Question

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 1

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

• 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:-

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