The outer diameter of a hollow cylindrical iron rod is 8.4 cm. The rod is 28 mm thick and is 30 cm long. If the weight of 1 cm3 of iron in the rod is 0.5 kg, then find the total weight of the rod.
739.2 kg
Given: outer diameter 8.4 cm, thickness 28 mm, length 30 cm, and 1 cm3 of iron weighs 0.5 kg.
The outer radius is \(R = \frac{8.4}{2} = 4.2\ \text{cm}\), and the thickness 28 mm equals 2.8 cm.
The inner radius is \(r = R - \text{thickness} = 4.2 - 2.8 = 1.4\ \text{cm}\).
The iron volume is that of a hollow cylinder: \(V = \pi (R^2 - r^2) h\).
Compute the radii squared: \(R^2 - r^2 = 4.2^2 - 1.4^2 = 17.64 - 1.96 = 15.68\).
So \(V = \frac{22}{7} \times 15.68 \times 30 = 22 \times 2.24 \times 30 = 1478.4\ \text{cm}^3\).
The total weight is \(1478.4 \times 0.5 = 739.2\ \text{kg}\).
Hence the total weight of the rod is 739.2 kg.
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