Question

can 6 dm in diameter and 12 dm high is full of grape juice. If the full content is to be transferred to smaller container 500 mL in capacity , how many such container are needed? Use π = 3.14 Note: 1 dm^3=1 L=1000 cc=1000 mL

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the Concept of Volume of Cylinder has been used. We see that can is cylindrical in shape. So firstly we can find volume of the can which will be its capacity. Now we can change this capacity into litre. Then we are given the capacity of smaller container which also can be changed in litre. Now then we can divide the capacity of can by capacity of smaller container to find the number of smaller container required this is because volume is fixed for a substance.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Volume\;of\;Cylinder\;=\;\bf{\pi r^{2}h}}}}$

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★ Solution :-

Given,

» Diameter of can = 6 dm

» Radius of can = ½ × 6 = 3 dm

» Height of can = 12 dm

» Capacity of smaller container = 500 mL

Firstly, let's find the volume of can. This is given as,

$\\\;\sf{:\rightarrow\;\;Volume\;of\;Cylinder\;=\;\bf{\pi r^{2}h}}$

By applying values, we get

$\\\;\sf{:\Longrightarrow\;\;Volume\;of\;Cylinder_{(Can)}\;=\;\bf{3.14\:\times\:(3)^{2}\:\times\:12}}$

$\\\;\bf{:\Longrightarrow\;\;Volume\;of\;Cylinder_{(Can)}\;=\;\bf{\orange{339.12\;\;dm^{3}}}}$

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~ For the change in capacities of different containers in Litre ::

We know that,

$\\\;\sf{\rightarrow\;\;\green{1\;dm^{3}\;=\;1\;L}}$

Then,

$\\\;\bf{\odot\;\;\red{339.12\;dm^{3}\;=\;339.12\;L}}$

Also,

$\\\;\sf{\rightarrow\;\;\green{1\;mL\;=\;\dfrac{1}{1000}\:L}}$

$\\\;\sf{\rightarrow\;\;500\;mL\;=\;\dfrac{500}{1000}\:L}$

$\\\;\sf{\rightarrow\;\;50\;mL\;=\;\dfrac{1}{2}\:L}$

$\\\;\bf{\rightarrow\;\;\red{50\;mL\;=\;0.5\:L}}$

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~ For the Number of smaller containers ::

$\\\;\sf{\mapsto\;\;\blue{Number\;of\;Containers\;=\;\bf{\dfrac{Volume\;of\;Can}{Volume\;of\;smaller\;Container}}}}$

By applying values, we get

$\\\;\sf{\mapsto\;\;Number\;of\;Containers\;=\;\bf{\dfrac{339.12}{0.5}}}$

$\\\;\bf{\mapsto\;\;Number\;of\;Containers\;=\;\bf{\purple{678.24\;\approx\;678}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\:number\;\:of\;\:cans\;\:needed\;=\;\bf{678\;cans}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;CSA\;of\;Cylinder\;=\;2\pi rh}$

$\\\;\sf{\leadsto\;\;TSA\;of\;Cylinder\;=\;2\pi rh\;+\;2\pi r^{2}}$

$\\\;\sf{\leadsto\;\;Volume\;of\;hollow\;cylinder\;=\;\pi (R^{2}\:-\:r^{2})h}$

$\\\;\sf{\leadsto\;\;CSA\;of\;hollow\;cylinder\;=\;2\pi (R\:-\:r)h}$

$\\\;\sf{\leadsto\;\;TSA\;of\;hollow\;cylinder\;=\;2\pi (R\:-\:r)h\;+\;2\pi (R\:-\:r)}$

Answer 2

\begin{gathered}\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}\end{gathered}

UnderstandingtheConcept

Here the Concept of Volume of Cylinder has been used. We see that can is cylindrical in shape. So firstly we can find volume of the can which will be its capacity. Now we can change this capacity into litre. Then we are given the capacity of smaller container which also can be changed in litre. Now then we can divide the capacity of can by capacity of smaller container to find the number of smaller container required this is because volume is fixed for a substance.

Let's do it !!

