Question

A solid iron rectangular block of dimension is cast into a hollow cylindrical pipe of internal radius 30cm and thickness 5cm. Find the length of the pipe.
A.120m
B.124m
C.112m
D.108m

Answer 1

Hint: The volume of a solid remains unchanged when it is casted in an object of any shape or size. The volume of a rectangular block is given by the product of its length, breadth and height. A hollow cylindrical pipe has two radii internal and external one. The volume of the hollow cylinder will be equal to the volume of the solid iron rectangular block.

Complete step-by-step answer:
We are given with a solid iron rectangular box of dimension \[4.4m,{\text{ }}2.6m,{\text{ }}1m\] and we have to find the length or height of the hollow cylindrical pipe of internal radius \[30cm\] and thickness of pipe is \[5cm\] .
Let \[{V_1}\] be the volume of the rectangular slab, \[{V_2}\] be the volume of the hollow pipe,
Volume of the slab, \[{V_1}{\text{ }} = {\text{ }}4.4\, \times {\text{ }}2.6\, \times {\text{ }}1{\text{ }}{m^3}\]
Now let \[{r_1}\] be the internal radius of the pipe, \[{r_2}\] be the external radius and \[h\] be the length of the pipe
We will get the values of radii in meters as they should be in the same unit, preferable in S.I. unit.
Internal radius, \[{r_1}{\text{ }} = {\text{ }}30cm\] for converting it in metre we have to divide it by \[100\] so we get
 \[{r_1}{\text{ }} = {\text{ }}\dfrac{{30}}{{100}}{\text{ }}m{\text{ }} = {\text{ }}0.3m\]
External radius, \[{r_2}{\text{ }} = {\text{ }}internal{\text{ }}radius{\text{ }} + {\text{ }}thickness{\text{ }}of{\text{ }}the{\text{ }}pipe\]
 \[{r_2}{\text{ }} = {\text{ }}30{\text{ }} + {\text{ }}5cm\]
 \[{r_2}{\text{ }} = {\text{ }}35cm\]
For converting it into metre divide it by \[100\] so we get
 \[{r_2}{\text{ }} = {\text{ }}\dfrac{{35{\text{ }}}}{{100}}{\text{ }}m{\text{ }} = {\text{ }}0.35m\]
Now, Volume of the hollow cylindrical pipe is given by
 \[{V_2}{\text{ }} = {\text{ }}\pi \left( {{r_2}^2{\text{ }}-{\text{ }}{r_1}^2} \right){\text{ }}h\]
Putting the values we get,
 \[{V_2}{\text{ }} = {\text{ }}\dfrac{{22}}{7}\left( {{{0.35}^2}{\text{ }}-{\text{ 0}}{\text{.}}{{\text{3}}^2}} \right){\text{ }}h\]
On simplifying we get,
 \[ \Rightarrow {V_2}{\text{ }} = {\text{ }}\dfrac{{22}}{7}\left( {{\text{0}}{\text{.1225 }}-{\text{ 0}}{\text{.09}}} \right){\text{ }}h\]
 \[{V_2}{\text{ }} = {\text{ }}\dfrac{{22}}{7}\left( {0.0325} \right){\text{ }}h\]
As the volume of a solid remains unchanged when it is casted in object of any shape or size
So, \[{V_1}{\text{ }} = {\text{ }}{V_2}\]
Substituting the values we got,
 \[\dfrac{{22}}{7}\, \times \,0.0325\, \times \,h = 4.4\, \times 2.6\, \times 1\]
Simplifying for \[h\]
 \[ \Rightarrow h = \dfrac{{4.4 \times \,2.6 \times \,1\, \times 7}}{{22\, \times 0.0325}}\,\]
On multiplication and division we get,
 \[ \Rightarrow h = 112m\]
Hence the length of the pipe is \[112m\]
Therefore option C is correct.
So, the correct answer is “Option C”.

Note: The calculation must be done in one common unit. It may be solved either in meters or in centimeters. Questions may be different having different dimensions. Length of the pipe refers to the height of the hollow cylinder. The value of \[\pi \] is taken as \[\dfrac{{22}}{7}{\text{ }}\] to be more precise we may take it as \[3.14\] as well.