Volume Of Solids (Cube, Cuboid, Prism, etc.) With Example

In this post, we will be learning about volume of solids questions and answers. We will see many volume of solids questions and answers with tricks/shortcuts etc.

So, let’s start the learning now.

Getting Started

This post contains three sections –

Section A – This section contains important formulas, short tricks, concepts on volume of solids.
Section B – This section contains some examples that demonstrates how to solve questions on volume of solids using short tricks, formulas shown in section A.
Section C – Go to this section to solve some problems on volume of solids.

Now,

Where to start from?

We recommend go through section A, then section B, then section C. Even if you are familiar with this topic, do not just skip section A and section B.

Section A: (Terms & Formulas)

Volume: The Volume of a solid is the number of unit cubes it takes to fill up a solid. There are different types of solids for which volume can be found out.

These are mentioned below:

Shape	Volume
Cube $Tutorialwing math surface area questions and answers of cube$	Volume of Cube = s³ Where, s = Length of the side Example: Find the volume of cube whose side is 3 cm. Solution: Consider below diagram where each side is of 3 cm. $Tutorialwing math surface area questions and answers of cube$ So, Volume of cube (v) = s³ Where, s = 3 cm. v = (3)³ = 27 cm³
Cuboid $Tutorialwing math volume and surface area questions and answers of cuboid$	*Volume of cuboid = l w * h Where, l = Length of cuboid, w = Width of cuboid, h = Height of cuboid Example: Find the volume of cuboid having dimensions 3 cm, 4 cm and 7 cm. Solution:** $Tutorialwing math questions and answers of cuboid example$ Volume of cuboid = 3 * 4 * 7 = 84 cm³
Prism $Tutorialwing Math Solid triangular prism$ $Tutorialwing rectangular prism$	*Volume of Prism = (Area of base) height Where, Height = Height of prism, Area of base depends on type of prism. For rectangular prism,** Area of base = Length * Width For Triangular prism, Area of base = Area of triangular base = (1/2) * height of triangle * Length of base. Example: Find the volume of triangular prism given below – $Tutorialwing math volume and surface area questions and answers of prism example$ Solution: Note that sides of triangle( having sides 3 cm and 6 cm) are perpendicular to each other. So, area of triangle = (1 / 2) * base * height. So, Area of the triangle = 1/2 * 3 cm * 6 cm = 9 cm² Volume(v) of the prism = *(Area of base) height** Where, Area of base = area of triangular base. height = Height of prism. So, => v = 9 cm * 8 cm = 72 cm³
Cylinder $Tutorialwing math volume cylinder formula$	*Volume of cylinder = π r² * h Where, r = Radius of cylinder, h = Height of cylinder Example: Find the volume of cylinder of radius 3 cm and height 5 cm. Solution:** Volume of cylinder = π * r² * h, Here, r = 3 cm h = 5 cm. So, V = π * 3² * 5, = 3.14 * 45 = 141.37 cm³
Hollow Cylinder $Tutorialwing math hollow cylinder example of hollow cylinder$	Volume of Hollow cylinder = π * h * (R² – r²) Where, R = Radius of the outer cylindrical surface, r = Radius of the inner cylindrical surface, h = Height of the hollow cylinder Example: Find the amount(in cm³) of metal used in making a hollow cylinder having thickness 2 cm, outer radius 14 cm and height 70 cm. Solution: $Tutorialwing math hollow cylinder example of hollow cylinder example$ The thickness of the hollow metallic cylinder = 2 cm. Height of the cylinder = 70 cm Outer radius of the cylinder = 14 cm. Inner radius of the cylinder = 14 – 2 = 12 cm Volume of the metal used in making the cylinder = Volume of the hollow cylinder = π * (R² – r²) * h = π * (14² – 12²) * 70 = π * 52 * 70 = 3.14 * 52 * 70 = 11429.6 cm³
Cone $Tutorialwing math cone example of cone example$	*Volume of cone = (1 / 3) π * r² * h Where, r = Radius of circular base, h = Height of cone Example: Find the volume of cone having radius 3 cm and height 11 cm. Solution:** Below image clearly depicts the scenario – $Tutorialwing math cone example of cone example$ Volume (v) of cone = (1 / 3) * π * r² * h Where, r = 3 cm, h = 11 cmSo, v = (1 / 3) * 3.14 * 3² * 11 = (1 / 3) * 3.14 * 99 = 103.69 cm³
Pyramid $Tutorialwing math pyramid example of pyramid example$	*Volume of pyramid = (1 / 3) B * h Where, B = Area of base, h = height of pyramid Example: Find the volume of pyramid given below – $Tutorialwing math pyramid example of pyramid example$ Solution:** Volume of pyramid = (1 / 3) * B * h Where, B = Area of base, h = height of pyramid Area of base = Length * Width Given, Length of pyramid = 8 km, Width of pyramid = 7 km, Height of pyramid = 7 km So, Area of base = 8 * 7 = 56 km² Volume of pyramid = (1 / 3) * (Area of base * height) = (1 / 3) * 56 * 7 = 130.67 km³
Sphere $Tutorialwing math sphere example of sphere example$	*Volume of Sphere = (4 / 3) π * r³ Where, r = Radius of sphere Example: Find the volume of sphere having radius 7 inch. Solution:** $Tutorialwing math sphere example of sphere example$ Volume of sphere = (4 / 3) * π * r³ So, Volume = (4 / 3) * 3.14 * 7³ = 1436 in³
Hemisphere $Tutorialwing math hemisphere example of hemisphere example$	*Volume of Hemisphere = (2 / 3) π * r³ Where, r = Radius of Hemisphere Example: Find the volume of hemisphere having radius 15 inch. Solution: A hemisphere of radius 15 inch would look like below – $Tutorialwing math hemisphere example of hemisphere example$ As we know, Volume(v) of hemisphere = (2 / 3) * π * r³** Where, r = 15 inch So, v = (2 / 3) * π * 15³ = (2 / 3) * (22 / 7) * 15³ = 7,071.42 inch³

