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.
Answers
Volume of the metal = 704 cm³
Volume of the metal = Total volume of cylindrical tube including the metal - volume w.r.t the inner radius of the tube
Inner radius of the tube = r1 = 10.4/2 = 5.2 cm
Outer radius of the tube (radius including the metal) = r2 = (10.4 + 1.6)/2 = 6 cm
Length of the tube = l = 25 cm
Inner Volume of tube = π(r1)²l
= (22/7) × (5.2)² × 25
= 2124.57 cm³
Total volume of the cylindrical tube w.r.t outer radius = π(r2)²l
= (22/7) × (6)² × 25
= 2828.57 cm³
Volume of metal = 2828.57 - 2124.57
= 704 cm³
Step-by-step explanation:
Given 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.
- Given internal diameter of the tube = 10.4 cm
- So internal radius = d / 2 = 10.4 / 2 = 5.2 cm
- Height = 25 cm
- Thickness of metal = 8 mm = 8 x 0.1 cm = 0.8 cm
- Outer radius R = inner radius + thickness = 5.2 + 0.8 = 6 cm
- Volume of metal which is used = outer volume – internal volume
- = π h (R^2 – r^2)
- = 3.14 x 25 (6^2 – 5.2^2)
- = 3.14 x 25 x 8.96
- = 703.36 cm^3
- Therefore the volume of metal used will be 703.36 cm^3
Reference link will be
https://brainly.in/question/1889099