Question

If weight of an object in air $W_a$ is $(10.0\pm 0.1)$g and its weight in water $W_L$ is $(5.00\pm 0.2)$g, then percentage error in its relative density is?$(Relativedensity=\frac{W_a}{W_a-W_L})$

• A. $2\%$
• B. $5\%$
• C. $7\%$
• D. $1.5\%$

Answer 1

The correct option is C $7\%$

Given $W_a=10.00\pm 0.1;W_L=5.00\pm 0.2$

Loss of weight in water$=W_a-W_L$

=(10.00$\pm$0.1)-(5.00$\pm$0.2)

=(10-5)$\pm$(0.1+0.2)

=5$\pm$0.3

Now percentage error in relative density is given by$=\frac{W_a}{W_a-W_L}=\frac{10.00\pm 0.1}{5\pm 0.3}=\left(\frac{10}{5}\right)\pm (\frac{0.1}{10}+\frac{0.3}{5})\times 100=2\pm (1+6)=2\pm 7$

7 percent error