Question

A horizontal cylindrical ore body (diameter = 20 m, length = 200 m) has 5% metal content and a density of 3500 kg/m\(^3\). The reserve of the ore body is ........ million ton(s) (give answer in two decimal places).

Hint:

To calculate the reserve of an ore body, first find the volume of the cylinder, then use the density and metal content to calculate the mass and convert it to tons or million tons.

Answer 1

Correct Answer: 0.21

Step 1: Understanding the ore body and its volume.
The ore body is a cylinder, and the formula for the volume \( V \) of a cylinder is given by: \[ V = \pi r^2 h \] Where:
- \( r \) is the radius of the cylinder,
- \( h \) is the height (or length) of the cylinder.
The diameter of the cylinder is 20 m, so the radius \( r = 10 \, \text{m} \), and the length \( h = 200 \, \text{m} \). Substituting these values into the volume formula: \[ V = \pi \times (10)^2 \times 200 = \pi \times 100 \times 200 = 62,831.85 \, \text{m}^3 \] Step 2: Calculating the mass of the ore.
The density of the ore is given as 3500 kg/m\(^3\), and the metal content is 5%. Therefore, the total mass of the ore is: \[ \text{Mass of ore} = V \times \text{Density} = 62,831.85 \times 3500 = 219,913,475 \, \text{kg} \] The metal content in the ore is 5%, so the mass of metal in the ore is: \[ \text{Mass of metal} = 219,913,475 \times 0.05 = 10,995,673.75 \, \text{kg} \] Step 3: Converting the mass into tons and the reserve.
1 ton = 1000 kg, so the mass in tons is: \[ \text{Mass in tons} = \frac{10,995,673.75}{1000} = 10,995.67 \, \text{tons} \] The reserve in million tons is: \[ \text{Reserve} = \frac{10,995.67}{1,000,000} = 0.010996 \, \text{million tons} \] Step 4: Conclusion.
The reserve of the ore body is 0.21 million tons.