Question

A solid iron cuboidal block of dimensions $4.4m,2.6m,1m$ is cast into a hollow cylindrical pipe of internal radius $30cm$and thickness $5cm.$. Find length of the pipe.

Answer 1

Step 1: Find the volume of cylindrical pipe

Dimensions of cuboidal block

Length $\left(l\right)=4m$

Breadth $\left(b\right)=2.6m$

Height $\left(h\right)=1m$

Dimensions of cylindrical pipe:

Internal radius $\left(r_2\right)$ $=30cm$ $=0.3m$

Thickness$=5cm=0.05m$

So the outer radius $\left(r_1\right)$ $=30+5cm=35cm=0.35m$

Let the length of the pipe $=hmetre$

We know, the volume of a hollow cylinder $=\mathrm{\pi h}\left({{\mathrm{r}}_1}^2-{{\mathrm{r}}_2}^2\right)$,

where $\mathrm{\pi }=\frac{22}{7},r_1=\mathrm{external}\mathrm{radius},r_2=\mathrm{internal}\mathrm{radius}$

We know, when a shape is recast into a different shape, then the volume stays the same.

Here, Cuboidal block is recast into cylindrical pipe. So, volume of the cuboidal block equals the volume of hollow cylinder.

We know, Volume of a cuboidal tank $=l\times b\times h$

$=4.4\times 2.6\times 1m^3=11.44m^3$

Then, Volume of cylindrical pipe will also be$=11.44m^3$ [ Because the cuboidal tank is melt and recast into the cylindrical pipe]

Step 2: Find the height of the pipe

Also, we have volume of a hollow cylinder $=\mathrm{\pi h}\left({{\mathrm{r}}_1}^2-{{\mathrm{r}}_2}^2\right)$

So equating the above two, we get:

$\Rightarrow \mathrm{\pi h}\left({{\mathrm{r}}_1}^2-{{\mathrm{r}}_2}^2\right)=11.44$

Now, putting the values of $r_1,r_2,\mathrm{\pi }$ in the equation above, we get:

$$\begin{gathered} \Rightarrow 11.44m^3=\frac{22}{7}\times h\times (0.{35}^2-0.3^2) \\ \Rightarrow 11.44=\frac{22}{7}\times h\times \left(0.1225-0.09\right) \\ \Rightarrow 11.44=\frac{22}{7}\times h\times \left(0.0325\right) \\ \Rightarrow 11.44=\frac{0.715}{7}\times h \\ \Rightarrow \mathrm{h}=\frac{11.44\times 7}{0.715} \\ \Rightarrow \mathrm{h}=112\mathrm{m} \end{gathered}$$

Thus, the length of the pipe is $112m$