Question

The inner diameter of a cylindrical wooden pipe is $24cm$ and its outer diameter is $28cm.$ The length of the pipe is $35cm$.Find the mass of the pipe, if $1c{m}^3$of wood has a mass of $0.6g$.

Answer 1

Step 1: Find the volume wooden pipe

Let inner diameter of pipe $=d_1=24cm$

Thus, $r_1=\frac{d_1}{2}=\frac{24}{2}=12cm$

Let outer diameter of pipe $=d_2=28cm$

Thus, $r_2=\frac{d_2}{2}=\frac{28}{2}=14cm$

We have to find out the mass of the pipe if $1c{m}^3$ of wood has a mass of $0.6g$

So, volume used to make wood = volume of outer cylinder – volume of inner cylinder

$$\begin{gathered} =\pi \left({r_2}^2\right)h–\pi \left({r_1}^2\right)h \\ =\pi \left(\right({r_2}^2–{r_1}^2)h \\ =\frac{22}{7}\times ({14}^2–{12}^2)\times 35 \\ =\frac{22}{7}\times (52)\times 35 \\ =5720c{m}^3 \end{gathered}$$

Step 2: Find the mass

Now we have given that

$$\begin{gathered} 1c{m}^3=0.6g \\ 5720c{m}^3=0.6\times 5720g \\ =3432grams=3.432kg \end{gathered}$$

[Since, $1000g=1kg$]

Hence, the mass of the pipe, if $1c{m}^3$of wood has a mass of $0.6g=3.432kg$.