Question

milk powder comes in a cylindrical container whose base has a diameter of 14cm and height of 20 cm find its volume.
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Answer 1

$\frak{Given} \begin{cases} & \sf{Base\:Diameter\:of\: Cylindrical\:container = \bf{14\:cm}} \\ & \sf{Base\:radius\:of\: Cylindrical\:container = \bf{7\:cm}} \\ & \sf{Height\:of\: cylindrical\:container = \bf{20\:cm}} \end{cases}$

To find: Volume of milk powder in container?

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• Volume of Milk powder = Volume of container

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★ Now, Finding Volume of cylindrical container,

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$\dag\;{\underline{\frak{Volume\:of\:cylinder\:is\:given\:by}}}$

$\star\;{\boxed{\sf{\pink{Volume_{\;(cylinder)} = \pi r^2 h}}}}$

where,

• r & h are radius and height of cylinder respectively.

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$:\implies\sf Volume_{\;(container)} = \dfrac{22}{7} \times 7 \times 7 \times 20$

$:\implies\sf Volume_{\;(container)} = \dfrac{22}{ \cancel{7}} \times \cancel{7} \times 7 \times 20$

$:\implies\sf Volume_{\;(container)} = 22 \times 7 \times 20$

$:\implies\sf Volume_{\;(container)} = 154 \times 20$

$:\implies{\underline{\boxed{\frak{\purple{Volume_{\;(container)} = 3080\:cm^3}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Volume\:of\:milk\:powder\:in\:container\:is\: \bf{3080\:cm^3}.}}}$

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$\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\:Formula\:Related\:to\:cylinder\:\bigstar}}}}$

• $\sf Area\:of\:base\:of\:cylinder = \bf{\pi r^2}$

• $\sf Total\:Surface\:area\:of\:cylinder = \bf{2 \pi r(r + h)}$

• $\sf Curved\:Surface\:area\:of\:cylinder = \bf{2 \pi rh}$

• $\sf Volume\:of\:cone = \bf{ \dfrac{1}{3} \times Volume_{\:(cylinder)}}$

Answer 2

A N S W E R :

• Volume of the milk container is 3080 cm³.

$\underline{\underline{\sf \bigstar\qquad Given :\qquad \bigstar}}$

$\frak{\pink{We \: have}}\begin{cases} \textsf{\red{Base diameter of cylindrical container = \textbf{14 cm}}}\\ \textsf{\purple{Height of the cylindrical container = \textbf{ 20 cm}}}\end{cases}$

$\underline{\underline{\sf \bigstar\qquad To \:Find:\qquad \bigstar}}$

We have to find the volume of the cylindrical container.

$\underline{\underline{\sf \bigstar\qquad Solution : \qquad \bigstar}}$

$\ddag \: \underline{\frak{To \: find \: the \: volume \: of \: cylindrical \: tank \: first \: we \: have \: to \: find \: the \: base \: radius \: of \: cylindrical \: \: container}} \: \ddag$

$:\implies \sf Base \: radius \: of \: cylindrical \: container = \dfrac{Base \: diameter \: of \: cylindrical \: container}{2}$

$:\implies \sf Base \: radius \: of \: cylindrical \: container = \dfrac{14}{2}$

$:\implies \underline{ \boxed{\sf Base \: radius \: of \: cylindrical \: container =7 \: cm}}$

$\therefore\:\underline{\textsf{The base radius of cylindrical container is \textbf{7 cm}}}.$

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$\ddag \: \underline{\frak{Now, let's \: find \: the \: volume \: of \: cylindrical \: container :}} \: \ddag$

$\dashrightarrow\:\:\sf Volume \: of \: cylindrical \: container = \pi r^2 h$

$\dashrightarrow\:\:\sf Volume \: of \: cylindrical \: container = \dfrac{22}{7} \times 7 \times 7 \times 20$

$\dashrightarrow\:\:\sf Volume \: of \: cylindrical \: container = 22\times 7 \times 20$

$\dashrightarrow\:\: \underline{ \boxed{\sf Volume \: of \: cylindrical \: container = 3080 \: {cm}^{3}}}$

$\therefore\:\underline{\textsf{The volume of cylindrical container is \textbf{3080 cm ^{\text3} }}}.$

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$\underline{\underline{\sf \bigstar\qquad Some \: important \: formulas \: related \: to \: it :\qquad \bigstar}}$

$\boxed{\bigstar{\sf \ Cylinder :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cylinder= \pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ cylinder= 2\pi r h\\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ cylinder= 2\pi r (h+r)$

$\boxed{\bigstar{\sf \ Cone :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Cone= \dfrac{1}{3}\pi r^2 h \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Cone = \pi r l \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Cone = \pi r (l+r) \\ \\ \\ \sf {\textcircled{\footnotesize4}} Slant \ Height \ of \ cone (l)= \sqrt{r^2+h^2}$

$\boxed{\bigstar{\sf \ Hemisphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Hemisphere= \dfrac{2}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Curved \ surface\ Area \ of \ Hemisphere = 2 \pi r^2 \\ \\ \\ \sf {\textcircled{\footnotesize3}} Total \ surface \ Area \ of \ Hemisphere = 3 \pi r^2$

$\boxed{\bigstar{\sf \ Sphere :- }}\\ \\\sf {\textcircled{\footnotesize1}} Volume \ of \ Sphere= \dfrac{4}{3}\pi r^3 \\ \\ \\ \sf {\textcircled{\footnotesize2}}\ Surface\ Area \ of \ Sphere = 4 \pi r^2$