Question

The density of gold is $19.3\times {10}^3{\text{kg/m}}^3$.
Suppose the weight of the crown in air that was offered to the king was $53.2\text{N}$. What would be its weight in water if it was genuine?

Answer 1

Let mass of crown be $\text{M}$
Density of gold be ${\rho }_g$​
Density of water be ${\rho }_w$​
The loss of weight of the crown in water will be same as buoyant force given by the water.
Now, Buoyant force $F_B=\text{V}{\rho }_{w}g$
Now, $\text{V}=\frac{\text{M}}{{\rho }_g}$

So, $F_B=\frac{\text{M}}{{\rho }_g}\times 1000$
($\because {\rho }_w=1000{\text{kg/m}}^3$)
Putting the values $F_B=2.8\text{N}$
$\therefore \text{The weight of the crown in water will be}(53.2-2.8)\text{N}=50.4\text{N}$