Question

Question 9
A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

Answer 1

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) = 2.6m and height of cuboidal pipe (h) = 1 m
So, volume of a solid iron cuboidal block $=l\times b\times h$
$=44\times 2.6\times 1=11.44m^3$
Also, internal radius of hollow cylindrical pipe $\left(r_1\right)=30cm=0.3m$ (1m= 100cm)
And thickness of hollow cylindrical pipe = 5 cm = 0.05 m
So, external radius of hollow cylindrical pipe $\left(r_e\right)=r_1+$ Thickness
= 0.3 + 0.05
= 0.35 m

Let 'h' be the height of the cylinder
$\therefore$ Volume of hollow cylindrical pipe = volume of cylindrical pipe with external radius – volume of cylindrical pipe with internal radius.
$=\pi r_e^{2}h-\pi r_i^{2}h=\pi (r_e^2-r_i^2)h=\frac{22}{7}\left[\right(0.35{)}^2-\left(0.3{)}^2\right].h=\frac{22}{7}\times 0.65\times 0.05\times h=0.715\times h/7$
Now, by given condition.
Volume of solid iron cuboidal block = Volume of hollow cylindrical pipe
$\Rightarrow 11.44=0.715\times h/7\therefore h=\frac{11440}{715}\times 7=112m$
Hence, required length of pipe is 112 m.