 # A solid iron cuboidal block of dimensions 4.4 m x 2.6m x lm

Question:

A solid iron cuboidal block of dimensions 4.4 m x 2.6m x lm is recast into a hollow cylindrical pipe of

internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

Solution:

Given that, a solid iron cuboidal block is recast into a hollow cylindrical pipe,

Length of cuboidal pipe (l) = 4.4 m

Breadth of cuboidal pipe (b) = 2.6 m and height of cuboidal pipe (h) = 1m

So, volume of a solid iron cuboidal block $=l \cdot b \cdot h$

$=4.4 \times 2.6 \times 1=11.44 \mathrm{~m}^{3}$

Also, internal radius of hollow cylindrical pipe $\left(r_{i}\right)=30 \mathrm{~cm}=0.3 \mathrm{~m}$

and thickness of hollow cylindrical pipe $=5 \mathrm{~cm}=0.05 \mathrm{~m}$

So, external radius of hollow cylindrical pipe $\left(r_{e}\right)=r_{i}+$ Thickness

$=0.3+0.05$

$=0.35 \mathrm{~m}$

$\therefore$ Volume of hollow cylindrical pipe $=$ Volume of cylindrical pipe with external radius

- Volume of cylindrical pipe with internal radius

$=\pi r_{e}^{2} h_{1}-\pi r_{i}^{2} h_{1}=\pi\left(r_{\theta}^{2}-r_{i}^{2}\right) h_{1}$

$=\frac{22}{7}\left[(0.35)^{2}-(0.3)^{2}\right] \cdot h_{1}$

$=\frac{22}{7} \times 0.65 \times 0.05 \times h_{1}=0.715 \times h_{1} / 7$

where, $h_{1}$ be the length of the hollow cylindrical pipe.

Now, by given condition,

Volume of solid iron cuboidal block = Volume of hollow cylindrical pipe

$\Rightarrow \quad 11.44=0.715 \times h / 7$

$\therefore$ $h=\frac{11.44 \times 7}{0.715}=112 \mathrm{~m}$

Hence, required length of pipe is 112 m.