Question

Find the volume of oil container which is in the shape of a cylinder with hemispherical ends The total length of the container is 23m and the radius is 3m


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Answer 1

Given :

• Oil container in cylindrical shape with hemispherical ends.
• Total length of the container is 23 m
• Radius of the container is 3 m.

To Find :

• Volume of the container.

Solution :

We have radius and total length of the container.

✪ Height :

Height of the cylinder will be difference of the radius. Since, it is a cylinder it has two ends, which means, height will be difference of total length and sum of radii.

✪ Height = Total length - radii

Height = 23 - (3 + 3)

Height = 23 - 6

Height = 17 m.

Volume of cylinder :

Using the formula,

$\large{\boxed{\mathtt{\red{Volume_{cylinder}\:=\:\:\pi\:r^2\:h}}}}$

$\longrightarrow$ $\mathtt{\dfrac{22}{7}\:\times\:3\:\times\:3\:\times\:17}$

$\longrightarrow$$\mathtt{\dfrac{22\:\times\:9\:\times\:17}{7}}$

$\longrightarrow$ $\mathtt{\dfrac{22\:\times\:153}{7}}$

$\longrightarrow$ $\mathtt{\dfrac{3366}{7}}$

$\longrightarrow$ $\mathtt{480.85}$

$\large{\boxed{\rm{\blue{Volume_{Cylinder}\:=\:480.85\:m^3}}}}$

Volume of hemisphere :

✪ Radius of hemisphere = 3 m.

We can find the volume using the formula,

$\large{\boxed{\mathtt{\red{Volume_{hemisphere}\:=\:{\dfrac{2}{3}\:\pi\:r^3}}}}}$

Block in the values,

$\longrightarrow$ $\mathtt{\dfrac{2}{3}\:\times\:{\dfrac{22}{7}\:\times\:3\:\times\:\times\:3}}$

$\longrightarrow$ $\mathtt{\dfrac{44}{21}\:\times\:27}$

$\longrightarrow$ $\mathtt{\dfrac{44}{7}\:\times\:9}$

$\longrightarrow$ $\mathtt{\dfrac{396}{7}}$

$\longrightarrow$ $\mathtt{56.57}$

$\large{\boxed{\rm{\purple{Volume_{Hemisphere}\:=\:56.57\:m^3}}}}$

✪ Since, we have two ends in a cylinder, we will therefore have two hemisphere formed each on either ends.

$\mathtt{Volume\:of\:2\:hemisphere\:=\:2\:\times\:56.57}$

$\mathtt{Volume\:of\:2\:hemisphere\:=\:113.14\:cm^3}$

Volume of container :

✪ The volume of container will be the sum of volume of cylinder and volume of 2 hemisphere.

$\mathtt{Volume\:of\:cylinder\:+\:Volume\:of\:two\:hemisphere}$

$\longrightarrow$ $\mathtt{480.85\:+\:113.14}$

$\longrightarrow$ $\mathtt{593.14\:m^3}$

$\large{\boxed{\rm{\green{Volume\:of\:container\:=\:593.14\:m^3}}}}$

Answer 2

★Reference of attached image as diagram :

AnswEr :

$\:\bullet$ Let's break the question into three parts.As given in question oil container is made up of Cylinder with two hemispherical ends.So, the total volume is the sum of volume of cylinder, two hemispheres.

$\:\bullet$ Total height of container = 23m

$\rule{100}2$

Volume of CylinDer :

$\:\bullet\sf\ Height \: of \: cylinder = [Height \: of \: container - 2 \times\ Height \: of \: hemisphere]$

$\normalsize\ : \implies\sf\ Height = 23 - 6 = 17m$

$\:\bullet\sf\ Radius \: of \: cylinder = 3m$

$\normalsize\ : \implies{\boxed{\sf{Volume_{cylinder} = \pi r^2h}}} \\ \\ \normalsize\ : \implies\sf\ V_{c}= \frac{22}{7} \times\ 3 \times\ 3 \times\ 17 \\ \\ \normalsize\ : \implies\sf\ V_{c} = \frac{22 \times\ 3 \times\ 3 \times, 17}{7} \\ \\ \normalsize\ : \implies\sf\ V_{c} = \frac{198 \times\ 17}{7} \\ \\ \normalsize\ : \implies\sf\ V_{c} = \frac{\cancel{3366}}{\cancel{7}} \\ \\ \normalsize\ : \implies\sf\ V_{c} = 480.85$

$\normalsize\ : \implies{\underline{\boxed{\sf \green{Volume_{cylinder} = 480.85m^{3} }}}}$

$\rule{100}2$

Volume of HemiSphere :

$\:\bullet\:\sf\ Radius = Height = 3m$

$\normalsize\ : \implies{\boxed{\sf{Volume_{hemisphere} = \frac{2}{3} \pi r^3}}} \\ \\ \normalsize\ : \implies\sf\ V_{h} = \frac{2}{3} \times\ \frac{22}{7} \times\ 3 \times\ 3 \times\ 3 \\ \\ \normalsize\ : \implies\sf\ V_{h} = \frac{ 2 \times\ 22 \times\ 3 \times\ 3 \times\ \cancel{3}}{\cancel{3} \times\ 7} \\ \\ \normalsize\ : \implies\sf\ V_{h} = \frac{44}{7} \times\ 9 \\ \\ \normalsize\ : \implies\sf\ V_{h} = \frac{396}{7} \\ \\ \normalsize\ : \implies\sf\ V_{h} = 56.57$

$\normalsize\ : \implies{\underline{\boxed{\sf \blue{Volume_{hemisphere} = 56.57cm^{3} }}}}$

$\rule{100}{1}$

$\normalsize\ : \implies\sf\ Volume \: of \: 2 \: hemispheres = 2 \times\ Volume \: of \: hemisphere \\ \\ \normalsize\ : \implies\sf\ V_{2 \times\ h} = 2 \times\ 56.57 \\ \\ \normalsize\ : \implies\sf\ V_{2 \times\ h} = 113.14$

$\normalsize\ : \implies{\underline{\boxed{\sf \blue{Volume_{2 \times\ hemispheres} = 113.14m^{3} }}}}$

$\rule{100}2$

Volume of Container :

$\normalsize\ : \implies{\boxed{\sf{Volume_{container} = Volume_{cylinder} + Volume_{2 \times\ hemisphere} }}} \\ \\ \normalsize\ : \implies\sf\ V_{cont.} = 480.85 + 113.14 \\ \\ \normalsize\ : \implies\sf\ V_{cont.} = 593.99$

$\normalsize\ : \implies{\underline{\boxed{\sf \red{Volume_{container} = 593.99m^3}}}}$