Question

A solid weighs $1.5\text{kgf}$ in air and $0.9\text{kgf}$ in a liquid of density $1.2\times {10}^3{\text{kg/m}}^3$. Calculate relative density of the solid. $\left[3\text{marks}\right]$

Answer 1

Weight of the fluid displaced = Apparent decrease in the weight $=1.5-0.9=0.6\text{kgf}$
Mass of fluid displaced $=0.6\text{kgf}$
Volume of solid = mass/density $=\frac{0.6}{1.2\times {10}^3}=0.5\times {10}^{-3}{\text{m}}^3$

Weight of the solid in the air $=1.5\text{kgf}$

Mass of the solid in air $=1.5\text{kg}$ $\left[1\text{mark}\right]$

Density of the solid $=\frac{\text{Mass}}{\text{Density}}=\frac{1.5}{0.5\times {10}^{-3}}=3\times {10}^3{\text{kg/m}}^3$
Thus,
Relative density $=\frac{3\times {10}^3{\text{kg/m}}^3}{1\times {10}^3{\text{kg/m}}^3}=3$ $\left[2\text{marks}\right]$