Studyledge

Density: Question 8

Syllabus 1.4

Structured Core 5 marks

A workshop technician turns a solid metal cylinder on a lathe. The cylinder has a diameter of 4.0 cm4.0\text{ cm} and a height of 6.0 cm6.0\text{ cm}.

(a) Calculate the volume of the cylinder, using V=πr2hV = \pi r^2 h. Give your answer to 3 significant figures. [2]

(b) The cylinder has a mass of 670 g670\text{ g}. Calculate the density of the metal. [2]

(c) A second, identical cylinder of the same metal is welded end-to-end onto the first, forming one long cylinder of twice the length. State the density of this combined cylinder, giving a reason. [1]

Show worked solution Hide worked solution

Worked solution

Part (a): Volume of the cylinder

The radius is half the diameter:

r=4.02=2.0 cmr = \frac{4.0}{2} = 2.0\text{ cm}

Substitute into the volume formula for a cylinder:

V=πr2h=π×(2.0)2×6.0V = \pi r^2 h = \pi \times (2.0)^2 \times 6.0

V=24π=75.39822… cm3V = 24\pi = 75.39822\ldots\text{ cm}^3

To 3 significant figures:

V=75.4 cm3V = \boxed{75.4\text{ cm}^3}

Part (b): Density of the metal

ρ=mV=670 g75.4 cm3\rho = \frac{m}{V} = \frac{670\text{ g}}{75.4\text{ cm}^3}

ρ=8.886… g/cm3\rho = 8.886\ldots\text{ g/cm}^3

To 3 significant figures:

ρ=8.89 g/cm3\rho = \boxed{8.89\text{ g/cm}^3}

Part (c): Welding an identical cylinder onto it

Welding on a second, identical cylinder doubles both the total mass and the total volume:

ρnew=2m2V=mV=ρ\rho_{\text{new}} = \frac{2m}{2V} = \frac{m}{V} = \rho

So the combined cylinder has exactly the same density, 8.89 g/cm38.89\text{ g/cm}^3, because density depends only on the material, not on how much of it there is.

Final answers

  • (a) Volume == 75.4 cm375.4\text{ cm}^3
  • (b) Density == 8.89 g/cm38.89\text{ g/cm}^3
  • (c) Density stays 8.89 g/cm38.89\text{ g/cm}^3. Mass and volume both double, so their ratio is unchanged.