Question

A solid iron rectangular block of dimension 4.4m, 2.6m,1m cast into a hollow cylindical pipe of inner radius of 30cm and the thickness 5cm find the length of the pipe

Answer 1

Volume of iron = $(440 * 260 * 100) cm^3.$ 
Internal radius of the pipe = $30 cm.$ 

External radius of the pipe = $(30 + 5) cm = 35 cm$. 

Let the length of the pipe be $'h' cm$. 

Volume of iron in the pipe = (External volume) – (Internal volume)

$[ (\pi * (35)^2 * h) - (\pi * (30)^2 * h)]cm^3 = \pi h * [(35)^2 - (30)^2]cm^3$
$[(65 * 5) \pi h]cm^3 = (325 \pi h)cm^3$
$h = ( \frac{440*260*100}{325} * \frac{7}{22} )cm$
$h = ( \frac{11200}{100} )m = 112m$

So, the length of the pipe is $112 m$

Answer 2

Hi there !!

Length , l = 4.4 metres

Breadth , b = 2.6 metres

Height , h = 1 metre

Volume of a cuboid = l b h

= 4.4 × 2.6 × 1

= 11.44 m³

The cuboid is melted and is made into a hollow cylindrical pipe.

But ,

Even if shape changes ,Volume of both the shapes remains same.

Volume of cuboid = Volume of hollow cylindrical pipe = 11.44 m³

==================================

Radius ,r = 30 cm = 0.3 metres

Thickness = 5 cm = 0.05 metre

Outer radius, R = inner radius + thickness 
= 0.3 + 0.05
= 0.35 metres

Height of the pipe = Length of the pipe

Volume of a cylindrical pipe = π [ R² - r ² ] × h
=  \frac{22}{7}722​ × [ (0.35)² - (0.3)² ] ×h


   \frac{22}{7}722​  × [ (0.35)² - (0.3)² ] ×h = 11.44 m³

   \frac{22}{7}722​  × [ 0.1225 - 0.09 ] ×h = 11.44 m³

   \frac{22}{7}722​  × 0.0325 × h = 11.44 m³

  
h = \frac{11.44*7}{22*0.0325}22∗0.032511.44∗7​ 

  = \frac{80.08}{0.715}0.71580.08​ 

  = 112 metres

Length of the pipe = 112 metres