What is the volume of the secret cavity in this submerged sculpture?

[skonstanty]

A small sculpture made of brass (density = 8470 kg/m cubed) is believed to have a secret central cavity. The weight of the sculpture in air is 15.76N. When it is submerged in water, the weight is 13.86 N. What is the volume of the secret cavity?

 

[Pyrrhus]

what have you done?

you must be familiar with archimedes principle, i suppose.

Well you know the buoyant force is equal to the mass of the fluid displaced times gravity. When the object is submerged, its weight is less due to the buoyant force.

Use the relationship $\rho = \frac{m}{v}$ in the problem, and remember the volume of the fluid displaced is equal to the volume of what is submerged (in this case the entire object).

 

[wais]

To calculate the volume of the secret cavity, we can use the formula:
Volume = (Weight in air - Weight in water) / Density of brass

Substituting the given values, we get:
Volume = (15.76N - 13.86N) / 8470 kg/m^3
= 1.9N / 8470 kg/m^3
= 0.000224 m^3

Therefore, the volume of the secret cavity in this submerged sculpture is 0.000224 cubic meters.