Answer
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Hint: In this question we have to find Volume of the boiler. First step is to draw a diagram. It will give us a clear picture of what to find out, considering the given quantity, break the diagram into 3 parts, two hemispheres and one cylinder. Now using the formula of volume of a cylinder and hemisphere you will get the final answer.
Complete step-by-step answer:
According to the question, a boiler is in the form of a cylinder 2 m long with hemispherical ends each of 2 metre diameter.
The height of the cylinder is h and the radius of the cylinder is r.
Height of the cylinder = h = 2 m
Diameter of the hemisphere = 2 m
As we know that Radius = $\dfrac{{{\rm{Diameter}}}}{2}$
∴ Radius of the hemisphere = ${\rm{r}} = \dfrac{{{\rm{Diameter}}}}{2} = \dfrac{2}{2} = 1{\rm{m}}$
Since at ends of cylinder hemisphere are attached,
So, Radius of the Cylinder = Radius of the hemisphere = r = 1m
From above diagram,
Total volume of the boiler = Volume of the cylindrical portion + Volume of the two hemispheres
We know that,
Volume of the cylinder = ${\rm{\pi }}{{\rm{r}}^2}{\rm{h}}$
Volume of the hemisphere = $\dfrac{{2{\rm{\;\pi }}{{\rm{r}}^3}}}{3}$
Now,
Total volume of the boiler = ${\rm{V\;}} = {\rm{\;\pi }}{{\rm{r}}^2}{\rm{h\;}} + {\rm{\;}}2 \times \left( {\dfrac{{2{\rm{\;\pi }}{{\rm{r}}^3}}}{3}} \right)$
$ \Rightarrow {\rm{V\;}} = {\rm{\;\pi }}{{\rm{r}}^2}{\rm{h\;}} + \dfrac{4}{3}{\rm{\pi }}{{\rm{r}}^3}$
$ \Rightarrow {\rm{V\;}} = {\rm{\;\pi }}{{\rm{r}}^2}\left( {{\rm{h\;}} + \dfrac{{4{\rm{r}}}}{3}} \right)$
$ \Rightarrow {\rm{V\;}} = \dfrac{{22}}{7} \times {\rm{\;}}{1^2} \times \left( {{\rm{\;}}2{\rm{\;}} + \dfrac{4}{3} \times 1} \right)$
$ \Rightarrow {\rm{V\;}} = \dfrac{{22}}{7} \times \left( {\dfrac{{6 + 4}}{3}} \right)$
$ \Rightarrow {\rm{V\;}} = \dfrac{{22}}{7} \times \dfrac{{10}}{3}{\rm{\;}} = \dfrac{{220}}{{21}}$
Therefore, Volume of the boiler is $\dfrac{{220}}{{21}}$ m³
Note: Whenever we face such types of problems the key concept is to draw the pictorial representation of the given problem and then split into parts such that we can use the formula of standard quantities in order to get the volume of the boiler.
Complete step-by-step answer:
According to the question, a boiler is in the form of a cylinder 2 m long with hemispherical ends each of 2 metre diameter.
The height of the cylinder is h and the radius of the cylinder is r.
Height of the cylinder = h = 2 m
Diameter of the hemisphere = 2 m
As we know that Radius = $\dfrac{{{\rm{Diameter}}}}{2}$
∴ Radius of the hemisphere = ${\rm{r}} = \dfrac{{{\rm{Diameter}}}}{2} = \dfrac{2}{2} = 1{\rm{m}}$
Since at ends of cylinder hemisphere are attached,
So, Radius of the Cylinder = Radius of the hemisphere = r = 1m
From above diagram,
Total volume of the boiler = Volume of the cylindrical portion + Volume of the two hemispheres
We know that,
Volume of the cylinder = ${\rm{\pi }}{{\rm{r}}^2}{\rm{h}}$
Volume of the hemisphere = $\dfrac{{2{\rm{\;\pi }}{{\rm{r}}^3}}}{3}$
Now,
Total volume of the boiler = ${\rm{V\;}} = {\rm{\;\pi }}{{\rm{r}}^2}{\rm{h\;}} + {\rm{\;}}2 \times \left( {\dfrac{{2{\rm{\;\pi }}{{\rm{r}}^3}}}{3}} \right)$
$ \Rightarrow {\rm{V\;}} = {\rm{\;\pi }}{{\rm{r}}^2}{\rm{h\;}} + \dfrac{4}{3}{\rm{\pi }}{{\rm{r}}^3}$
$ \Rightarrow {\rm{V\;}} = {\rm{\;\pi }}{{\rm{r}}^2}\left( {{\rm{h\;}} + \dfrac{{4{\rm{r}}}}{3}} \right)$
$ \Rightarrow {\rm{V\;}} = \dfrac{{22}}{7} \times {\rm{\;}}{1^2} \times \left( {{\rm{\;}}2{\rm{\;}} + \dfrac{4}{3} \times 1} \right)$
$ \Rightarrow {\rm{V\;}} = \dfrac{{22}}{7} \times \left( {\dfrac{{6 + 4}}{3}} \right)$
$ \Rightarrow {\rm{V\;}} = \dfrac{{22}}{7} \times \dfrac{{10}}{3}{\rm{\;}} = \dfrac{{220}}{{21}}$
Therefore, Volume of the boiler is $\dfrac{{220}}{{21}}$ m³
Note: Whenever we face such types of problems the key concept is to draw the pictorial representation of the given problem and then split into parts such that we can use the formula of standard quantities in order to get the volume of the boiler.
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