Question

a shere and a cube have same surface are . find the ratio of volume of sphere to that of cube​

Answer 1

Answer:

Volume of sphere : Volume of cube = √6 : √π.

Step-by-step explanation:

Let r and a be the radius of the sphere and edge of the cube respectively.

Given, Surface area of sphere = Surface area of cube

4πr^2 = 6a^2

(r/a)^2 = 3 / 2π

r / a = √(3/2π)

Volume of sphere / Volume of cube = (4/3)πr^3 / a^3 = (4π/3)(r/a)^3

= (4π/3)(√(3/2π))^3

= (4π/3)(3/2π)(√(3/2π))

= 2√(3/2π)

= √(4x3/2π)

= √(6/π)

Thus, Volume of sphere : Volume of cube = √6 : √π.

Answer 2

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{TSA_{(sphere)} = TSA_{(cube)} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \:To\:find - \begin{cases} &\sf{Volume_{(sphere)} : Volume_{(cube)} }  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$1. \: \: \: \boxed{ \sf{ \: TSA_{(sphere)} = 4\pi \: {r}^{2} }}$

$2. \: \: \: \boxed{ \sf{ \: TSA_{(cube)} = 6 {(edge)}^{2} }}$

$3. \: \: \: \boxed{ \sf{ \: Volume_{(sphere)} = \dfrac{4}{3} \pi \: {r}^{3} }}$

$4. \: \: \: \boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\bf \:Let - \begin{cases} &\sf{radius \: of \: sphere = \: r} \\ &\sf{edge \: of \: cube \: = \: a} \end{cases}\end{gathered}\end{gathered}$

According to statement,

$\rm :\longmapsto\:TSA_{(sphere)} = TSA_{(cube)}$

$\rm :\longmapsto\:4\pi \: {r}^{2} = 6 {a}^{2}$

$\rm :\longmapsto\: {a}^{2} = \dfrac{2}{3} \pi \: {r}^{2}$

$\bf\implies \:a = \dfrac{ \sqrt{2} }{ \sqrt{3}} \times \sqrt{\pi} \times r - - (1)$

Now,

$\rm :\longmapsto\:Volume_{(sphere)} : Volume_{(cube)}$

$\rm :\longmapsto\: = \: \dfrac{4}{3} \pi \: {r}^{3} : {a}^{3}$

$\rm :\longmapsto\: = \: \dfrac{4}{3} \pi \: {r}^{3} : {\bigg(\dfrac{ \sqrt{2} }{ \sqrt{3} } \times \sqrt{\pi} \times r \bigg) }^{3}$

$\rm :\longmapsto\: = \: \dfrac{4}{\cancel 3} \cancel \pi \: \cancel{{r}^{3}} : \dfrac{2 \sqrt{2} }{\cancel 3 \sqrt{3} } \times\cancel \pi \: \sqrt{\pi} \times \cancel { {r}^{3}}$

$\rm :\longmapsto\: = \: \cancel 2\times\cancel {\sqrt{2}} \times \sqrt{2} \times \sqrt{3} \: \: : \: \: \cancel 2 \times \cancel {\sqrt{2}} \times \sqrt{\pi}$

$\rm :\longmapsto\: = \: \sqrt{6} \: : \: \sqrt{\pi}$

$\rm :\implies\:Volume_{(sphere)} : Volume_{(cube)} = \sqrt{6} : \sqrt{\pi}$

Additional Information :-

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$

$\boxed{ \sf{ \: CSA{(cube)} = 4 \times {(edge)}^{2} }}$

$\boxed{ \sf{ \: CSA{(cuboid)} = 2(l + b) \times h}}$

$\boxed{ \sf{ \: TSA{(cylinder)} = 2\pi \: r(h + r)}}$

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$

$\boxed{ \sf{ \: TSA{(cone)} = {6(edge)}^{2} }}$