Answer to Question #216344 in Molecular Physics | Thermodynamics for Dullah

Question

11. A solid gold vase of mass 20.0 kg is suspended by a rope at sea bottom. What is the Buoyant Force of the seawater on the vase? 

Answer 1

The Buoyant Force is given as follows:

$$F = \rho_w g V_v$$

where $\rho_w =1024kg/m^2$ is the density of sea water, $g = 9.8N/kg$ is the gravitational acceleration, and $V_v$ is the volume of the vase.

By definition, the volume is:

$$V_v = \dfrac{m}{\rho_g}$$

where $\rho_g = 19300kg/m^3$ is the density of gold, and $m = 20kg$ is the mass of the vase.

Finally, obtain:

$$F = \rho_w g\dfrac{m}{\rho_g}\\ F = 1024\cdot 9.8\cdot \dfrac{20}{19300}\approx 10.4N$$

Answer. 10.4N.