Question

A metallic sphere of radius 4.2 cm is metaled and recast into the shape of a cylinder of radius 6cm. Find the height of the cylinder.

Answer 1

Solution :-

Sphere :-

Radius = 4.2 cm

$\star\bf Volume \ of \ sphere = \dfrac{4}{3} \pi r^3$

$: \implies \sf \dfrac{4}{3} \times \pi \times 4.2 \times 4.2 \times 4.2$

Cylinder :-

Radius = 6 cm

$\star\bf Volume \ of \ cylinder = \pi r^2h$

$: \implies \sf \pi \times 6 \times 6 \times h$

According to the question :-

Volume of cylinder = Volume of sphere

$: \implies \sf \dfrac{4}{3} \times \pi \times 4.2 \times 4.2 \times 4.2 = \pi \times 6 \times 6 \times h$

$: \implies \sf 4 \times 1.4 \times 4.2 \times 4.2 = 6 \times 6 \times h$

$: \implies \sf 36h = 98.784$

$: \implies \sf h = \dfrac{98.784}{36}$

$: \implies \bf h = 2.744$

Height of the cylinder :- 2.744 cm

Answer 2

★Given:-

• Radius of sphere =4.2 cm
• Radius of cylinder =6 cm
• Volume of cone = Volume of cylinder

★To find:-

• Height of the cylinder .

●Solution:-

• First come to sphere :

Diagram:-

$\setlength{\unitlength}{1.2cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\qbezier(-2.3,0)(0,-1)(2.3,0)\qbezier(-2.3,0)(0,1)(2.3,0)\thinlines\qbezier (0,0)(0,0)(0.2,0.3)\qbezier (0.3,0.4)(0.3,0.4)(0.5,0.7)\qbezier (0.6,0.8)(0.6,0.8)(0.8,1.1)\qbezier (0.9,1.2)(0.9,1.2)(1.1,1.5)\qbezier (1.2,1.6)(1.2,1.6)(1.38,1.9)\put(-0.1,1){\bf 4.2\:cm }\end{picture}$

• Radius(r)=4.2 cm.

As we know that in a Sphere :

$\bigstar\underline{\boxed{\bf Volume=\dfrac {1}{3}\pi r^3}}\bigstar$

• Substitute the values :

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Volume_{(Sphere)}=\dfrac {4}{3}\pi (4.2)^3$

$\\\qquad\quad\displaystyle\tt {:}\longrightarrow Volume_{(Sphere )}=\dfrac {4}{3}\pi \times 4.2\times 4.2\times 4.2$

________________

Now come to cylinder

• Radius (r) = 6 cm.

Diagram:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\footnotesize{7cm}}\put(9,17.5){\sf{h}}\end{picture}$

As we know that in a Cylinder :

$\bigstar{\boxed{\bf Volume=\pi r^2h}}\bigstar$

• Substitute the values

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Volume_{(Cylinder )}=\pi (6)^2h$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow Volume_{(Cylinder )}=\pi \times6\times 6\times h$

___________________

ATQ,

$\bf {:}\longrightarrow Volume_{(Sphere )}=Volume_{(Cylinder)}$

• Substitute the values :

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {4}{3}\bcancel {\pi} \times 4.2\times 4.2\times 4.2=\bcancel {\pi} \times 6\times 6\times h$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac{4}{\cancel {3}}\times \cancel {4.2}\times 4.2\times 4.2=36h$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 36h=4\times (4.2)^2\times 1.4$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 36h=98.784$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow h =\cancel{\dfrac {98.784}{36}}$

$\\\qquad\quad\displaystyle {:}\longrightarrow\large {\boxed{\bf{ h=2.74\:cm}}}$

$\\\\\therefore{\underline{\bf {Height\:of\:Cylinder\:is\:2.74\:cm.}}}$