Question

A spherical ore body (diameter = 40m) has 7% metal content and density of 3300 kg/m³. The reserve (in tonne) of the ore body is .............

Hint:

To find the reserve of an ore body, calculate the volume of the sphere, the total mass using the density, and then determine the amount of metal content by applying the given percentage.

Answer 1

Correct Answer: 110528 - 110629

1. Volume Calculation

The ore body is a sphere with a diameter of $40\text{ m}$.

The radius ($r$) is:

$$r = \frac{40\text{ m}}{2} = 20\text{ m}$$

The volume ($V$) of the spherical ore body is:

$$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (20\text{ m})^3 = \frac{32,000}{3}\pi \text{ m}^3$$

2. Mass Calculation (in kg)

The mass ($M$) of the ore body is found using the volume ($V$) and the given density ($\rho = 3300\text{ kg/m}^3$):

$$M = V \times \rho$$

$$M = \left( \frac{32,000}{3}\pi \text{ m}^3 \right) \times \left( 3300\text{ kg/m}^3 \right)$$

$$M = 32,000\pi \times \frac{3300}{3} \text{ kg}$$

$$M = 32,000\pi \times 1100 \text{ kg}$$

$$M \approx 110,528,000 \text{ kg}$$

3. Conversion to Tonne

The total mass is converted from kilograms to tonne ($1\text{ tonne} = 1000\text{ kg}$):

$$\text{Total Mass (tonne)} = \frac{110,528,000 \text{ kg}}{1000}$$

$$\text{Total Mass (tonne)} \approx 110,528 \text{ tonne}$$

Conclusion

The total mass (tonnage) of the ore body, which the question refers to as the reserve, is approximately $\mathbf{110,528\text{ tonne}}$.