A cubical block of density rB and with sides of length L floats in a liquid of greater density rL.

Question

(a) What fraction of the block’s volume is above the surface of the liquid?
(b) The liquid is denser than water (density rW) and does not mix with it. If water is poured on the surface of the liquid, how deep must the water layer be so that the water surface just
rises to the top of the block? Express your answer in terms of L, rB, rL, and rW.
(c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of
iron, and the side length is 10.0 cm.
I solved the first part but I do not know if it is right
V cubic/V submerged =density cubic/ density liquid = rb/rl

Answer 1

Force up on block = rhoL *L * L * d * g
where d is the draft or distance submerged

that has to equal the weight of the block
which is rhoB L^3 g
so

rhoB L^3 g = rhoL L^2 d g

d = (rhoB/rhoL) L
d/L = rhoB/rhoL
but we want (L-d)/L which is 1 - d/L
1 - d/L = fraction above
= 1-rhoB/rhoL

Answer 2

B)
rhoL L^2 d + rhoW L^2 (L-d) = rhoB L^3

Answer 3

again that d is the part in the heavy liquid
for the water depth use L-d