Question

The relative density of a material is found by weighing the body first in air and then in water. If the weight in air is $(10.0 \pm 0.1)$ gf and the weight in water is $(5.0 \pm 0.1)$ gf then the maximum permissible percentage error in relative density is

• A. 1
• B. 2
• C. 3
• D. 5

Answer 1

Answer: (D)

Relative density $=\frac{W_{a}}{W_{a}-W_{w}}, \rho=\frac{W_{a}}{w}$
where $\rho$ is relative density, $W_{a}$ is weight in air, and $w$ is loss in weight.
$\frac{\Delta \rho}{\rho}=\frac{\Delta W_{a}}{W_{a}}-\frac{\Delta w}{w}$
For maximum error, $\frac{\Delta \rho}{\rho}=\frac{\Delta W_{a}}{W_{a}}+\frac{\Delta w}{w}$
For maximum percentage error,
$\frac{\Delta \rho}{\rho} \times 100=\frac{\Delta W_{a}}{W_{a}} \times 100+\frac{\Delta w}{w} \times 100$
Given $\Delta W_{a}=0.1 \,gf$ and $W_{a}=10.0 \,gf$
$w=10.0-5.0=5.0 gf$
$\Delta w=\Delta W_{a}+\Delta W_{w}=0.1+0.1=0.2 \,gf$
$ \frac{\Delta \rho}{\rho} \times 100=\left(\frac{0.1}{10.0}\right) \times 100+\left(\frac{0.2}{5.0}\right) \times 100 $
$=1+4=5 $