Question

A solid metallic cylinder of radius 12 cm and height
20 cm is melted into 8 cylinders of radius 6 cm. Find
the height of the small cylinders.

Only mods/stars are allowed to ans xD ​

Answer 1

✥ Question :–

Q) A solid metallic cylinder of radius 12 cm and height 20 cm is melted into 8 cylinders of radius 6 cm. Find the height of the small cylinders.

✥ Answer :–

$\frak{Given}\begin{cases}\sf{Radius\:of\:big\:Cylinder=\bf{12\:cm.}}\\\sf{Height\:of\:big\:Cylinder=\bf{20\:cm.}}\\\sf{\bf{8}\:\sf new\:cylinders\:are\:formed.}\\\sf{Radius\:of\:new\:Cylinder=\bf{6\:cm.}}\end{cases}$

$\underline{\bigstar\;\bf To\:Find:-}$

• The height of the new cylinder formed.

$\underbrace{\underline{\bf\bigstar\:Required\:Solution:-}}$

» Let's denote the dimensions of the original cylinder by capital letters and the new cylinder by small letters.

• R = 12 cm
• H = 20 cm
• r = 6 cm
• h = ?

» Since 8 new cylinders are formed by melting the original cylinder so, the volume of the 8 new cylinders combined would be equal to the volume of big/original cylinder.

$\odot \: \boxed{ \sf volume _{ \: orignal \: cyl} =8 \times volume _{ \: new \: cyl}}$

Using the formula, to find the volume of a cylinder -

$\odot \: \boxed{ \sf volume = \pi {r}^{2} h}$

So,

$\colon\implies\sf\:\cancel\pi R^2H = 8\times\cancel\pi r^2 h\\\\\colon\implies\sf (12)^2\times 20=(6)^2 \times h\\\\\sf\colon\implies \: \cancel{12}\times\cancel{12}\times20 =8\times\cancel6\times\cancel6\times h\\\\\sf\colon\implies\:\cancel2\times\cancel2\times\cancel{20}=\cancel8\times h\\\\\colon\dashrightarrow\underline{\boxed{\frak{\pink{h=10\:cm}}}}$

$\red\therefore\;\underline{\sf The\:height\:of\:the\:small\:cylinders=\bf{10\:cm.}}$

_____________________________

Answer 2

Answer:

• Height of the small cylinders is 10 cm.

Step-by-step explanation:

Given :-

• Radius of big solid cylinder is 12 cm.
• Height of big solid cylinder is 20 cm.
• Big cylinder is melted in 8 small cylinders.
• Radius of one smaller cylinder is 6 cm.

To find :-

• Height of small cylinders.

Solution :-

We know,

Volume of cylinder = πr²h

Volume of big cylinder :

$\longrightarrow$ Volume = πr²h

$\longrightarrow$ Volume = 22/7 × (12)² × 20

$\longrightarrow$ Volume = 22/7 × 144 × 20

$\longrightarrow$ Volume = 22/7 × 2880

$\longrightarrow$ Volume = 63360/7

$\longrightarrow$ Volume = 9051.42

Volume of big cylinder is 9051.42 cm³.

Volume of 8 small cylinders :

Let, Height of one small cylinder be h.

So,

$\longrightarrow$ Volume = πr²h × 8

$\longrightarrow$ Volume = 22/7 × (6)² × h × 8

$\longrightarrow$ Volume = 22/7 × 36 × h × 8

$\longrightarrow$ Volume = 6336/7 × h

$\longrightarrow$ Volume = 905.142 h

Volume of 8 small cylinders is 905.142 h cm³.

• Now, If we are melting one big cylinder in 8 small cylinders than Volume of big cylinder will equal to the total volume of 8 cylinders.

So,

$\sf \star \bold{Volume \: of \: big \: cylinder = Volume \: of \: 8 \: small \: cylinders}$

$\longrightarrow$ 9051.42 = 905.142 h

$\longrightarrow$ 9051.42/905.142 = h

$\longrightarrow$ h = 10

Therefore,

Height of small cylinders is 10 cm.