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A solid iron cone consists of cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole given that 1cm³ of iron is equal to 8g. (use pie =3.14)?
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Volume of the Iron Cone
To find the mass of the iron cone, we first need to calculate its volume. The solid iron cone is made up of two parts: a cylinder and a smaller cylinder on top of it.
1. Volume of the Larger Cylinder
- Height (H1): 220 cm
- Radius (R1): Diameter/2 = 24 cm / 2 = 12 cm
Using the formula for the volume of a cylinder, V = πR²H:
- V1 = 3.14 × (12)² × 220
- V1 = 3.14 × 144 × 220
- V1 = 3.14 × 31680 = 99379.2 cm³
2. Volume of the Smaller Cylinder
- Height (H2): 60 cm
- Radius (R2): 8 cm
Using the same formula:
- V2 = 3.14 × (8)² × 60
- V2 = 3.14 × 64 × 60
- V2 = 3.14 × 3840 = 12065.6 cm³
Total Volume of the Iron Cone
- Total Volume (V) = V1 + V2
- V = 99379.2 cm³ + 12065.6 cm³ = 111444.8 cm³
Mass Calculation
To find the mass, we multiply the total volume by the density of iron:
- Density of Iron: 8 g/cm³
- Mass (M) = Volume × Density
- M = 111444.8 cm³ × 8 g/cm³ = 891558.4 g
Final Mass in Kilograms
- Mass in kg = 891558.4 g / 1000 = 891.56 kg
Thus, the mass of the solid iron cone is approximately 891.56 kg.
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A solid iron cone consists of cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole given that 1cm³ of iron is equal to 8g. (use pie =3.14)?
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