Question

A piece of metal weighs $50.0N$ in air, $36.0N$ in water, and $41.0N$ in an unknown liquid. Find the density of the unknown liquid.

Answer 1

Water exerts a buoyant force to the metal $=50-36=14N$, which equals the weight of water displaced.
The mass of water displaced is $\frac{14}{10}=1.4kg$.
Since the density of water is $1kg/l$, the volume displaced is $1.4l$.
Therefore the density of the metal is $\frac{\left(\frac{50}{10}\right)}{1.4}=3.57$$kg/l$
The unknown liquid exerts a buoyant force of $50-41=$$9N$. The density of this liquid is $\frac{\left(\frac{9}{10}\right)}{1.4}=0.642kg/litre=6.42kg/m^3$