Question

Iron has a density of 7.9 g/cm3. The volume of a regular cylinder is V =πr2h. An iron cylinder has a height of 3.00 m and a mass of 2.00 kg. Using the value 3.1416 for the constant π, the radius in cm of the iron cylinder is?

Answer 1

The radius in cm of the iron cylinder is 0.52 cm.

Determination of the volume of the iron

• Mass of iron = 2 Kg = 2 × 1000 = 2000 g

• Density of iron = 7.9 g/cm³

• Volume of iron =?

Volume = mass / density

Volume of iron = 2000 / 7.9

Volume of iron = 253.16 cm³

Determination of the radius of the cylinder

• Volume (V) = 253.16 cm³

• Height (h) = 3 m = 3 × 100 = 300 cm

• Pi (π) = 3.1416

• Radius (r) =?

V = πr²h

253.16 = 3.1416 × r² × 300

253.16 = 942.48 × r²

Divide both side by 942.48

r² = 253.16 / 942.48

Take the square root of both side

r = √(253.16 / 942.48)

r = 0.52 cm

Thus, the radius of the iron cylinder is 0.52 cm

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