Finding the density of gold crown using archimedes principle Question:  As shown in diagram below the crown has a mass of 14.7kg when measured above water and 13.4kg when measured in water. Is the crown made of gold?

I have this following solution provided:
The apparent weight of the submerged crown, $w'$,equals the actual weight, $w$, minus the buoyant force, $F_B$.
$w' = w - F_B = w - \rho_{fluid}gV $
$ w - w' = \rho_{fluid}gV$
$ w = \rho_{object}gV$
I still agree with the answer up to this point as the only difference before submerging and after is the upthrust that is equivalient to the amount of water being displaced
However, it takes the following ratio between $w$ and $w-w'$ like this:
$w/(w-w') = (\rho_{object}gv)/(\rho_{fluid}gv) = \rho_{object}/\rho_{fluid} = 14.7 kg/ (14.7kg - 13.4kg) = 11.3 $
And it use this value 11.3 as the density of the object. How does this ratio of weight before and after(weight loss because upthrust) is equivalent to the density of the object?Can anyone please enlighten me?
 A: Consider $w$;  it's the volume of the object times its density times $g$.
Consider the buoyant force;  it's the volume of the object times the density of the fluid times $g$
So, what's the ratio of the weight to the buoyant fore, when you cancel out the constant factors?
A: Basically, displacement applies.
Although the weight of the crown remains the same, the force (or heft) doesn't.  What happens is that the weight displaces a volume of water, and this displacement of water is what makes the force lesser.  So it needs fewer weights to balance it on the other side.
If $C_a$ is the weight in air, and $C_w$ in water, the difference is the weight of the displaced water, ie $W_d = C_a - C_w$  One can derive the volume of water directly.  Since the specific gravity of the crown is C_a / W_d,  one might then compare the specific gravity of the object against various gold standards.
A value of 11.3 is nearly what you might expect of silver or iron with a mixture of gold into it.  Gold itself has a density of 19.3, silver as 10.5.  Here we would expect of a silver-gold alloy, that the 8.8 extra by gold appears in the 0.8 extra over of silver.  This means that putting 10 parts of silver (10.5) gives 105, and one part of gold at 19.3 gives 124.3, which divides by 11 to give 11.3.
A: What you have here is the relative density of the substance with the fluid. 
We can have standard value for relative density of pure gold with that fluid.
And easily compare. 
Here I think the fluid could be water, standard relative density of gold is 19.3, we got it 11.3,hence not gold !  
