# Can the buoyant force on an object be seen as reduction in weight on a scale?

Consider a container with some fluid of density $\rho_l$ and volume $V_l$. This is kept on a measuring device and has weight $\rho_l V_lg$. Now, consider a block of density $\rho_b$ and volume $V_b$. This block is put into the fluid and here, its apparent weight equals $\rho_b V_bg - F_b$ , where $F_b$ equals the buoyant force which equals $V_b\rho_lg$.

Therefore, the apparent weight of block equals $gV_b(\rho_b - \rho_l)$.

What happens if you take the weight of this whole apparatus? Will it equal $$gV_b(\rho_b - \rho_l) + \rho_lV_lg,$$ or $$\rho_bV_bg + \rho_lV_lg?$$

It will actually weigh $F_b+\rho_lV_lg$ which is $\rho_lV_bg+\rho_lV_lg$, and not one of the two options you say.
Consider the liquid as the system and the block as an external body. Now we know that the liquid applies a buoyant force on the block. According to Newton's third law, the block will apply a reaction force on the liquid, equal in magnitude and opposite in direction. Thus total force on liquid is $F_b+F_g$ which gives $\rho_lV_bg+\rho_lV_lg$.
In the case when the block is resting on the floor, the weight will simply be $\rho_lV_bg+\rho_lV_lg$, because when you consider the block and liquid as a system, the buoyant force will become an internal force and cancel out on the whole system. So the only force responsible for the weight will be gravity.