Question

13. A cylindrical tube, open at both ends, is made of
metal. The internal diameter of the tube is 10.4 cm and
its length is 25 cm. The thickness of the metal is 8 mm
everywhere. Calculate the volume of the metal.​

Answer 1

Given :

• Inner diameter of tube = 10.4 cm
• Inner radius (r1) = 10.4/2 = 5.2 cm

• Thickness = 8 mm = 0.8 cm
• Height = 25 cm

Find :

Volume of the metal.

$\bold{\underline{\underline{Solution\::}}}$

Outer radius of tube = Inner radius + thickness

⇒ 5.2 + 0.8

⇒ 6 cm

Volume of metal used = Outer volume - Inner Volume

Let radius of outer tube be "r2".

And from above calculations r2 = 6 cm

⇒ π(r2)²h - π(r1)²h

Substitute the known values in above formula

⇒ π [(r2)² - (r1)²] h

⇒ 22/7 × [(6)² - (5.2)²] × 25

⇒ 22/7 × (36 - 27.04) × 25

⇒ 22/7 × 8.96 × 25

⇒ 704 cm³

∴ Volume of the metal is 704 cm³

Answer 2

$\huge\sf{Solution}$

Refer the above attachments.

Explanation:-

10.4 cm is the inner diameter of tube and let R be the radius as well as let R1 be the inner radius and R2 be the outer radius ,8mm is the thickness and 8mm is equal to 0.8mm and 25cm is the height.

Two Furmula's Used

1. [Inner Radius (+) The Thickness ⟹ Outer Radius]

1. [Outer Radius (-) Inner Radius ⟹ Volume of the Metal]