Physics, asked by prateek3611, 4 months ago

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Answered by BrainlyEmpire
5

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

Answered by Anonymous
0

Answer:

Diameter=14 cm

Radius=7cm

Height=20cm

Volume of cylinder=πr²h

= 22/7*7²*20

= 22/7*7*7*20

= 22*7*20

= 3080cm³

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