Question

A solid consisting of right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

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Answer 1

${ \large{ \sf{ \underbrace{\underline{\bigstar \: Answer }}}}}$

$\textbf{Volume of Water left in Cylinder =}$ $\sf\color{darkviolet}1131428.57 cm³$

$\underline{\bf{Explanation\::}}$

$\bold\pink{\fbox{\sf{Given: }}}$

$\textbf{Height\: of \:Cone}$

$\underline{\bf{h =}}$$\sf\color{darkviolet}120$

$\textbf{Radius\: of \:Cone,\: r \:= 60 \:cm}$

$\bold{\boxed{Radius\: of \:Hemisphere\:, r\: =\: 60 \:cm}}$

${ \large{ \boxed{ \red{ \underline{ \bf \:Height\: of \:Cylinder, }}}}}$

H =$\sf\color{red}180 cm$

Radius of Cylinder, r = $\sf\color{red}60 cm$

$\underline{\bf{To\: find:}}$ Volume of water left in cylinder

$\textbf{Volume of water left in cylinder = volume of cylinder - }$

$\textbf{(volume of cone + volume of hemisphere )}$

$= \pi r^2H-(\frac{1}{3}\pi r^2h+\frac{2}{3}\pi r^3)$

$= \pi r^2(H-(\frac{1}{3}h+\frac{2}{3}r))$

$= \frac{22}{7}\times60^2(180-(\frac{1}{3}\times120+\frac{2}{3}\times60))$

$= \frac{22}{7}\times3600(180-(40+40))$

$= \frac{22}{7}\times3600\times100 1131428.57 cm³$

${ \large{ \sf{ \underbrace{\underline{\bigstar\:Therefore,}}}}}$

$\boxed{\bf{Volume\: of\: Water\: left\: in \:Cylinder =}}$

$\mapsto\color{red}\bf{I1131428.57}$

Thankyou :)

Answer 2

Question:-

A solid consisting of right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Required Answer:-

Given:-

$\\ \sf{ \rightarrow{Height \: of \: the \: cone = 120cm}}$

$\sf{ \rightarrow{Radius \: of \: the \: cone = 60cm}}$

$\sf{ \rightarrow{Radius \: of \: the \: hemisphere= 60cm}}$

$\sf{ \rightarrow{Radius \: of \: the \: cylinder = 60cm}}$

$\sf{ \rightarrow{Height \: of \: the \: cylinder= 180cm}}$

To Find:-

$\\ \sf{ \rightarrow{The \: volume \: of \: water \: left \: in \: the \: cylinder}}$

♡Solution:-

First, we have to find the volumes of the cone, the hemisphere and the cylinder. After finding their volumes, we will remove the volumes of the solids from the volume of the cylinder. Then, we can find the volume of the water left in the cylinder.

Now,

We know that,

$\\ \underline{ \boxed{ \bold{Volume \: of \: cone} = \frac{1}{3} \pi {r}^{2} h}}$

According to the question,

$\\ \sf{Volume \: of \: the \: cone = \bold{(\frac{1}{3} \times \frac{22}{7} \times {60}^{2} \times 120})}cm {}^{3}$

$\sf{ \therefore{Volume \: of \: the \: cone = \bold{ 452571.43{cm}^{3} }}}$

Again,

$\\ \underline{ \boxed{ \sf{\bold{Volume \: of \: a \: hemisphere }= \frac{2}{3} \pi r {}^{3} }}}$

According to the question,

$\\ \sf{ Volume \: of \: the \: hemisphere = \bold{ ( \frac{2}{3} \times \frac{22}{7} \times 60 {}^{3} {cm}^{3} }}$

$\sf{ \therefore{Volume \: of \: the \: hemisphere = \bold{452571.43cm {}^{3} }}}$

Again,

$\\ \underline{ \boxed{ \sf{ \bold{Volume \: of \: cylinder} = \pi r {}^{2} h}}}$

According to the question,

$\\ \sf{volume \: of \: the \: cylinder = \bold{ \frac{22}{7} \times 60 {}^{2} \times 180cm {}^{3} }}$

$\sf{ \therefore{Volume \: of \: the \: cylinder = \bold{2036571.43cm {}^{3} }}}$

Therefore,

$\\ \sf{Volume \: of \: the \: left \: water = (2036571.43 - 452571.43 - 452571.43) {cm}^{3} }$

$\sf{ \therefore{\sf{Volume \: of \: the \: left \: water = \red{1131428.573 {cm}^{3} }}}}$

$\\ \underline{ \boxed{ \rm{ \green{Hence \: the \: volume \: of \: the \: left \: water \: \bold{1131428.573 {cm}^{3} }}}}}$

N. B. :- Kindly see the attachment for more information.