.1

Question 5.
Solution:
Answer = (c)
Length of wall (l) = 8m = 800 cm
Breadth (b) = 22.5 cm
Height (h) = 6 m
= 600 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 5.1

Question 6.
Solution:
Answer = (c)
Edge of cube = 10 cm
Volume = a3 = (10)3 = 1000 cm3
Edge of box = 1 m = 100 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 6.1

Question 7.
Solution:
Answer = (a)
Ratio in sides of a cuboid = 1 : 2 : 3
Surface area = 88 cm2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 7.1

Question 8.
Solution:
Ratio in the two volumes = 1 : 27
Let volume of first volume = x3
and volume of second volume = 27x3
Side of first cube = x
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 8.1

Question 9.
Solution:
Surface area of a brick of measure 10 cm x 4 cm x 3 cm
= 2 (l x b + b x h + h x l)
= 2 [10 x 4 + 4 x 3 + 3 x 10] cm2
= 2 [40 + 12 + 30]
= 82 x 2
= 164 cm2 (c)

Question 10.
Solution:
Length of beam (l) = 9 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 10.1

Question 11.
Solution:
Water in rectangular reservoir = 42000
Volume = \(\\ \frac { 42000 }{ 1000 } \) = 42 m3
Length (l) = 6 m
Breadth (b) = 3.5 m
Depth = \(\\ \frac { volume }{ l\times b } \)
= \(\\ \frac { 42 }{ 6\times 3.5 } \)
= 2 m (c)

Question 12.
Solution:
Dimensions of a room are 10 m, 8 m, 3.3 m
Volume of air in it = lbh
= 10 x 8 x 3.3 = 264 m3
Air required for one man = 3 m3
No. of men = \(\\ \frac { 264 }{ 3 } \)
= 88 (b)

Question 13.
Solution:
Length of water tank (l) = 3 m
Width (b) = 2 m
and height (h) = 5 m
Volume = lbh = 3 x 2 x 5 = 30 m3
Water in it = 30 x 1000
= 30000 (a)

Question 14.
Solution:
Size of box = 25 cm, 15 cm, 8 cm
Surface area = (lb + bh + hl)
= 2 ( 25 x 15 + 15 x 8 + 8 x 25) cm2
= 2 (375 + 120 + 200) cm2
= 2(695)
= 1390 cm(b)

Question 15.
Solution:
Diagonal of cube = 4√3
Side = \( \frac { 4\sqrt { 3 } }{ \sqrt { 3 } } \)
= 4 cm
Volume = a3 = (4)3
= 64 cm3 (d)

Question 16.
Solution:
Diagonal of cube = 9√3 cm
Side = \( \frac { 9\sqrt { 3 } }{ \sqrt { 3 } } \)
= 9 cm
Surface area = 6a2
= 6 (9)2 = 6 x 81 cm2
= 486 cm2 (b)

Question 17.
Solution:
Let side of cube in first case = a
Then volume = a3
If side of cube is doubled, then side = 2a
Volume (2a)3 = 8a3
Becomes 8 times (d)

Question 18.
Solution:
Let side of cube in first case = a
Then surface area = 6a2
and side of second cube = 2a
Surface area = 6 (2a)2 = 6 x 4a2 = 24a2
Ratio = \(\frac { { 24a }^{ 2 } }{ { 6a }^{ 2 } } \) = 4
Becomes 4 times (b)

Question 19.
Solution:
Sides (edges) of 3 cubes are 6 cm, 8 cm, and 10 cm respectively
Volume of first cube = (6)3 = 216 cm3
Volume of second cube = (8)3 = 512 cm3
and volume of third cube
= (10)3 = 1000 cm3
Sum of volumes of 3 cubes = 216 + 512 + 1000
= 1728 cm3
Volume of new single cube = 1728 cm3
Edge = \(\sqrt [ 3 ]{ 1728 } \)
\(\sqrt [ 3 ]{ { \left( 12 \right) }^{ 3 } } \)
= 12 cm (a)

Question 20.
Solution:
Each edge of 5 cubes = 5 cm
Placing than adjacent to each other
Length of new cuboid (l)
= 5 x 5 = 25 cm
Breadth (b) = 5 cm
and height (h) = 5 cm
Volume of new cuboid = lbh
= 25 x 5 x 5 cm3
= 625 cm3 (d)

Question 21.
Solution:
Diameter of circular well = 2n
Radius = \(\\ \frac { 2 }{ 2 } \) = 1 m
Depth(h) = 14 m
Volume of earth dug out = πr2h
= \(\\ \frac { 22 }{ 7 } \) x 1 x 1 x 14
= 44 m (d)

Question 22.
Solution:
Capacity of cylindrical tank = 1848 m3
Diameter = 14 m
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 22.1

Question 23.
Solution:
Radius of a cylinder (r) = 20 cm
and height (h) = 60 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 23.1

Question 24.
Solution:
Radius of each coin (r) = 0.75 cm
and thickness (h) = 0.2 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 24.1

Question 25.
Solution:
Volume of silver = 66 cm3
Diameter of wire = 1 mm = \(\\ \frac { 1 }{ 10 } \)
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 25.1

Question 26.
Solution:
Diameter of cylinder = 10 cm
Radius (r) = \(\\ \frac { 10 }{ 2 } \) = 5 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 26.1

Question 27.
Solution:
Diameter of cylinder = 7 cm
Radius (r) = \(\\ \frac { 7 }{ 2 } \) cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 27.1

Question 28.
Solution:
Curved surface area of a cylinder = 264 cm3
Height (h) = 14 cm
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 28.1

Question 29.
Solution:
Diameter of cylinder = 14 cm
Radius (r) = 7 cm
Curved surface area = 220 cm2
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 29.1

Question 30.
Solution:
Ratio in radii of two cylinder = 2 : 3
and ratio in their height = 5 : 3
Let radii of two cylinder = 2x and 3x
and corresponding heights = 5y, 3y
RS Aggarwal Class 8 Solutions Chapter 20 Volume and Surface Area of Solids Ex 20C 30.1

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