Volume questions and answers are available in an easily understandable format along with required formulas. Students can practise questions on finding the volume of various solids provided here, and verify their solutions with the available answers. This is the best way to learn about various problem-solving techniques of solids in geometry.
What is volume?
Volume is the amount of space occupied by a solid shape in a three-dimensional plane or region. Some examples of solids include cubes, cuboids, spheres, cones, cylinders, etc.
- Volume of cube = a3, where a is the edge of the cube.
- Volume of cuboid = lbh, where l = length, b = breadth and h = height.
- Volume of sphere = (4/3) πr3, where r is the radius of the sphere.
- Volume of cylinder = πr2h, where r is the radius of the circular bases and h is the height.
- Volume of cone = (⅓) πr2h, where r is the radius of the circular base and h is the height.
- Volume of hemisphere = (2/3) πr3, where r is the radius of the sphere.
- Volume of frustum = πh/3 (R2 + r2 + Rr), where ‘R’ and ‘r’ are the radii of the base and top of the frustum.
- Volume of prism = Base Area x Height.
- Volume of pyramid = ⅓ (Area of base) (Height).
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Volume Questions and Answers
1. Find the volume of a cuboidal box with dimensions 11 cm × 8 cm × 13 cm.
Solution:
Given, the dimensions of a cuboidal box: 11 cm × 8 cm × 13 cm
Here,
Length = l = 11 cm
Breadth = b = 8 cm
Height = h = 13 cm
As we know, volume of a cuboid = lbh
= 11 × 8 × 13
= 1144
Thus, the volume of the cuboidal box is 1144 cm3.
2. What is the volume of a sphere of diameter 21 units?
Solution:
Given,
Diameter of the sphere = 21 units
Radius of the sphere = (21/2) units
Volume of a sphere = (4/3) πr3
= (4/3) × (22/7) × (21/2) × (21/2) × (21/2)
= 4851
Therefore, the volume of the sphere is 4851 cubic units.
3. A cuboidal block of wood was cut into eight equal cubes of edges 4 cm. Find the volume of the initial block of wood.
Solution:
Given,
Edge of a cubical wood = 4 cm
Volume of the cube = a3
= 43
= 4 × 4 × 4
= 64 cm3
Volume of 8 such cubes = 8 volume of one cube
= 8 × 64
= 512 cm3
Hence, the volume of the initial cuboidal block of wood is 512 cm3.
4. If three solid spherical beads of radii 3 cm, 4 cm, and 5 cm, respectively, are melted into one spherical bead, then find its radius in cm.
Solution:
Given,
Radii of three solid spherical beads: 3 cm, 4 cm and 5 cm
As we know, the volume of a sphere = (4/3) πr3
Let R be the radius of the new spherical bead, which is made by melting three spherical beads.
Volume of a new spherical bead = Sum of volumes of three spherical beads
(4/3)πR3 = (4/3) × 3 × 3 × 3 + (4/3) × 4 × 4 × 4 + (4/3) × 5 × 5 × 5
(4/3)πR3 = (4/3)π (27 + 64 + 125)
R3 = 216
R3 = 63
R = 6
Hence, the radius of the spherical bead is 6 cm.
5. How many bricks, each measuring 25 cm x 11.25 cm x 6 cm, will be needed to build a wall of 8 m x 6 m x 22.5 cm?
Solution:
Given,
Dimensions of a brick = 25 cm x 11.25 cm x 6 cm
Dimensions of a wall = 8 m x 6 m x 22.5 cm
= 800 cm x 600 cm x 22.5 cm
Number of bricks = volume of the wall/ volume of a brick
= (800 x 600 x 22.5) / (25 x 11.25 x 6)
= 6400
6. Find the depth of the cylindrical tank if its capacity is 1848 m3 and the diameter of the base is 14 m.
Solution:
Let h be the depth of the cylindrical tank.
