Question

A hollow cylindrical iron pipe of length 28cm has inner radius 6cm and outer radius 8cm. Find the volume of the pipe and the weight of the pipe if the weight of 1cu.cm of iron is 7gm.

Answer 1

Hint: Find the volume of the inner cylinder and the volume of the outer cylinder using the formula for the volume of a cylinder of height h and radius r, i.e. $V=\pi {{r}^{2}}h$. Use the fact that the volume of the pipe is the volume of the inner cylinder subtracted from the volume of the outer cylinder, i.e,. ${{V}_{p}}={{V}_{o}}-{{V}_{i}}$, where ${{V}_{p}}$ is the volume of the pipe, ${{V}_{o}}$ is the volume of the outer cylinder and ${{V}_{i}}$ the volume of the inner cylinder. Hence find the volume of the pipe. Using $\rho =\dfrac{m}{V}$, where $\rho $ is the density, m the mass and V the volume, determine the mass of the pipe.

Complete step-by-step answer:

A cross-sectional view of the pipe is shown above.
Here BC = r = 6cm and BD = R = 8cm.

A lateral view of the pipe is shown above
Here AB = h = 28cm.
Now, we know
The radius of inner cylinder = r = 6cm and the height of the inner cylinder = h= 28cm.
We know that the volume of a cylinder of radius r and height h is given by
$V=\pi {{r}^{2}}h$
Hence, we have
${{V}_{i}}=\pi {{r}^{2}}h=\dfrac{22}{7}\times {{6}^{2}}\times 28=3168c{{m}^{3}}$
Also, we have
The radius of the outer cylinder = R = 8cm and the height of the outer cylinder = h = 28cm
We know that the volume of a cylinder of radius r and height h is given by
$V=\pi {{r}^{2}}h$
Hence, we have
${{V}_{o}}=\pi {{R}^{2}}h=\dfrac{22}{7}\times {{\left( 8 \right)}^{2}}\times 28=5632c{{m}^{3}}$
We know that the volume of the pipe is the volume of inner cylinder subtracted from the volume of the outer cylinder, i.e. ${{V}_{p}}={{V}_{o}}-{{V}_{i}}$
Hence, we have the volume of the pipe $={{V}_{p}}={{V}_{o}}-{{V}_{i}}=5632-3168=2464c{{m}^{3}}$
Hence the volume of the pipe is equal to 2464 cubic centimetres.
Now, we have
Mass of 1 cubic centimetre of iron = 7gm
Hence, we have the density of iron $\rho =7g/c{{m}^{3}}$
Let m be the mass of the pipe
We know that $\rho =\dfrac{m}{V}$
Hence, we have
$7=\dfrac{m}{2464}$
Multiplying both sides by 2464, we get
m = 17248gm
Hence the weight of the pipe is 17.248 kg.

Note: In mensuration problems, special care should be taken about the units. When comparing things or finding area/volume, units should be kept the same. Many students do not change the units or make them the same and end up with incorrect results.