Question

A cylindrical pipe,open at both ends has internal radius 8cm and its length is 21cm. if it's thickness is 2cm, find the volume of the metal used in making the vessel​

Answer 1

Given :

• Internal Radius → 8 cm
• Thickness of the Cylindrical pipe → 2 cm
• Length of the pipe → 21 cm

What we have to do ?

Here, We have to find out the outer radius of the cylindrical pipe first. For that we will simply add up the thickness and the Inner radius.

Outer Radius = Inner Radius + Thickness

Then, we have to find out the volume of the metal used. For that we will subtract the Volume of inner part from that of the volume of outer cylindrical portion.

For that, Formula :

Volume of the Cylinder → πr²h

$\underline{\sf Solution :}$

Let's we consider the inner radius be " r " and outer radius be " R " respectively.

Now,

$:\implies\sf Outer \: Radius = (Internal Radius + Thickness)$

$:\implies\sf (8 + 2)$

$:\implies\sf 10$

Volume of the metal used :

• Internal Volume = πr²h
• External Volume = πR²h

External Volume - Internal Volume

↬ πR²h - πr²h

Taking " π " and " h " common :

→ π(R² - r²)h

Substituting the respective values, we get :

$\dashrightarrow\:\:\sf {\dfrac{22}{7} \times {10}^{2} - {8}^{2} \times 21}$

$\dashrightarrow\:\:\sf{\dfrac{22}{7} \times (100 - 64) \times 21}$

$\sf \longrightarrow \: Volume \: = {\dfrac{22}{ \cancel{7}^{ \: \: 1} } \: \times 36 \times \: \cancel{21}^{ \: \: 3} \: cm}$

$\dashrightarrow\:\:\sf {22 \times 36 \times 3}$

$\dashrightarrow\:\: \underline{ \boxed{\sf Volume = 2376 cm³}}$

Answer 2

Answer:

=>Outer radius=(internal radius+Thickness)

=>(8cm+2cm)

=>10 cm

Volume of metal used:

$= > internal \: volume = \pi {r}^{2} h$

$= > external \:volume = \pi \: a {}^{2}h$

Formula:-

External volume-Internal value

$=> \pi(a {}^{2} - \pi \: r {}^{2})h$

Value of pie is 22/7

$volume = 22 \: by \: 7 \times 36 \times 21 \: cm \\ therefore \: volume = 2376 \: cm {}^{</strong><strong>3</strong><strong>}$