Question

A tabular ore body of 9 km\(^2\) area and an average thickness of 9 m has a density of 3000 kg/m\(^3\). The tonnage (in million tonnes) of the ore body is _________.

Hint:

The tonnage of an ore body can be calculated by multiplying the volume of the ore by its density and converting the mass to tonnes.

Answer 1

Correct Answer: 243

Step 1: Calculate the volume of the ore body.
The volume of the ore body is the product of its area and thickness. Given the area of 9 km\(^2\) (which is 9,000,000 m\(^2\)) and an average thickness of 9 m, the volume \(V\) is: \[ V = \text{Area} \times \text{Thickness} = 9,000,000 \, \text{m}^2 \times 9 \, \text{m} = 81,000,000 \, \text{m}^3. \] Step 2: Calculate the mass of the ore body.
The mass \(m\) of the ore body is calculated by multiplying the volume by the density: \[ m = V \times \text{Density} = 81,000,000 \, \text{m}^3 \times 3000 \, \text{kg/m}^3 = 243,000,000,000 \, \text{kg}. \] Step 3: Convert mass to tonnage.
Since 1 tonne = 1000 kg, the tonnage is: \[ \text{Tonnage} = \frac{243,000,000,000 \, \text{kg}}{1000} = 243,000,000 \, \text{tonnes} = 243 \, \text{million tonnes}. \] Step 4: Conclusion.
The tonnage of the ore body is 243 million tonnes.