How is the density of a floating object ($\rho$) related to the density of the fluid (${\rho }_{\mathrm{fluid}}$) it is floating in?

• A. The density of the object is unrelated to the density of the fluid.
• B. The density of the object is less than the density of the fluid: $\rho <{\rho }_{\mathrm{fluid}}$
• C. The density of the object is equal to the density of the fluid: $\rho ={\rho }_{\mathrm{fluid}}$
• D. The density of the object is greater than the density of the fluid: $\rho >{\rho }_{\mathrm{fluid}}$

Answer: (B)

Verified step by step guidance

1. Recall that an object floats in a fluid when the buoyant force equals the weight of the object. The buoyant force depends on the volume of fluid displaced and the fluid's density.
2. Express the buoyant force using Archimedes' principle: \(F_{buoyant} = \rho_{fluid} \times V_{displaced} \times g\), where \(\rho_{fluid}\) is the fluid density, \(V_{displaced}\) is the volume of fluid displaced, and \(g\) is acceleration due to gravity.
3. Express the weight of the object as \(W = \rho_{object} \times V_{object} \times g\), where \(\rho_{object}\) is the object's density and \(V_{object}\) is its volume.
4. For the object to float, set the buoyant force equal to the object's weight: \(\rho_{fluid} \times V_{displaced} \times g = \rho_{object} \times V_{object} \times g\).
5. Since the object floats partially submerged, \(V_{displaced} < V_{object}\), which implies \(\rho_{object} < \rho_{fluid}\). Therefore, the density of the floating object is less than the density of the fluid.