Section B: (Examples)

Now, we will see how to approach any questions on volume of solids. What are the key factors we should look into before trying to solve such questions. Please do not move to next section without going through each volume of solids questions and answers.

Example 1

A rectangular prism has a volume of 4200 mm³ . It has a rectangular base of length 10 mm and width 6 mm. Find the height of the prism.

Solution

Below image clearly depicts the scenario,

$Tutorialwing math prism example$

Given,
Volume of prism = 4200 mm³ ,
Length of prism = 10 mm,
Width of prism = 6 mm

Volume of prism = (Area of Base * height)
Area of Base = Length * Width

So,
=> 4200 = (l * w) * h
=> 4200 = (10 * 6) * h
=> 4200 = 60 * h
h = 70 mm

Example 2

The length of a side of a cube is equal to 8 cm. Find the Volume of the cube.

Solution

Volume(v) of cube = s³Where, s is side of a cube

Side(s) of cube = 8 cm
So, volume = 8³
= 8 * 8 * 8

Volume = 512 cm³

Example 3

Find the thickness of the hollow cylinder of height 100 cm if the volume between the inner and outer cylinders is 13000π mm³ and outer diameter is 14 mm.

Solution

At first, let’s convert dimensions into common unit. So, we will convert it into mm unit (10 mm. = 1 cm.)
Height of cylinder = 100 cm = 1000 mm
Outer diameter (d) = 14 mm
So, Outer radius (R) = 7 mm (Diameter = 2 x Radius)
Below diagram depicts the scenario.

Here,

r = Inner radius of cylinder,
R = Outer radius of cylinder,
h = Height of cylinder

So, we have to find R – r .

$Tutorialwing math hollow cylinder example$

Volume (V) of Cylinder = 13000π mm³
V = π * h * ( R² – r² )
13000π = π * 1000 * (7² – r² )
7² – r² = 13
49 – r² = 13
r² = 49 – 13
r² = 36
r = 6 mm

So, Thickness of Cylinder = R – r
= 7 – 6

= 1 mm

Example 4

Find the volume of cone with radius 5 cm and height 10 cm.

