A right circular metalic cone of length 20 CM and base radius is melte

1.	A right circular metalic cone of length 20 CM and base radius is melted and recost in to a sphere. find the radius of sphere
Answer» $\bigstar\large \purple{ \underline{ \pmb{ \sf question... }}}$ A RIGHT circular metalic cone of LENGTH 20 cm and base radius 5 cm is melted and recost in to a SPHERE. find the radius of sphere $\bigstar\large \purple{ \underline{ \pmb{ \sf concept... }}}$ in the question we are GIVEN that there is a cone of height 20 cm and base radius 5 cm which is melted and casted into a sphere. We are asked to find the radius of the sphere so formed. Here, we need to UNDERSTAND that if cone is melted and sphere is formed then there Volumes will be same. We can equate volumes of both cone and sphere and find the value of radius. $\bigstar\large \purple{ \underline{ \pmb{ \sf formulas... }}}$ $\longrightarrow$ $\underline {\boxed{ \sf volume \: of \: cone = \frac{1}{3} \pi r ^{2} h }}$ here, r and h represents radius and height respectively. $\longrightarrow$ $\underline {\boxed {\sf volume \: of \: sphere = \frac{4}{3} \pi r ^{3} }}$ $\bigstar\large \purple{ \underline{ \pmb{ \sf solution... }}}$ $\red \star \: \sf \pink{volume \: of \: cone = volume \: of \: sphere} \: \red \star$ $: \implies \sf \dfrac{1}{3} \pi r ^{2} h = \dfrac{4}{3} \pi r ^{3}$ $: \implies \sf \dfrac{1}{3} \cancel \pi \times (5) ^{2} \times 20= \dfrac{4}{3} \cancel\pi r ^{3}$ $: \implies \sf \dfrac{1}{ \cancel3} \times \dfrac{ \cancel3}{ \cancel4} \times 25 \times \cancel{20} ^{5} = {r}^{3}$ $:\implies \sf r = \red{5 cm}$ So, the required value of radius is 5cm. ADDITIONAL INFORMATION:- $\boxed{\begin{minipage}{7 cm}More Formulas\\ \\$ \sf CSA \: of \: cone = \pi rl \\ \\TSA \:of\:cone=2\pi r(r+l) \\ \\CSA \:of\:cylinder=2\pi rh\\\\TSA\:of\:cylinder=2\pi r(r+h)\\\\volume\:of\:cyliner=\pi r^{2} h\\\\TSA\:of\:sphere=4\pi r^{3} \\\\TSA\:of\:hemisphere=3\pi r^{3} \\\\CSA\:of\:hemisphere=2\pi r^{3} $\end{minipage}}$

A right circular metalic cone of length 20 CM and base radius is melted and recost in to a sphere. find the radius of sphere

Answer»

$\bigstar\large \purple{ \underline{ \pmb{ \sf question... }}}$

A RIGHT circular metalic cone of LENGTH 20 cm and base radius 5 cm is melted and recost in to a SPHERE. find the radius of sphere

$\bigstar\large \purple{ \underline{ \pmb{ \sf concept... }}}$

in the question we are GIVEN that there is a cone of height 20 cm and base radius 5 cm which is melted and casted into a sphere. We are asked to find the radius of the sphere so formed.
Here, we need to UNDERSTAND that if cone is melted and sphere is formed then there Volumes will be same.
We can equate volumes of both cone and sphere and find the value of radius.

$\bigstar\large \purple{ \underline{ \pmb{ \sf formulas... }}}$

$\longrightarrow$ $\underline {\boxed{ \sf volume \: of \: cone = \frac{1}{3} \pi r ^{2} h }}$

here, r and h represents radius and height respectively.

$\longrightarrow$ $\underline {\boxed {\sf volume \: of \: sphere = \frac{4}{3} \pi r ^{3} }}$

$\bigstar\large \purple{ \underline{ \pmb{ \sf solution... }}}$

$\red \star \: \sf \pink{volume \: of \: cone = volume \: of \: sphere} \: \red \star$

$: \implies \sf \dfrac{1}{3} \pi r ^{2} h = \dfrac{4}{3} \pi r ^{3}$

$: \implies \sf \dfrac{1}{3} \cancel \pi \times (5) ^{2} \times 20= \dfrac{4}{3} \cancel\pi r ^{3}$

$: \implies \sf \dfrac{1}{ \cancel3} \times \dfrac{ \cancel3}{ \cancel4} \times 25 \times \cancel{20} ^{5} = {r}^{3}$

$:\implies \sf r = \red{5 cm}$

So, the required value of radius is 5cm.

ADDITIONAL INFORMATION:-

$\boxed{\begin{minipage}{7 cm}More Formulas\\ \\$ \sf CSA \: of \: cone = \pi rl \\ \\TSA \:of\:cone=2\pi r(r+l) \\ \\CSA \:of\:cylinder=2\pi rh\\\\TSA\:of\:cylinder=2\pi r(r+h)\\\\volume\:of\:cyliner=\pi r^{2} h\\\\TSA\:of\:sphere=4\pi r^{3} \\\\TSA\:of\:hemisphere=3\pi r^{3} \\\\CSA\:of\:hemisphere=2\pi r^{3} $\end{minipage}}$