Given,
Diameter of the base = 14 m
Radius of the circular base = r = 14/2 = 7 m
Volume of the cylindrical tank = πr2h
So, πr2h = 1848 m3
(22/7) × 7 × 7 × h = 1848
h = 1848/(22 × 7)
= 12
Therefore, the depth of the cylindrical tank is 12 m.
7. What is the ratio of the volume of a cone, a sphere and a cylinder if each has the same radius and height?
Solution:
Let r be the radius of the cone, sphere and cylinder.
Let h be the height of the cone and cylinder.
Also, r = h
We know that,
Volume of cone = (1/3)πr2h = (1/3)πr3
Volume of sphere = (4/3)πr3
Volume of cylinder = πr2h = πr3
Volume of cone : Volume of sphere : Volume of cylinder
= (1/3)πr2h : (4/3)πr3 : πr2h
= (1/3)πr3 : (4/3)πr3 : πr3
= (1/3) : (4/3) : 1
= 1 : 4 : 3
Therefore, the required ratio is 1 : 4 : 3.
8. Find the water weight in a conical vessel that is 21 cm deep and 16 cm in diameter.
Solution:
Given,
Depth of height of a conical vessel = h = 21 cm
Diameter of the circular base = 16 cm
Radius = r = 16/2 = 8 cm
Volume of a cone = (1/3)πr2h
= (1/3) × (22/7) × 8 × 8 × 21
= 1408 cm3
= (1408/1000) kg
= 1.408 kg
Therefore, the weight of the water in a conical vessel is 1.408 kg.
9. A vessel is in the form of a frustum of a cone. Its radius at one end and the heights are 8 cm and 14 cm, respectively. If its volume is 5676/3 cm3, find the radius at the other end.
Solution:
Given,
Radius of one circular end = r1 = 8 cm
Height of the frustum = h = 14 cm
Let r2 be the radius of the other circular end.
Volume of the frustum = πh/3 (r12 + r22 + r1r2)
= (1/3) × (22/7) × 14 × [(8)2 + (r2)2 + 8r2]
According to the given,
(1/3) × (22/7) × 14 × [(8)2 + (r2)2 + 8r2] = 5676/3 cm3
64 + r22 + 8r2 = 5676/44
r22 + 8r2 = 129 – 64
r22 + 8r2 – 65 = 0
r22 + 13r2 – 5r2 – 65 = 0
r2(r2 + 13) – 5(r2 + 13) = 0
(r2 – 5)(r2 + 13) = 0
r2 = 5, r2 = -13
Hence, the radius of the other circular end is 5 cm.
10. If the curved surface area of a right circular cone is 10010 cm2 and its radius is 35 cm, find the volume of the cone.
Solution:
Given,
Radius of the circular base of a cone = r = 35 cm
Let h and l be the height and slant height of the right circular cone.
Also, given that the curved surface area = 10010 cm2
πrl = 10010
(22/7) × 35 × l = 10010
l = (10010 × 7)/ (22 × 35)
= 91 cm
As we know, l2 = r2 + h2
h2 = l2 – r2
= (91)2 – (35)2
= 8281 – 1225
= 7056
h = 84 cm
Volume of the cone = (1/3) πr2h
= (1/3) × (22/7) × 35 × 35 × 84
= 107800
Hence, the volume of the right circular cone is 107800 cm3.
Practice Questions on Volume
- The height of the wall is six times its width, and the length of the wall is seven times its height. If the volume of the wall is 16128 cm3, what is the width of the wall?
- A fish tank has a length of 45 cm, a width of 25 cm, and a depth of 10 cm. Calculate the volume of the fish tank.
- Find the volume of the cylinder with a radius of 5.4 units and a height of 16 units.
- Find the number of small cubes with edges of 10 cm that can be accommodated in a cubical box with a 1 m edge.
- The radius of a sphere is increased by 10%. Prove that the volume will be increased by 33.1% approximately.
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