Question

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

Answer 1

$\huge \fcolorbox{black}{pink}{Solution.}$

$\bf{For \: Sphere}$

$\rm Radius \: (r) = 4.2 \: cm$

$\therefore \sf Volume \: of \: sphere = \frac{4}{3}\pi r {}^{2}$

$\: \: \: \: \: \: \sf = \frac{4}{3}\pi (4.2) {}^{3} \: cm {}^{3}$

$\bf {For \: Cylinder}$

$\rm Radius(R) = 6 \: cm$

$\sf{Let \: the \: height \: of \: the \: cylinder \: be \: H \: cm.}$

$\rm Then, \:$

$\sf{Volume \: of \: Cylinder = \pi r {}^{2} H = \pi(6) {}^{2} H \: cm {}^{3} }$

$\rm According \: to \: the \: Question,$

$\sf{Volume \: of \: the \: metallic \: sphere \: must \: be}$

$\rm = Volume \: of \: the \: cylinder$

$\implies \: \: \: \: \sf\frac{4}{3} \: \pi(4.2) {}^{3} = \pi(6) {}^{2} h$

$\rm Dividing \: both \: sides \: by \: \pi \: and \: cross \: multiplying,$

$\: \: \: \: \: \: \: \: \sf{3(6) {}^{2} H = 4(4.2) {}^{3} }$

$\implies \: \: \: \: \: \: \sf{ H = \frac{4(4.2) {}^{3}} {3(6) {}^{2}}}$

$\implies \: \: \rm H = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6}$

$\: \: \: \: \: \: \: \sf = 4 \times 1.4 \times 0.7 \times 0.7$

$\: \: \: \: \: \: \: \: \: \rm = 4 \times \frac{14}{10} \times \frac{7}{10} \times \frac{7}{10}$

$\: \: \: \: \: \: \sf = \frac{56 \times 49}{1000} = \frac{2744}{1000}$

$\implies \rm \: \: \: \: \: \: \: \: H = 2.744$

$\sf Hence, \: the \: height \: of \: the \: cylinder \: is \: 2.744 \\ \: \sf \:cm.$