Question

A right circular metallic cone of height 10cm and base of radius

2cm is melted and recast into a solid sphere. Find the radius of the

sphere.​

Answer 1

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$\small\boxed{\underline{\sf{Thies\:sum\:is\:from\:ch\:-\:\bf{Surface\:area\:and\:volume}}}}$

$\underline{\textsf{\textbf{Question\::}}}$

A right circular metallic cone of height 10 cm and base of radius 2 cm is melted and recast into a solid sphere. Find the radius of the sphere.

$\underline{\textsf{\textbf{Diagram\::}}}$

$\qquad\qquad$$\setlength{\unitlength}{1.2mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(18,1.6){\sf{2\:cm}}\put(9.5,10){\sf{10\:cm}}\end{picture}$

$\underline{\textsf{\textbf{Given\::}}}$

• Height(h) of cone = 10 cm
• Radius(r) of base of cone = 2 cm
• Cone is melted and recast into a sphere

$\underline{\textsf{\textbf{To\:find\::}}}$

• Radius of the sphere ?

$\underline{\textsf{\textbf{Solution\::}}}$

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

★ As we know that cone is recasted to ㅤsphere :

Therefore,

$\qquad:\implies\:\sf Volume_{(cone)} = Volume_{(sphere)}$

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

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

$\qquad:\implies\:\sf r^2 h = 4 r^3$

$\qquad:\implies\:\sf \dfrac{2\:\times\:2\:\times\:10}{4} = r^3$

$\qquad:\implies\:\sf r = \sqrt[3]{\dfrac{2\:\times\:2\:\times\:10}{2\:\times\:2}}$

$\qquad:\implies\:\sf r = \sqrt[3]{\dfrac{\cancel{2\:\times\:2}\:\times\:10}{\cancel{2\:\times\:2}}}$

$\qquad:\implies\:\sf r = \sqrt[3]{10}$

$\qquad:\implies\:\bf{ r = \red{2.15\:cm}}$

This is the required answer.

$\large\underline{\boxed{\sf{Radius\:of\:sphere\:\approx\:\rm\purple{2.2\:cm}}}}$

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

★ Formulae that we used above :

~ Volume of cone = 1/3πr²h

~ Volume of sphere = 4/3πr³

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

★ Formulae related to surface area and ㅤvolume :

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area\: formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$

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Answer 2

Answer:

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$\red{\small\boxed{\underline{\sf{Thies\:sum\:is\:from\:ch\:-\:\bf{Surface\:area\:and\:volume}}}}}$

$\orange{\underline{\textsf{\textbf{Question\::}}}}$

A right circular metallic cone of height 10 cm and base of radius 2 cm is melted and recast into a solid sphere. Find the radius of the sphere.

$\blue{\underline{\textsf{\textbf{Diagram\::}}}}$

$\qquad\qquad \setlength{\unitlength}{1.2mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(18,1.6){\sf{2\:cm}}\put(9.5,10){\sf{10\:cm}}\end{picture}$

$\green{\underline{\textsf{\textbf{Given\::}}}}$

• Height(h) of cone = 10 cm
• Radius(r) of base of cone = 2 cm
• Cone is melted and recast into a sphere

$\underline{\textsf{\textbf{To\:find\::}}}$

• Radius of the sphere ?

$\underline{\textsf{\textbf{Solution\::}}}$

ㅤㅤㅤㅤㅤ ━━━━━━━━━━

★ As we know that cone is recasted to sphere

Therefore,

$\qquad:\implies\:\sf Volume_{(cone)} = Volume_{(sphere)}$

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

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

$\qquad:\implies\:\sf r^2 h = 4 r^3$

$\qquad:\implies\:\sf \dfrac{2\:\times\:2\:\times\:10}{4} = r^3$

$\purple{\qquad:\implies\:\sf r = \sqrt[3]{\dfrac{2\:\times\:2\:\times\:10}{2\:\times\:2}}}$

$\pink{\qquad:\implies\:\sf r = \sqrt[3]{\dfrac{\cancel{2\:\times\:2}\:\times\:10}{\cancel{2\:\times\:2}}}}$

$\qquad:\implies\:\sf r = \sqrt[3]{10}$

$\purple{\qquad:\implies} \pink{\:\bf{ r =} \: \red{2.15\:cm}}$

This is the required answer.

$\large\underline{\boxed{\sf{Radius\:of\:sphere\:\approx\:\rm\purple{2.2\:cm}}}}$

━━━━━━━━━━

★ Formulae that we used above :

~ Volume of cone = 1/3πr²h

~ Volume of sphere = 4/3πr³

ㅤㅤㅤㅤㅤㅤ━━━━━━━━━━

★ Formulae related to surface area and ㅤvolume :

$\begin{gathered}\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area\: formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}\end{gathered}$

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