Question

A cylindrical hollow metal pipe is 16 cm long.

It has an external diameter of 10 cm and an internal diameter of 8 cm.
The density of the metal from which the pipe is made is 8 grams per cm3

What is the mass of the pipe in grams?

Answer 1

3619.1 grams (nearest tenth)

Explanation:

Volume of a cylinder

`V=\pi r^2 h`

where:

• r is the radius.
• h is the height.

The radius of a circle is half its diameter.

Find the volumes of the external and internal cylinders by substituting the given values into the formula for volume of a cylinder:

`\implies \textsf{Volume of the external cylinder}&=\pi \cdot 5^2 \cdot 16=400 \pi \; \sf cm^3`

`\implies \textsf{Volume of the internal cylinder}&=\pi \cdot 4^2 \cdot 16=256 \pi \; \sf cm^3`

Find the volume of the cylindrical hollow metal pipe by subtracting the volume of the internal cylinder from the volume of the external cylinder:

`\begin{aligned}\implies \textsf{Volume of the metal pipe}&=\sf Volume_(external)-Volume_(internal)\\&=400 \pi - 256 \pi\\& = 144 \pi \sf \; cm^3\end{aligned}`

Density formula

`\rho=(m)/(V)`

where:

• ρ = density
• m = mass
• V = volume

Given:

• ρ = 8 g/cm³
• V = 144π cm³

Substitute the given values of density and the found volume into the density formula and solve for m:

`\begin{aligned}\implies \rho&=(m)/(V)\\8&=(m)/(144 \pi)\\m&=8 \cdot 144\pi\\m&=1152 \pi \\m&=3619.11473...\\m&=3619.1\; \sf g \;(nearest\;tenth)\end{aligned}`

Therefore, the mass of the pipe is 3619.1 grams (nearest tenth).

Note: I have used the exact value of π (not a rounded value).

Answer 2

• 3617.28 grams

Explanation:

Find the volume of the metal as following:

• 
  `V_(metal)=V_(external)-V_(internal)`

• 
  `V_(external) = \pi d_(external)^2h/4=\pi *10^2*16/4=1256` cm³

• 
  `V_(internal) = \pi d_(internal)^2h/4=\pi *8^2*16/4=803.84` cm³

• 
  `V_(metal)=1256-803.84 = 452.16` cm³

Find the mass:

• 
  `mass = volume*density`

• 
  `mass = 452.16*8=3617.28` grams