Solution
Volume (V) of cone = (1 / 3) * π * r² * h
Where,
r = radius of base,
h = height of code

Given,
r = 5 cm,
h = 10 cm,
π = 22 / 7

$Tutorialwing math cone example$

So, Volume = (1 / 3) * (22 / 7) * ( 5² ) * 10

Volume of cone = 261.90 cm³

Section C: (Exercises)

In this section, we will see different volume of solids questions and answers (with short tricks, if available).

Mark against the correct answer:

At first, please try to solve question by yourself, then move to see it’s answer.

Question 1

The radius of a sphere is 3r, then it’s volume will be –
a. (4 / 3)πr³
b. 36πr³
c. (8 / 3)πr³
d. (32 / 3)πr³

Solution (b)

Explanation
Volume (v) of sphere = (4 / 3) * π * R³
Given,
R = 3r

So, volume(v) of sphere = (4 / 3) * π * (3r)³
v = 36πr³

Question 2

The total surface area of a cube is 96 cm². The volume of cube is –

a. 16 cm³
b. 512 cm³
c. 64 cm³
d. 27 cm³

Solution (c)

Explanation
Given,
Surface area of cube = 96 cm²

As per formula,
Surface area of cube = 6 * s²,
Where, s is side of a cube

6s² = 96
s² = (96 / 6)
s² = 16
s = 4

Volume (v) of cube = s³ ,
Where, s is side of a cube.

So, v = 4³
v = 64 cm³

Question 3

A cube of side 4 cm contains a sphere touching its sides. Find the volume of gap in between.

a. 30.48 cm³
b. 54 cm³
c. 46 cm³
d. 29 cm³

Solution (a)

Explanation
Side(s) of cube = 4 cm
Diameter(d) of sphere = 4 cm
Radius(r) of sphere = 2 cm
Volume of gap in between = Volume of cube(v) – Volume of sphere(s)
= s³ – (4 / 3) * π * r³
= 4³ – (4 / 3) * (22 / 7) * 2 * 2 * 2 , (Taking π = 22 / 7)
= 64 – 33.52
= 30.48 cm³

Question 4

Find the amount of water displaced by a solid spherical ball of diameter 4.2 cm, when it is completely immersed in water.

a. 25.6 cm³
b. 38.80 cm³
c. 43 cm³
d. 56.60 cm³

Solution (b)

Explanation
Diameter(d) of spherical ball = 4.2 cm
So, Radius(r) of spherical ball = 2.1 cm
Amount of water displaced = Volume of ball
= (4 / 3) * πr³
= (4 / 3) * (22 / 7) * 2.1 * 2.1 * 2.1 (Since, π = 22 / 7)
= 38.808 cm³

Question 5

A shopkeeper has one jamun of radius 5 cm. With the same amount of material, how many jamun of radius 2.5 cm can be made?

a. 5
b. 10
c. 8
d. 6

Solution (c)

Explanation
Radius(R) of big jamun = 5 cm
Volume of big jamun = (4 / 3) * π * R³ ,
Where, R is radius of big jamun.

= (4 / 3) * π * (5)³
= (500 / 3)π cm³

Radius(r) of small jamun = 2.5 cm

Volume of small Jamun = (4 / 3) * π * r³ ,
Where, r is radius of small jamun.

= (4 / 3) * π * (2.5)³
= (62.5 / 3)π cm³

Total no. of jamun = Volume of big jamun / Volume of small jamun
= [(500 / 3)π / (62.5 / 3)π]
= 500 / 62.5
= 8

Question 6

A right triangle with sides 6 cm, 8 cm, and 10 cm is revolved about the side 8 cm. Find the volume of the solid.
a. 330.6 cm³
b. 185.9 cm³
c. 200 cm³
d. 301.7 cm³

Solution (d)

Explanation

Below picture clearly depicts the scenario –

$Tutorialwing Volume triangle$

Rotating triangle about 8 cm side

You can see a cone shape is formed with dimensions –
Radius (r) = 6 cm,
Height (h) = 8 cm,

