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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 $$\rho_bV_bg + \rho_lV_lg?$$

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up vote 1 down vote accepted

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 $\rho_lV_bg+\rho_lV_lg$.

Note that this is the weight when the block is attached to the spring balance, and suspended in the liquid. If the block is kept on the floor not attached to the spring balance, the reading weight will be different.


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.

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Actually, I wanted to consider the situation where the block isn't suspended. Just resting on the floor. But, I'm voting this up anyway, since I had this question in mind as well. –  Gerard Aug 3 '13 at 16:14
@user1305192, I added the case you wanted as an edit. –  udiboy Aug 3 '13 at 16:22
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