Question

Plastic drum of cylindrical shape is made by melting spherical solid plastic balls of radius 1cm.find the number of balls required to make a drum of thickness 2cm,height 90cm and outer radius 30cm(drum has no lid)

Answer 1

here is your answer

Answer 2

$\underline { \blue { Outer \: dimensions \:of \:drum :}}$

$Height (H) = 90 \:cm$

$Radius (R) = 30 \:cm$

$\blue { Thickness \:of \:of \:the \:drum (w)=2\:cm}$

$\underline { \blue { Inner \: dimensions \:of \:drum :}}$

$height (h) = H - w \\= 90 - 2 \\= 88 \:cm$

$Radius (r) = R - w\\ = 30 - 2 \\= 28 \:cm$

$and \\ \orange { radius \:of \: a \:each \:ball = 1\:cm }$

/* According to the problem given */

If solid plastic spherical balls melted and cast into a lid less cylidercal drum */

$Let \:number \: of \: plastic \:balls \: required =n$

$n = \frac{ Volume_{(outer \: cylinder)} - Volume_{(Inner \:cylider) }}{Volume_{(sphere)}} </p><p>$

$= \frac{\pi R^{2} H - \pi r^{2} h }{ \frac{4}{3} \pi (radius)^{3}}$

$= \frac{\pi (R^{2} H - r^{2} h) }{ \frac{4}{3} \pi (radius)^{3}}$

$= \frac{ R^{2} H - r^{2} h }{ \frac{4}{3} (radius)^{3}}$

$= \frac{30^{2} \times 90 - 28^{2} \times 88 }{\frac{4}{3} \times 1^{3}}$

$= \frac{81000 - 68992}{\frac{4}{3}}\\= 12008 \times \frac{3}{4} \\= 9006$

Therefore.,

$\red { Required \:plastic \:balls } \green {=9006 }$

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