Slant height (l) = 10 cm
Volume of cone = (1/3)πr²h
= (1/3) * (22/7) * 6 * 6 * 8
= 301.7 cm³

Question 7

A cylindrical tube which is opened at both the ends is made of iron sheet which is 2 cm thick. If the outer diameter is 16 cm and length is 100 cm, find how many cubic centi meters of iron has been used in making tube ?

a. 8800 cm³
b. 9000 cm³
c. 5000 cm³
d. 3000 cm³

Solution (a)

Explanation
Outer diameter (D) of cylindrical tube = 16 cm
So, Outer radius (R) of tube = (16 / 2) = 8 cm

Thickness of iron sheet = 2 cm
So, inner radius (r) = 8 – 2 = 6 cm
Height (h) of iron sheet = 100 cm

Volume (v) of metal = Outer Volume – Inner Volume
= π * h * (R² – r²)
= (22/7) * 100 * (8² – 6²)
= 8800 cm³

Question 8

A steel vessel has a base of length 30 cm and breadth 15 cm. Water is poured in the vessel. A cubical steel box having edge of 15 cm is immersed in the vessel. How much will the water rise?

a. 7.5 cm
b. 10 cm
c. 15 cm
d. 30 cm

Solution (a)

Explanation
Dimensions of steel vessel are – Length (l) = 30 cm, Breadth (w) = 15 cm. Height (h) = ?

Length of cubical steel = 15 cm.
When an object is immersed in water, the liquid level rises.
Volume (v) of cube = s³
Where, S = Side of cube.

Volume of cubical steel = l * w * h
Where, l = length of cubical steel,
w = breadth of cubical steel,
h = height of cubical steel.

So, volume of liquid displaced = Volume of cubical steel
l * w * h = s³
=> 30 * 15 * h = 15³
=> 30 * h = 15 * 15
=> h = ( 15 * 15) / 30
=> h = 7.5 cm
So, water will rise upto 7.5 cm

Question 9

A cylinder is having radius 1 m and height 5 m is completely filled with juice. In how many conical flasks can this juice be filled if the flask radius and height is 50 cm each?

a. 50
b. 600
c. 120
d. 90

Solution (c)

Explanation

Volume of cone = (1 / 3) π r² h
Where, r = radius of cone,
h = height of cone.
Here, r = 50 cm = 0.5 m , h = 50 cm = 0.5 m

Volume of Cylinder = π R² H
Where, R = radius of cylinder,
H = height of cylinder.
Here, R = 1 m, H = 5 m

No. of cones = Volume of Cylinder / Volume of one conical flask
= (π R² H) / ((1 / 3) π r² h)
= (π * 1² *5) / ((1/3) * π * (0.5)² * 0.5)
= 5 / ((1 / 3) * (0.125))
= 15 / 0.125
= 120

Question 10

2 cm thick 15 metal plates are kept exactly one above the other. On top of the metal plate, a hemisphere of diameter 6 cm is kept. This just covers the top plate. What is the volume of the entire object?

a. 360π cm³
b. 144π cm³
c. 81 cm³
d. 288π cm³

Solution (d)

Explanation
15 plates on each other form a cylinder

Hence, the height of the cylinder = (15*2) = 30 cm

Diameter(D) of the hemisphere = 6 cm
Radius(r) of hemisphere = 3 cm
This will also be radius of plate.

$Tutorialwing math volume hemisphere example$

Volume of the entire object = Volume of cylinder + Volume of Hemisphere
= πr²h + (2/3)πr³

= π * 3² * 30 + (2/3) * π * 3³

= 270π + 18π
= 288π cm³

Question 11

A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each corner of the sheet, a square is cut off so as to make an open box. If the length of the square is 10 m, the volume of the box (in m³) is:
a. 4480
b. 5120
c. 6420
d. 8960

Solution (a)

Explanation

The scenario is depicted by the fig below

$Tutorialwing math volume rectangle example$

For the box, if the four sides are cut down,

Clearly, l = (48 – 20) m = 28 m,

b = (36 -20) m = 16 m,

h = 10 m.

