Question

A cylinder is open at both ends and made of 3 cm thick metal , the external diameter is 24 cm and height is 168 cm. What is the volume of metal used in making the cylinder ? Find the weight of cylinder . If 2 cm cube of the metal weight 15g.​

Answer 1

${\huge{\underbrace{\rm{Question}}}}$

A cylinder is open at both ends and made of 3 cm thick metal , the external diameter is 24 cm and height is 168 cm . What is the volume of metal used in making the cylinder? find the weight of cylinder . If 1 cm cube of the metal weight 15 g.

${\huge{\underbrace{\rm{Answer}}}}$

⠀⠀

Given:

⠀⠀

• Thickness of the metal is 3 cm

• the external diameter of the cylinder is 24 cm

• Height of the cylinder is = 168 cm

⠀⠀

To find:

⠀⠀

• What is the volume of metal used in making the cylinder

• find the weight of cylinder . If 2 cm cube of the metal weight 15 g.

⠀⠀

Solution:

⠀⠀

External diameter of the cylinder is 24

External radius of the cylinder is= $\sf{\dfrac{24}{2}\:cm}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= $\sf{\cancel{\dfrac{24}{2}}\:cm}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= $\sf{12\:cm}$

⠀⠀

Thickness of the cylinder is = 3 cm

⠀⠀

.°. Internal radius of the cylinder is

$\sf{External\:radius\:of\:the\:cylinder- thickness\:of\:the\:cylinder}$

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= $\sf{12-3\:cm}$

⠀⠀

= $\sf{9\:cm}$

⠀⠀

We know that,

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⠀⠀⠀⠀$\boxed{\bf{\pink{Volume\:of\:cylinder=πr^{2}h}}}$

⠀⠀

Where,

⠀⠀

• r = radius of the cylinder

• h = height of the cylinder

• π = $\sf{\dfrac{22}{7}}$

⠀⠀

$\sf{Volume\:of\:the\:external\:part\:of\:the\:cylinder\:is=πr^{2}h}$

⠀⠀

= $\sf{\dfrac{22}{7}×12×168\:cm^{3}}$

⠀⠀

= $\sf{22×12×24\:cm^{3}}$

⠀⠀

= $\sf{6336\:cm^{3}}$

⠀⠀

$\sf{Volume\:of\:the\:internal\:part\:of\:the\:cylinder\:is=πr^{2}h}$

⠀⠀

= $\sf{\dfrac{22}{7}×9×168\:cm^{3}}$

⠀⠀

= $\sf{22×9×24\:cm^{3}}$

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= $\sf{4752\:cm^{3}}$

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.°. The volume of metal used in making the cylinder is

⠀⠀

$\sf{\implies (Volume\:of\:external\:part)-(Volume\:of\:internal\:part)}$

⠀⠀

$\sf{\implies (6336-4752)\:cm^{3}}$

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$\sf{\implies 1584\:cm^{3}}$

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Now,

given that, 1 cm³ metal = 15 g

.°. 1584 cm³ metal= 5 × 1584 cm³ = 23760 cm³

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Therefore,

⠀⠀

• Volume of metal used in making the cylinder is 1584 cm³

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• the weight of cylinder is 23760 If 2 cm cube of the metal weight 15 g.

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Answer 2

Provided Data :

• ⠀A cylinder is open at both ends and made of 3 cm thick metal , the external diameter is 24 cm and height is 168 cm. ⠀

Problem :

• What is the volume of metal used in making the cylinder ?
• Find the weight of cylinder.If 2 cm cube of the metal weight 15 g.

⠀⠀⠀⠀⠀⠀⠀$\boxed{\bf{\pink{Working\:Rule}}}$

Here we have to find out the external radius of the cylinder by using the given external data .After this, to get the inner radius we have to subtract the thickness of the metal from the external radius. Then , for volume of the external cylinder , we will simply substitute the respective values of r, π and h. After that , we will find the volume of the inner part of the cylindrical cylinder. We will subtract the internal part volume of cylinder from the external part volume of cylinder in order to get the volume of the metal used in making the cylinder. As from the given data we know the weight for 2 cm³ , so we will find the weight of 1cm³ then multiply it by the volume.

⠀⠀

Appropriate Answer :

For External Part Volume :

External radius of the cylinder

$\Rightarrow \bf{Radius = \dfrac{Diameter}{2}}$

Given , Diameter = 24 cm.

Putting the value in the diameter , we get :

$\implies \bf{Radius = \dfrac{Diameter}{2}} \\ \\ \implies \bf{Radius = \dfrac{24}{2}} \\ \\ \implies \bf{Radius = 12 cm}$

Radius of the External part → 12 cm.

Finding Volume now ,

$\Rightarrow \bf{Volume = {π r^{2}h}}$

Putting up the values :

= $\sf{\dfrac{22}{7}×12×168\:cm^{3}}$

⠀⠀

= $\sf{22×12×24\:cm^{3}}$

⠀⠀

= $\sf{6336\:cm^{3}}$

⠀⠀

-------------------------------

For internal Part Volume :

Internal radius of the cylinder is (External Radius - Thickness )⠀⠀

= $\sf{12-3\:cm}$

⠀⠀

= $\sf{9\:cm}$

Finding Volume now ,

$\Rightarrow \bf{Volume = {π r^{2}h}}$

Putting up the values :

= $\sf{\dfrac{22}{7}×9×168\:cm^{3}}$⠀⠀

= $\sf{22×9×24\:cm^{3}}$

⠀⠀

= $\sf{4752\:cm^{3}}$

-------------------------------⠀⠀

For the Volume of metal used :

{External Volume - Internal Volume}⠀⠀

= $\sf{(6336-4752)\:cm^{3}}$

⠀⠀

= $\sf{1584\:cm^{3}}$

For weight of the Metal :

2 cm³ metal → 15 g

So 1 cm³ → 15/2 = 7.5 g

For weight → 1 cm³ { given weight } × Volume of metal

= $\sf{1584 × 7.5\:cm^{3}}$

= $\sf{11880\:cm^{3}}$

Hence,

⠀

• Volume of metal is 1584 cm³

• The weight of metal is 11880 cm³

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