Question

1)

A homogeneous solid object floats on water with 78% of its volume below the surface. The same object when placed in a second liquid floats on that liquid with 71% of its volume below the surface.
Determine the density of the object.

2)

Determine the specific gravity of the liquid.

Answer 1

The buoyant force is the weight of liquid displaced which means the volume of liquid displaced times the density of the liquid displaced (volume*mass/volume = mass) times g For water the density is
Dw=1kg/liter. So one liter of water weighs g Newtons
If the density of the solid object is Dx and it's Volume is Vx then Vx*Dx*g =0.78*Vx*Dw*g
That is, the weight of the object weighs the same as 0.78 times it's Volume of water
Vx is on both sides of the equation so it drops out. Dw = 1, g is on both sides of the equation so it drops out and we are left with:
Dx = 0.78kg/liter = 0.78kg/1000cm^3*10^6cm^3/m^3 = 780kg/m^3

Specific gravity is a ratio of densities with the reference desity being water: 1000kg/m^3
We know the density of the block. We can calculate the density of the unknown liquid since we know that a volume of the unknown liquid = to 71% of the blocks volume weighs what the block does. We can write: Vx*780*g = 0.71*Vx*Dl*g where Dl is the density of the unknown liquid. Vx and g both appear on both sides of the equation so they drop out leaving
780=0.71*Dl => Dl = 780/0.71 = 1098.59 kg/m^3 which means it has higher density than water because the block floats even higher in this liquid than it does in water. So this density makes sense
The specific gravity is the ratio of the unknown liquid to that of water:
1098.56/1000 = 1.0985