V would be equal to the volume of the liquid displaced by the submerged part. Now that you have added an iron block, the total mass of the iron block and the wooden block must have gone up and resulted in vertical disturbance since the buoyant force now has to change so as to be able to sustain the new weight. The mass has come a bit more submerged in the water, so now the buoyancy will increase. The V will get a bit large also. The thing to note is that V can be defined as the volume of the solid submerged under water as well as the volume of the liquid displaced. That's obvious though-if a box has volume of 100m^3, then upon fully submerging it into the water (forcefully or let it go naturally: how it goes isn't but it goes is important), only 100m^3 of water (any liquid or gas rather that you put it in-the volume of the displaced liquid doesn't depend on the density).
The volume of the displaced liquid will be equal to both
the liquid's volume which we can replace where the object meant to be inside liquid or is it the object's partial volume
if by "partial volume", you mean NOTHING ELSE but volume of the solid now submerged
For the second part, if the new submerged volume be V, then the buoyant force will be V(rho)g The negative sign that you have used will come if you take the usual down as positive.
EDIT:
I am not sure what we have misinterpreted. If in the present picture you have shown, you have let the system settle down for a while, then the buoyant force is:ρAg*l (small l). EDIT 2: I just saw that it is observed that the body sinks half. SO keep the half.
Now, there will be a DIFFERENCE IN BUOYANT FORCE if we consider that this setup as you have shown, is just at the instant that the iron block is placed. This is because iron has its own mass, adding to the weight of the total system, pushing it further down to displace a little more water so as to have a new submerged volume (or volume of displaced liquid: they are the same) so as to support the weight of the new "system". 815
Therefore, we wouldn't be correct t use either of L or l-since the iron will force the wooden block surface to sink deeper. L submerging may be done when the iron block is way too heavy.
Say the new length of the box that is submerged becomes L' .
Then, L'-l is the "extra" distance that the iron block forced the wooden block to move. (since the wooden block has become more heavier with the iron block introduction, it becomes necessary for the block to sink just enough that there can be extra water displaced so as to compensate for the iron block mass)
Now, we need to assume one more thing for a simpler calculation. That is: the extra length L'-l isnt greater than L-l. The reason: so the iron block can always be kept at the top and we can take a value of "A" that is constant throughout with the height.
If you fancy though, then you can also discard the assumption. And say A(iron block)(submerged length of iron block)+ A(wooden block)100% of length (which is L from your picture), (because for the iron block to be wet, the wooden block has to first sink completely) for the calculation of volume submerged.
And L-l is exactly that buffer or "extra" length that can be submerged further before the iron block gets wet. (l=small L). If it is equal to L'-l, then the iron block surface will just touch the water surface, but not get wet fully. There's like 0.00000000000000000000000001% of length submerged, basically 0%
Else, assuming that L' is the new submerged length, and that L'>l, and that (L'-l)<=(L-l) {which ultimately is L'<=L, but I like to think of it in differences}, the after some time (required for the system to settle), new submerged length*area of wooden block (I assume L'=L, so iron block is yet to be submerged) is volume.
Mathematical part: New submerged volume= AL' Buoyant force=rhoAL'g If however, L'-l>L-l (i.e. the iron block gets submerged too)
Let's say iron block has total length of h, with x% submerged
New volume (submerged) =V'=AL + A(iron block)(hx%)
Now V'rhog will be the buoyant force on the system.
And to calculate the exact dip, I am afraid I find no success for a generalized formula. 815