Question

The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is $W_{1}=8.00 \pm 0.05 \,N$ and the weight in water is $W_{2}=6.00 \pm 0.05\, N$, then the relative density $\rho_{r}=W_{1} /\left(W_{1}-W_{2}\right)$ with the maximum permissible error is

• A. $4.00 \pm 0.62 \%$
• B. $4.00 \pm 0.82 \%$
• C. $4.00 \pm 3.2 \%$
• D. $4.00 \pm 5.62 \%$

Answer 1

Answer: (D)

Relative density $p_{r}=\frac{W_{1}}{W_{1}-W_{2}}=\frac{8.00}{8.00-6.00}=4.00$
$\frac{\Delta \rho_{r}}{\rho_{r}} \times 100=\frac{\Delta W_{1}}{W_{1}} \times 100+\frac{\Delta\left(W_{1}-W_{2}\right)}{W_{1}-W_{2}} \times 100$
$=\frac{0.05}{8.00} \times 100+\frac{0.05+0.05}{2} \times 100=5.62 \%$
$\therefore \rho_{r}=4.00 \pm 5.62 \%$