Question

A solid weighs 80gm in air, 68gm in water and 60gm in oil. Calculate the relative density of solid and oil.

Answer 1

Hint: The weight of a solid is dependent on the medium in which it is weighed. We can relate the densities of different media given the weight of the same in them. The relative density can be found from the given information as they are related to the weights.

Complete answer:
We know from the Archimedes’ Principle that the weight of liquid displaced by the solid is equal to its weight. Comparing the weights in water and that of it the air and oil can give the relative density of the solid in the oil.
The ratio of difference in weights in air and oil of the solid to that of the difference in the weights in air and water of the solid can give the relative density of the solid and the oil. This is the direct application of the Archimedes’ Principle. The idea is that the weight of the solid displaced is dependent on the liquid in which it is immersed. The denser the liquid or when it has a higher relative density, the weight experienced by the solid will be lesser.
Now, let us calculate the relative density of the solid and oil by finding the ratio between the weights displaced by the two liquid media as –
\[\begin{align}
  & \text{Relative Density of solid and oil is - } \\
 & {{\text{r}}_{so}}=\dfrac{\text{Weight displaced by solid in oil}}{\text{Weight displaced by soild in water}} \\
 & {{r}_{os}}=\dfrac{{{W}_{air}}-{{W}_{oil}}}{{{W}_{air}}-{{W}_{water}}} \\
 & \Rightarrow \text{ }{{r}_{os}}=\dfrac{80-60}{80-68} \\
 & \Rightarrow \text{ }{{r}_{os}}=\dfrac{20}{12}=1.67 \\
\end{align}\]
So, we have found the relative density of solid with respect to the oil as 1.67.

Note:
The Archimedes’ principle is the underlying property for a major portion of the phenomena based on the relative densities, buoyancy and other liquid-solid relations when the solid is immersed in the liquid. The relative density of water is almost 1 with respect to air.