Question

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 $(5.00\pm 0.05)$N and the weight in water is $(4.00\pm 0.05)$N. Find the relative density along with the maximum permissible percentage error.

Answer 1

Weight in air = $(5.00\pm 0.05)$N
Weight in water = $(4.00\pm 0.05)$N
Loss of weight in water = $(1.00\pm 0.1)$N
Now relative density = $\frac{weightinair}{weightlossinwater}$
i.e., $R.D=\frac{5.00\pm 0.05}{1.00\pm 0.1}$
Now relative density with maximum permissible error
$\frac{5.00}{1.00}\pm (\frac{0.05}{5.00}+\frac{0.1}{1.00})\times 100$
= $5.0\pm (1+10)\%=5.0\pm 11\%$