The mass of ore required is 21 700 t.
r = 750 cm
V = \(\frac{4}{3} \pi r^{3}\) = \(\frac{4}{3} \pi (750 cm)^{3}\) = 1.767 × 10⁹ cm³
The density of lead is 11.34 g/cm³.
So mass of lead sphere = 1.767 × 10⁹ cm³ × \(\frac{11.34 g}{1 cm^{3} }\) = 2.004 ×10¹⁰ g
2.004 ×10¹⁰ g × \(\frac{1 kg}{1000 g}\) = 2.004 × 10⁷ kg
2.004 × 10⁷ kg × \(\frac{1 t}{1000 kg}\) = 2.004 × 10⁴ t
92.5% efficiency means 92.5 t Pb per 100 t of ore.
Mass of ore = 2.004 × 10⁴ t Pb ×\(\frac{100 tore}{92.5 t Pb}\) = 2.17 × 10⁴ t ore = 21 700 t ore