Volume of the box = (28 x 16 x 10) m³ = 4480 m³

Question 12

The surface of a cube is 24 cm² . Find its volume ?
a. 8 cm³
b. 16 cm³
c. 32 cm³
d. 40 cm³

Solution (a)

Explanation

Surface area of cube = 6 * s²
Where, s = side of a cube

=> 6 * s² = 24
=> s²= 24 / 6 = 4
=> s = 2 cm

Now, volume of cube = s³
=> s³ = 2³
= 8 cm³

Question 13

The areas of the two spheres are in the ratio 1 : 16. The ratio of their volume is ?
a. 1 : 4
b. 1 : 16
c. 1 : 8
d. 1 : 64

Solution (d)

Explanation

Area of sphere = 4 π r²
Where, r = radius of sphere

As per question,
Area of sphere 1 : Area of sphere 2 = 1 : 16
=> 4 π r₁² : 4 π r₂² = 1 : 16
=> r₁ : r₂= 1 : 4

Now, volume of sphere = (4 / 3) π r³So, ratio of volume = (4 / 3) π r₁³: (4 / 3) π r₂³=> r₁³ : r₂³ = 1 : 64

Question 14

A cuboidal garage has a volume of 480 m³ a length of 10 m and a width of 8 m. What is the height of the garage?
a. 4 m
b. 5 m
c. 6 m
d. 7 m

Solution (c)

Explanation

Volume of cuboidal garage = l * w * h
Where, l = Length of cuboid
w = Width of cuboid
h = Height of cuboid

As per question,
=> l * w * h = 480
=> 10 * 8 * h = 480
=> h = 480 / 80
= 6 m

Question 15

Which set of dimensions belongs to a right rectangular prism with a volume of 550 cm³ ?

a. 6 cm , 12 cm , 5 cm
b. 10 cm , 5 cm , 11 cm
c. 7 cm , 9 cm , 6 cm
d. 12 cm , 2 cm , 5 cm

Solution (b)

Explanation

Volume of prism = l * w * h
Where, l = Length of prism
w = Width of prism
h = Height of prism

There is only option (b) that satisfy the given condition.
=> 550 = 10 * 5 * 11

Question 16

A standard dice is a cube having sides of 10 mm. What is the volume of the dice?
a. 30 mm³
b. 45 mm³
c. 425 mm³
d. 1000 mm³

Solution (d)

Explanation

Volume of cube = s³
Where, s = Side of cube

= 10³

= 1000 mm³

Question 17

A cylindrical bowl has a radius of 4.3 cm and a height of 11.6 cm. What is the volume of the bowl to the nearest tenth of a cubic centimetre?

a. 49.9 cm³
b. 168.4 cm³
c. 673.5 cm³
d. 1816.8 cm³

Solution (c)

Explanation

Volume of cylindrical bowl = π r² h
Where, r = radius of bowl,
h = Height of bowl

= π * (4.3)² * 11.6
= (22 / 7) * (4.3)² * 11.6
= 673.5 cm³

Question 18

A metallic rod has a diameter of 17 cm and has 72 cm between the two ends. What is the volume of the space between the ends?
a. 10300.4 cm³
b. 3589.5 cm³
c. 67890.4 cm³
d. 16334.2 cm³

Solution (d)

Explanation

$Tutorialwing math volume metal rod example$

Diameter of rod= 17 cm

Hence the radius = 17 / 2 = 8.5 cm

Height of the rod= 72 cm

Volume of cylindrical rod= π × r ² × h

V = 3.14 × 8.5 × 8.5 × 72

V = 16334.28 cm³.

