Question

A cylinder is open at both ends and is made of metal 2cm thick . It's external diameter is 10cm and height is 40cm . Find the volume of metal used in making the cylinder and weight if cylinder , if 1cm^3 of metals weight 8g . (Use π as 3.4)

Answer 1

Answer:

volume = 2009.6 cm3

weight = 16076.8 g

Step-by-step explanation:

Volume =  π(R^2-r^2)h

= 3.14 x ( 5^2 - 3^2) x 40

= 3.14 x 16 x 40

=2009.6 cm^3

1cm^3 = 8g

therefore, weight = 2009.6 x 8 = 16076.8 g

Answer 2

❏ Formula Used:-

✦ CYLINDER✦

For a right circular cylinder of base radius r and height h,

$\sf\longrightarrow\boxed{ L.S.A=2\pi r h}$

$\sf\longrightarrow\boxed{ T.S.A.=2\pi r (r+h)}$

$\sf\longrightarrow \boxed{Volume=\pi r{}^{2}h}$

Where, •L.S.A.=Curved Surface area.

•T.S.A.=Total Surface Area.

❏ Question:-

@ A cylinder is open at both ends and is made of metal 2cm thick . It's external diameter is 10cm and height is 40cm . Find the volume of metal used in making the cylinder and weight if cylinder , if 1cm^3 of metals weight 8g .

(Use π as 3.4)

❏ Solution:-

➝Given:-

• thickness (t)=2 cm

• $\bf D_{ex}$=10 cm.

• $\bf R_{ex}$= 5 cm.

• Height (h)=40 cm

• density($\bf\rho$)=8 gm/cm³.

➝To Find:-

• volume(V) of the material used=?

• Weight(W)of the metal cylinder=?

∴ Inner diameter($\bf D_{In}$)=$\bf D_{Ex}-2t$

=(10-2×2)cm=6 cm.

∴ Inner Radius($\bf R_{In}$)= 3 cm.

Now, therefore the volume of the material is

$\sf\bf \longrightarrow V_{\red{material}}=\pi \times (R^{2}_{ex}-R^{2}_{in})\times h$

$\sf\bf \longrightarrow V_{\red{material}}=[3.14\times (5^{2}-3^{2})\times40]\:\:cm^{3}$

$\sf\bf \longrightarrow V_{\red{material}}=[3.14\times16\times40]\:\:cm^{3}$

$\sf\bf \longrightarrow \boxed{V_{\red{material}}=[2009.6]\:\:cm^{3}\:\:\:(approx)}$

∴ Volume of the material is=2009.6 cm³.

∴ weight of the material

=(8×2009.6) gm.

= 16076.8 gm.

= 16.0768 kg.

= 16.1 kg (almost).

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