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

(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

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

B)

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

again that d is the part in the heavy liquid

for the water depth use L-d

Well, let me play around with the question a bit and provide you with my own unique insight!

(a) What fraction of the block's volume is above the surface of the liquid? Ah, the floating block, seeking freedom from the clutches of gravity by embracing the buoyant forces! To find the fraction above the surface, we divide the volume of the submerged portion by the total volume. So, our fraction would be V_submerged / V_total. You got that part right!

(b) Ah, the addition of water to the mix! Water, the universal solvent, trying to assert its dominance in the realm of densities. To find the depth of the water layer that makes it rise to the top of the block, we need to consider the densities of the liquids involved. Let's call the depth of the water layer "d."

Now, when the water surface just reaches the top of the block, the total mass of the displaced liquid (block included) will equal the mass of the block. In other words, the densities of the block and the combined liquid (rL + rW) must be equal. Using this knowledge, we can set up an equation:

rB * L^3 = (rL * (L - d)^2 * d) + (rW * (L - d)^2 * d)

Solving this equation will give you the depth of the water layer!

(c) Ah, specific values! How splendid! Let us substitute the values for this particular scenario. If the liquid is mercury, the block is made of iron, and the side length is 10.0 cm, we will have to use their respective densities (rL and rB). Now, without the values for rL and rW, I can't give you an exact numerical answer, but I can help you set up the equation:

rB * (10.0 cm)^3 = (rL * (10.0 cm - d)^2 * d) + (rW * (10.0 cm - d)^2 * d)

Solve this equation, and you will find the depth of the water layer in all its numerical glory!

Remember, science can be fun and enlightening! Do let me know if you need any further assistance or a laughter-inducing explanation.

To solve part (a) of the question, you correctly used the concept of relative densities. The fraction of the block's volume above the surface of the liquid can be determined by comparing the density of the block with the density of the liquid.

Let's assume the volume of the block is represented by V_cube, and the volume of the submerged part is represented by V_submerged. The density of the block is denoted as r_B, and the density of the liquid is denoted as r_L.

The equation you used is correct:

V_submerged / V_cube = r_B / r_L

To find the fraction of the block's volume above the surface of the liquid, you need to subtract the fraction submerged from 1:

Fraction above surface = 1 - V_submerged / V_cube

So, to solve part (a), you need to calculate the fraction above the surface by using the equation mentioned above.

For part (b), you need to find the depth of the water layer required for the water surface to just rise to the top of the block. To do this, you can use the concept of buoyancy.

When the water surface just rises to the top of the block, the buoyant force on the block is equal to the weight of the block itself. This buoyant force can be calculated using the volume of the block submerged in water and the density of water:

Buoyant force = Weight of the block
ρ_W × V_submerged × g = ρ_Block × V_cube × g

Since ρ_Block = r_B and V_cube = L^3 (because it is a cube), the equation becomes:

ρ_W × V_submerged = r_B × L^3

To find the depth of the water layer, we divide both sides by the area of the water surface, which is L^2:

ρ_W × V_submerged / A = r_B × L^3 / A
Depth of water layer = r_B × L

So, to solve part (b), you need to multiply the density of the block (r_B) by the length of the side of the cube (L).

For part (c), the density of mercury (r_L) and the side length of the block (L) are given as 13.6 g/cm^3 and 10.0 cm, respectively. To find the depth of the water layer, you can use the formula mentioned in part (b) and substitute the given values:

Depth of water layer = r_B × L
Depth of water layer = (given density of iron) × (given side length)
Depth of water layer = (density of iron) × 10.0 cm

Please note that the actual values of densities and lengths need to be provided for accurate calculations in part (b) and part (c).