______________________________________________

★ Formula Used :-

\begin{gathered}\\\;\boxed{\sf{\pink{Volume\;of\;Cylinder\;=\;\bf{\pi r^{2}h}}}}\end{gathered}

VolumeofCylinder=πr

2

h

______________________________________________

★ Solution :-

Given,

» Diameter of can = 6 dm

» Radius of can = ½ × 6 = 3 dm

» Height of can = 12 dm

» Capacity of smaller container = 500 mL

Firstly, let's find the volume of can. This is given as,

\begin{gathered}\\\;\sf{:\rightarrow\;\;Volume\;of\;Cylinder\;=\;\bf{\pi r^{2}h}}\end{gathered}

:→VolumeofCylinder=πr

2

h

By applying values, we get

\begin{gathered}\\\;\sf{:\Longrightarrow\;\;Volume\;of\;Cylinder_{(Can)}\;=\;\bf{3.14\:\times\:(3)^{2}\:\times\:12}}\end{gathered}

:⟹VolumeofCylinder

(Can)

=3.14×(3)

2

×12

\begin{gathered}\\\;\bf{:\Longrightarrow\;\;Volume\;of\;Cylinder_{(Can)}\;=\;\bf{\orange{339.12\;\;dm^{3}}}}\end{gathered}

:⟹VolumeofCylinder

(Can)

=339.12dm

3

______________________________________________

~ For the change in capacities of different containers in Litre ::

We know that,

\begin{gathered}\\\;\sf{\rightarrow\;\;\green{1\;dm^{3}\;=\;1\;L}}\end{gathered}

→1dm

3

=1L

Then,

\begin{gathered}\\\;\bf{\odot\;\;\red{339.12\;dm^{3}\;=\;339.12\;L}}\end{gathered}

⊙339.12dm

3

=339.12L

Also,

\begin{gathered}\\\;\sf{\rightarrow\;\;\green{1\;mL\;=\;\dfrac{1}{1000}\:L}}\end{gathered}

→1mL=

1000

1

L

\begin{gathered}\\\;\sf{\rightarrow\;\;500\;mL\;=\;\dfrac{500}{1000}\:L}\end{gathered}

→500mL=

1000

500

L

\begin{gathered}\\\;\sf{\rightarrow\;\;50\;mL\;=\;\dfrac{1}{2}\:L}\end{gathered}

→50mL=

2

1

L

\begin{gathered}\\\;\bf{\rightarrow\;\;\red{50\;mL\;=\;0.5\:L}}\end{gathered}

→50mL=0.5L

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~ For the Number of smaller containers ::

\begin{gathered}\\\;\sf{\mapsto\;\;\blue{Number\;of\;Containers\;=\;\bf{\dfrac{Volume\;of\;Can}{Volume\;of\;smaller\;Container}}}}\end{gathered}

↦NumberofContainers=

VolumeofsmallerContainer

VolumeofCan

By applying values, we get

\begin{gathered}\\\;\sf{\mapsto\;\;Number\;of\;Containers\;=\;\bf{\dfrac{339.12}{0.5}}}\end{gathered}

↦NumberofContainers=

0.5

339.12

\begin{gathered}\\\;\bf{\mapsto\;\;Number\;of\;Containers\;=\;\bf{\purple{678.24\;\approx\;678}}}\end{gathered}

↦NumberofContainers=678.24≈678

\begin{gathered}\\\;\underline{\boxed{\tt{Hence,\;\:number\;\:of\;\:cans\;\:needed\;=\;\bf{678\;cans}}}}\end{gathered}

Hence,numberofcansneeded=678cans

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★ More to know :-

\begin{gathered}\\\;\sf{\leadsto\;\;CSA\;of\;Cylinder\;=\;2\pi rh}\end{gathered}

⇝CSAofCylinder=2πrh

\begin{gathered}\\\;\sf{\leadsto\;\;TSA\;of\;Cylinder\;=\;2\pi rh\;+\;2\pi r^{2}}\end{gathered}

⇝TSAofCylinder=2πrh+2πr

2

\begin{gathered}\\\;\sf{\leadsto\;\;Volume\;of\;hollow\;cylinder\;=\;\pi (R^{2}\:-\:r^{2})h}\end{gathered}

⇝Volumeofhollowcylinder=π(R

2

−r

2

)h

\begin{gathered}\\\;\sf{\leadsto\;\;CSA\;of\;hollow\;cylinder\;=\;2\pi (R\:-\:r)h}\end{gathered}

⇝CSAofhollowcylinder=2π(R−r)h

\begin{gathered}\\\;\sf{\leadsto\;\;TSA\;of\;hollow\;cylinder\;=\;2\pi (R\:-\:r)h\;+\;2\pi (R\:-\:r)}\end{gathered}

⇝TSAofhollowcylinder=2π(R−r)h+2π(R−r)