The relative density of material of a body is found by weighing it first in air and then in water. If the weight in air is $\left(5.00\pm 0.05\right)\mathrm{N}$ and the weight in water is $\left(4.00\pm 0.05\right)\mathrm{N}$. Find the relative density along with the maximum permissible percentage error.

$\mathrm{Weight} \mathrm{in} \mathrm{air} =\left(5.00\pm 0.05\right)\mathrm{N}$

Weight in water $=\left(4.00\pm 0.05\right)\mathrm{N}$

Loss of weight in water $=\left(1.00\pm 0.1\right)\mathrm{N}$

Now relative density $=\frac{\mathrm{Weight} \mathrm{in} \mathrm{air}}{\mathrm{Weight} \mathrm{loss} \mathrm{in} \mathrm{water}}$

i.e. $\mathrm{RD}=\frac{5.00\pm 0.05}{1.00\pm 0.1}$ 

Now relative density with Maximum permissible error.

$$\begin{gathered} =\frac{5.00}{1.00}\pm \left[\frac{0.05}{5.00}+\frac{0.1}{1.00}\right]\times 100 \\ =5.0\pm \left(1+10\right)\%=5.0\pm 11\% \end{gathered}$$