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

Step 1: Calculate the volume of the spherical ore body.
The volume \( V \) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] where \( r \) is the radius of the sphere. Since the diameter is 40 m, the radius is: \[ r = \frac{40}{2} = 20 \, \text{m} \] Substitute the value of \( r \) into the volume formula: \[ V = \frac{4}{3} \pi (20)^3 = \frac{4}{3} \pi (8000) = 33510.32 \, \text{m}^3 \]

Step 2: Calculate the mass of the ore body.
The mass \( m \) of the ore body can be found using the formula: \[ m = \text{density} \times \text{volume} \] Substitute the given values: \[ m = 3300 \, \text{kg/m}^3 \times 33510.32 \, \text{m}^3 = 11078305.6 \, \text{kg} \]

Step 3: Calculate the amount of metal content.
The metal content is 7% of the total mass. Therefore, the metal mass is: \[ \text{Metal mass} = \frac{7}{100} \times 11078305.6 = 775481.4 \, \text{kg} \]

Step 4: Convert the mass to tonnes.
Since 1 tonne = 1000 kg, the reserve in tonnes is: \[ \text{Reserve} = \frac{775481.4}{1000} = 775.48 \, \text{tonnes} \] Thus, the reserve of the ore body is 775.48 tonnes.