SME
Grade 10 Math · Problem type
Mass of a cone from its density
Use this for “A solid aluminum cone (2.7 g/cm³) has radius 3 cm and height 4 cm. Find its mass.”
Diagram
V = 1/3 × π × r² × h
V
=
1
3
×
π
×
3
2
×
4
V = 12π ≈ 37.7 cm³
Density: 2.7 g/cm³
Mass: 101.79 g
This is an example. Type your own values to start a new problem.
V (cm³)
=
1
3
×
π
×
r
2
×
h
ρ (g/cm³)
=
m (g)
÷
V (cm³)
Formulas
V
=
π
×
r
2
×
h
3
37.7
=
π
×
3
2
×
4
3
ρ
=
m
V
2.7
=
101.79
37.7
Assumptions
•
Model the object as a cone: V = πr²h ÷ 3, a third of the cylinder on the same base.
•
Mass is density times volume: m = ρV.
•
Given the mass, work back to the volume first, then the height or the radius.
Step-by-step
WE KNOW
r = 3 cm, h = 4 cm, ρ = 2.7 g/cm³
FIND
volume (V), mass (m)
Step 1 · Find volume (V)
Use V = π × r² × h ÷ 3
A cone holds a third of the cylinder with the same base and height.
V
=
π
×
3
2
×
4
3
V
=
π
×
9
×
4
3
V
=
9π
×
4
3
V
=
36π
3
V
=
37.7
cm³
Step 2 · Find mass (m)
Use ρ = m ÷ V
Each
cubic
centimeter
has
ρ
grams:
multiply.
m
=
ρ
×
V
m
=
2.7
×
37.7
m
=
101.79
g
Check
Put the numbers back in. Both sides should match.
37.7
≈
π
×
3
2
×
4
3
✓
2.7
=
101.79
37.7
✓
Topic check
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