Question 19

A hockey puck has a diameter of 7.6 cm and a height of 3.4 cm. What is the volume of a cylindrical package containing 5 pucks?

a. 770.8 cm³
b. 480.7 cm³
c. 570.3 cm³
d. 690.4 cm³

Solution (a)

Explanation

Volume(V) of one puck = (π × r ²) × h
Where, r = radius of puck,
h = Height of puck

=> V = (3.14 × 3.8 × 3.8) × 3.4 , (π = 3.14)
=> V = 154.16144

Volume of 5 pucks = 5 × 154.16144 = 770.8 cm³

Question 20

A swimming pool 9 m wide and 10 m long is 1 m deep on the shallow side and 5 m deep on the deeper side. Its volume is:

a. 390 m³
b. 270 m³
c. 420 m³
d. 530 m³

Solution (b)

Explanation

The scenario is explained with the fig below:

$Tutorialwing math volume swimming pool example$

Volume of the pool = length * breadth * height,

In this case, there are two heights given. So we will take average,
Avg. height = (1 + 5) / 2 = 3 m

Volume = (10 ∗ 9 ∗ 3) m³

= 270 m³

Question 21

A circular well with a diameter of 2 meters, is dug to a depth of 21 meters. What is the volume of the earth dug out?

a. 46 m³
b. 79 m³
c. 66 m³
d. 28 m³

Solution (c)

Explanation

Well with given dimensions form a cylindrical shape.
And, volume of earth dug out = volume of cylindrical well
Volume of cylindrical well = π r² h
Where, r = radius of well,
h = Height of well

Volume = ((22/7) ∗ 1 ∗ 1 ∗ 21) m³ = 66 m³

Question 22

Two right circular cylinders of equal volumes have their heights in the ratio 1 : 4. Find the ratio of their radii.

a. 2 : 1
b. 4 : 1
c. 1 : 3
d. 1 : 2

Solution (a)

Explanation

Volume of cylinder = π r² h,
Where, r = radius of cylinder,
h = Height of cylinder

Let the heights of the cylinders be h and 4h and radii be r and R respectively. Then,

=> π r² h = π R² (4h)
=> r² / R² = 4h / h = 4 / 1
=> r / R = 2 / 1
=> r : R = 2 : 1

Question 23

The radius of the base of a cone is 4 cm and its volume is 68.32 cm³. (π = 3.14) What is the height of the cone?

a. 16.34 cm
b. 28.48 cm
c. 47.39 cm
d. 12.48 cm

Solution (a)

Explanation

Given,
Radius (r) of cone = 4 cm
Volume (v) of cone = 68.32 cm³

So,
Volume of cone (v) = (1 / 3) π r² h = 68.32 cm³

π r² h = 68.32 * 3
4² * h = (68.32 * 3) / 3.14
=> h = (68.32 * 3) / (3.14 * 4²)

= 4.07 cm

Conclusion

We have seen how to find volume of solids like cube, cuboid, prism, cylinder, hollow cylinder, cone, sphere and hemisphere.

List of important formulas for volume of solids –

Shape	Volume
Cube	Volume of Cube = s³ Where, s = Length of the side
Cuboid	*Volume of cuboid = l w * h** Where, l = Length of cuboid, w = Width of cuboid, h = Height of cuboid
Prism	*Volume of Prism = (Area of base) height Where, Height = Height of prism, Area of base depends on type of prism. For rectangular prism,** Area of base = Length * Width For Triangular prism, Area of base = Area of triangular base = (1/2) * height of triangle * Length of base.
Cylinder	*Volume of cylinder = π r² * h** Where, r = Radius of cylinder, h = Height of cylinder
Hollow Cylinder	Volume of Hollow cylinder = π * h * (R² – r²) Where, R = Radius of the outer cylindrical surface, r = Radius of the inner cylindrical surface, h = Height of the hollow cylinder
Cone	*Volume of cone = (1 / 3) π * r² * h** Where, r = Radius of circular base, h = Height of cone
Pyramid	*Volume of pyramid = (1 / 3) B * h** Where, B = Area of base, h = height of pyramid
Sphere	*Volume of Sphere = (4 / 3) π * r³** Where, r = Radius of sphere
Hemisphere	*Volume of Hemisphere = (2 / 3) π * r³** Where, r = Radius of Hemisphere

Getting Started

Section A: (Terms & Formulas)

Example:

Example:

Example:

Example:

Example:

Example:

Example:

Example:

Section B: (Examples)

Example 1

Example 2

Example 3

Example 4

Section C: (Exercises)

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Conclusion

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