A solid iron rectangular block with dimensions of $4.4 \mathrm{m},2.6 \mathrm{m}$and $1 \mathrm{m}$ is cast into a hollow cylindrical pipe with an internal radius of $30 \mathrm{cm}$and a thickness of $5 \mathrm{cm}$Determine the length of the pipe.

To find the answer to this question, keep in mind that whenever a solid is transformed into another solid, the volume remains constant. So, the volume of an iron rectangular block equals the volume of a hollow cylindrical pipe, where the volume of the rectangular block is$lbh$ and the volume of the hollow cylindrical is $h\left(r_1^2-r_2^2\right).$

The volume of the cuboid  $=lbh$and the volume of the cylinder $=\pi r^{2}h$

Volume of iron block = volume of a cuboid $=(440\times 260\times 100){\mathrm{cm}}^3$

Internal radius of the pipe $=30 \mathrm{cm}$

External radius of the pipe =internal radius + thickness  $=(30+5)\mathrm{cm}=35 \mathrm{cm}.$

Let the length of the pipe be $\mathrm{hcm}$.

Volume of hollow cylindrical pipe $\text{=(external volume)-(Internal volume)}=\left[\pi \times {\left(35\right)}^2\times \text{h}-\pi \times {\left(30\right)}^2\times \text{h}\right]{\text{cm}}^3$

$=\pi \mathrm{h}(35{)}^2-\pi \mathrm{h}(30{)}^2{ \mathrm{cm}}^3$

$=\left({35}^2-{30}^2\right)\pi \mathrm{hcm}{}^3=\left(325\pi \mathrm{h}\right){\mathrm{cm}}^3$

As per question,

Volume of cylindrical pipe =Volume of iron bar

$\therefore 325\pi \mathrm{h}=440\times 260\times 100$

$\Rightarrow \mathrm{h}=\left(\frac{440\times 260\times 100}{325}\times \frac{7}{22}\right)\mathrm{cm}$

$\Rightarrow \mathrm{h}=\left(\frac{11200}{100}\right)\mathrm{m}=112 \mathrm{m}$.

Hence, the length of the pipe is $112 \mathrm{m}$.