Question

The density of gold is $19.3\times {10}^3{\text{kg/m}}^3$.
A plate made of gold was kept in a museum stating that it is of pure gold. The weight of this plate was $83.2\text{N}$. To test its purity it was immersed in water. If it was genuine real gold then how much it should weigh in water?

Answer 1

Let mass of plate be $\text{M}$
Density of gold be ${\rho }_g$​
Density of water be ${\rho }_w$​
The loss of weight of the plate 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, $F_B=\frac{\text{M}}{{\rho }_g}\times 1000\times g$
($\because {\rho }_w=1000{\text{kg/m}}^3$)
$F_B=\frac{\text{8.32}}{19.3\times {10}^3}\times 1000\times g$
Putting the values $F_B=4.31\text{N}$
$\therefore \text{The weight of the crown in water will be}(83.2-4.31)\text{N}=78.89\text{N}$