
A hollow iron pipe is 21cm long and its external diameter is 8cm. If the thickness of the pipe is 1cm and iron weights $8\dfrac{{\text{g}}}{{{\text{c}}{{\text{m}}^3}}}$, then weight of the pipe is
$
{\text{A}}{\text{. 3}}{\text{.6kg}} \\
{\text{B}}{\text{. 3}}{\text{.696kg}} \\
{\text{C}}{\text{. 36kg}} \\
{\text{D}}{\text{. 36}}{\text{.9kg}} \\
$
Answer
612.3k+ views
Hint – To find the weight of the pipe, we calculate the external and internal radii of the pipe and determine the volume. Using the volume we determine the weight of the pipe.
Complete step by step answer:
Let the external diameter of the pipe be ${{\text{d}}_2}$and the internal diameter be ${{\text{d}}_1}$.
$ \Rightarrow {{\text{d}}_2} = 8{\text{cm and }}{{\text{d}}_1} = {\text{8 - 1 - 1 = 6cm}}$ -- (as the thickness is to be deducted on both sides of outside diameter to obtain the inner diameter).
${\text{radius = }}\dfrac{{{\text{diameter}}}}{2}$
$ \Rightarrow {{\text{r}}_2} = \dfrac{{{{\text{d}}_2}}}{2} = \dfrac{8}{2} = 4{\text{cm and }}{{\text{r}}_1} = \dfrac{{{{\text{d}}_1}}}{2} = \dfrac{6}{2} = 3{\text{cm}}$
Volume of a hollow cylindrical pipe with radii ${{\text{r}}_1}$,${{\text{r}}_2}$and a height h is given as
${\text{V = }}\pi \left( {{\text{r}}_2^2 - {\text{r}}_1^2} \right) \times {\text{h}}$
$
\Rightarrow {\text{V = }}\dfrac{{22}}{7}\left( {{4^2} - {3^2}} \right) \times 21 \\
\Rightarrow {\text{V = }}\dfrac{{22}}{7} \times 7 \times 21 \\
\Rightarrow {\text{V = 462c}}{{\text{m}}^3} \\
$
Given density of Iron =$8\dfrac{{\text{g}}}{{{\text{c}}{{\text{m}}^3}}}$
Weight of Iron can be computed as, weight = volume × density
$ \Rightarrow {\text{Weight of pipe = 462 }} \times {\text{ 8 = 3696g = 3}}{\text{.696kg}}$
Hence Option B is the correct answer.
Note – In order to solve problems of this type the key is to convert the outside diameter into inner diameter with the use of thickness. Also, it is important to identify the value of the iron given in the question is its density, using it we compute the weight of the pipe. Also 1kg = 1000g.
Complete step by step answer:
Let the external diameter of the pipe be ${{\text{d}}_2}$and the internal diameter be ${{\text{d}}_1}$.
$ \Rightarrow {{\text{d}}_2} = 8{\text{cm and }}{{\text{d}}_1} = {\text{8 - 1 - 1 = 6cm}}$ -- (as the thickness is to be deducted on both sides of outside diameter to obtain the inner diameter).
${\text{radius = }}\dfrac{{{\text{diameter}}}}{2}$
$ \Rightarrow {{\text{r}}_2} = \dfrac{{{{\text{d}}_2}}}{2} = \dfrac{8}{2} = 4{\text{cm and }}{{\text{r}}_1} = \dfrac{{{{\text{d}}_1}}}{2} = \dfrac{6}{2} = 3{\text{cm}}$
Volume of a hollow cylindrical pipe with radii ${{\text{r}}_1}$,${{\text{r}}_2}$and a height h is given as
${\text{V = }}\pi \left( {{\text{r}}_2^2 - {\text{r}}_1^2} \right) \times {\text{h}}$
$
\Rightarrow {\text{V = }}\dfrac{{22}}{7}\left( {{4^2} - {3^2}} \right) \times 21 \\
\Rightarrow {\text{V = }}\dfrac{{22}}{7} \times 7 \times 21 \\
\Rightarrow {\text{V = 462c}}{{\text{m}}^3} \\
$
Given density of Iron =$8\dfrac{{\text{g}}}{{{\text{c}}{{\text{m}}^3}}}$
Weight of Iron can be computed as, weight = volume × density
$ \Rightarrow {\text{Weight of pipe = 462 }} \times {\text{ 8 = 3696g = 3}}{\text{.696kg}}$
Hence Option B is the correct answer.
Note – In order to solve problems of this type the key is to convert the outside diameter into inner diameter with the use of thickness. Also, it is important to identify the value of the iron given in the question is its density, using it we compute the weight of the pipe. Also 1kg = 1000g.
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