595) The volume flow rate out of the large container has to equal the sum of the volume flow rates out of a and b, such that Q = Qa + Qb (This means all the fluid leaving the large middle tube has to either flow through a or b), but there is no requirement that Qa=Qb. In fact from v = sqrt(rho*g*h) the velocities must be the same, so since the area of b is larger the flow rate (A*Qb) out of b is larger than out of a.

629) The clue is it says "According to the table" this means it must be a bulk modulus question. The bulk modulus of water is 2 GPa (there appears to be a typo in the table, I think it should be 2 and not 200) where as mercury is 27 GPa. So you have two things, a greater pressure at the bottom in mercury than in water because mercury is heavier (denser) and therefor pushing down more, and a higher bulk modulus so it compresses more. There is a greater change in volume for the mercury than for water, therefor at that point it becomes relatively more dense than it was before, and the SG > 13.6.

626) In an ideal non-compressible fluid buoyant force wouldn't change with depth, in the real world it does because the density of water (for example) increases with depth defined by its bulk modulus. However, the bulk modulus of lead is greater than water (again water is 2 not 200) so it compresses more than water for a given pressure. So the buoyant force, which is rho*V*g of water, the V goes down faster than rho goes up so the buoyant force decreases (again this is because the bulk modulus of lead is larger than the bulk modulus of water, if the two bulk modulus s were reversed it would increase).

Cheers

Mitchell