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.


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Answer 1

Answer :-

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of Volume of Sphere and Volume of Cylinder has been used. We see we are given that the metallic sphere is recasted into cylinder. So its volume is not gonna change due to this the because Volume is the amount of matter.

Let's do it !!

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★ Equations Used :-

$\\\;\large{\boxed{\sf{Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\:\times\:\pi r^{3}}}}}$

$\\\;\large{\boxed{\sf{Volume\;of\;Cylinder\;=\;\bf{Volume\;of\;Sphere}}}}$

$\\\;\large{\boxed{\sf{Volume\;of\;Cylinder\;=\;\bf{\pi r^{2} h}}}}$

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★ Solution :-

Given,

» Radius of the sphere = 4.2 cm

» Radius of the Cylinder = 6 cm

• Let the height of the Cylinder be 'h' .

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~ For the Volume of Sphere :-

$\\\;\quad\sf{:\Longrightarrow\;\;Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\:\times\:\pi r^{3}}}$

$\\\;\quad\sf{:\Longrightarrow\;\;Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\;\times\;\dfrac{22}{7}\;\times\;(4.2)^{3}}}$

$\\\;\quad\sf{:\Longrightarrow\;\;Volume\;of\;Sphere\;=\;\bf{\dfrac{4}{3}\;\times\;\dfrac{22}{7}\;\times\;74.1}}$

$\\\;\quad\sf{:\Longrightarrow\;\;Volume\;of\;Sphere\;=\;\underline{\underline{\bf{310.5\;\;cm^{3}}}}}$

This is an approximate answer by rounding off the decimal digits.

$\\\;\underline{\boxed{\tt{Volume\;\;of\;\;Sphere\;\;=\;\bf{310.5\;\;cm^{3}}}}}$

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~ For the Height of the Cylinder :-

From question itself :-

$\\\;\quad\sf{:\mapsto\;\;Volume\;of\;Cylinder\;=\;\bf{Volume\;of\;Sphere}}$

Also,

$\\\;\quad\sf{:\mapsto\;\;Volume\;of\;Cylinder\;=\;\bf{\pi r^{2} h}}$

Combining these equations we get,

$\\\;\quad\sf{:\mapsto\;\;\pi r^{2} h\;=\;\bf{\dfrac{4}{3}\:\times\:\pi r^{3}}}$

$\\\;\quad\sf{:\mapsto\;\;\pi r^{2} h\;=\;\bf{310.5}}$

$\\\;\quad\sf{:\mapsto\;\;\dfrac{22}{7}\;\times\;(6)^{2}\;\times\;h\;=\;\bf{310.5}}$

$\\\;\quad\sf{:\mapsto\;\;\dfrac{22}{7}\;\times\;36\;\times\;h\;=\;\bf{310.5}}$

$\\\;\quad\sf{:\mapsto\;\;h\;=\;\bf{\dfrac{310.5\;\times\;7}{22\;\times\;36}}}$

$\\\;\quad\sf{:\mapsto\;\;h\;=\;\bf{\dfrac{2173.5}{792}}}$

$\\\;\quad\sf{:\mapsto\;\;h\;=\;\underline{\underline{\bf{2.74\;\;cm}}}}$

$\\\;\large{\underline{\underline{\rm{Thus,\;the\;height\;of\;the\;cylinder\;is\;\;\boxed{\bf{2.74\;\;cm}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Hemisphere\;=\;\dfrac{2}{3}\;\times\;\pi r^{3}}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Cuboid\;=\;Length\:\times\:Breadth\:\times\:Height}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Cube\;=\;(Side)^{3}}$

$\\\;\sf{\leadsto\;\;\;Volume\;of\;Cone\;=\;\dfrac{1}{3}\;\times\;\pi r^{2} h}$

$\\\;\sf{\leadsto\;\;\;TSA\;of\;Sphere\;=\;CSA\;of\;Sphere\;=\;4\pi r^{2}}$

$\\\;\sf{\leadsto\;\;\;CSA\;of\;Cylinder\;=\;2\pi rh}$

$\\\;\sf{\leadsto\;\;\;TSA\;of\;Cylinder\;=\;2\pi rh\;+\;2\pi r^{2}}$

Answer 2

$\red{\:❥ᴀɴsᴡᴇʀ}$

see this attachment you will get your answer.

I hope it is helpful for you

@ Aman jha