Question

A hollow iron pipe of $21cm$ long and its external diameter is $8cm$. If the thickness of the pipes is $1cm$ and iron weights $8gm/c{{m}^{2}}$, then the weight of the pipe is equal to
A. 3.6 kg
B. 3.69 kg
C. 36 kg
D. 36.96 kg

Answer 1

Hint: Find the radius of inner surface using thickness and outer surface using thickness and outer surface and hence use the formula to find volume $\pi \left( {{R}^{2}}-{{r}^{2}} \right)h$ where $R,r$ are outer and inner radiuses of cylinder. Then multiply by $8g/c{{m}^{3}}$ to find weight.

Complete step-by-step answer:

In the question we are given a hollow iron pipe of $21cm$ long and its external diameter is $8cm$. If the thickness of the pipes is $1cm$ and also said that iron weighs $8g/c{{m}^{3}}$.
Now in the question we are asked to find the weight of pipe, so hence we have to find the volume which we can get using formula:
$\pi \left( {{R}^{2}}-{{r}^{2}} \right)h$ Where thickness can be represented as $\left( R-r \right)$ , here $R$ is external radius and $r$ is internal radius and $h$ is height.
As we know external diameter is $8cm$ so external radius is $4cm$. In the question thickness is given as $1cm$ so the inner radius will be $4cm-1cm=3cm$.
Hence the $R=4cm,r=3cm$ and height is $21cm$ and value of $\pi =\dfrac{22}{7}$.
$\Rightarrow$ The volume is $\pi \left( {{R}^{2}}-{{r}^{2}} \right)h$ which is $\dfrac{22}{7}\times \left( {{4}^{2}}-{{3}^{2}} \right)\times 21$
This can be simplified as,
$22\times 21=462c{{m}^{3}}$.
$\Rightarrow$ The volume of pipe is $462c{{m}^{3}}$.
We are given density iron which is $8g/c{{m}^{3}}$. So for $1c{{m}^{3}}$ of iron it weighs $8g$.
Hence for $462c{{m}^{3}}$, the iron will weighs,
$\left( 462\times 8 \right)g=3696g$ or $3.696kg$.
The correct option is ‘B’.

Note: Students should know to find the radius of either external or internal if one of them and thickness is given. They should also know about formulas related to it.
Another approach when the hollow pipes formula i.e., $\pi \left( {{R}^{2}}-{{r}^{2}} \right)h$, is not known. In this case, we can find out the volume of the outer cylinder and then subtract its volume from the inner cylinder. So using this volume of the hollow pipe will be,
$\pi \left( {{R}^{2}} \right)h-\pi \left( {{r}^{2}} \right)h$