Question

A rock has a volume of 2.24x10^-6 m^3. When it is completely submerged in a beaker of water, the water exerts a buoyant force on the rock equal to 0.021 newtons. What is the density of the rock?

a) 8000 kg/m^3
b) 10,000 kg/m^3
c) 7500 kg/m^3
d) 9000 kg/m^3

Answer 1

The density of the rock is approximately 8000 kg/m³.

Therefore, correct option is a) 8000 kg/m³.

Step-by-step explanation:

Find the mass of the rock using the buoyant force.

F = mg

`\[ m = (F)/(g) \]`

Given buoyant force, F = 0.021 N, and acceleration due to gravity, g = 9.8
`m/s^2`:

`\[ m = (0.021)/(9.8) \approx 0.00214 \ kg \]`

Use the density formula
`\( \rho = (m)/(V) \)` with the known volume of the rock.

`\[ \rho = (0.00214)/(2.24 * 10^(-6)) \approx 8035 \ kg/m^3 \]`

The buoyant force acting on the rock is essentially the weight of the water displaced by the submerged rock. Using the relationship F = mg , we find the mass of the rock. Then, applying the density formula
`\( \rho = (m)/(V) \)`, where V is the volume of the rock, we determine the density. In this case, the calculated density is approximately 8035 kg/m³, which rounds to 8000 kg/m³, and this is the density of the rock.

Therefore, correct option is a) 8000 kg/m³.