Question

The relative density of a material is found by weighing it first in airand then in water. If the weight of the body in air is W₁= 8.00 +- 0.05 N​and its weight in water is W₂= 6.00 +- 0.05 N, then the relative density q_r= W₁/(W₁- W₂) with maximum permissibleerror is
(a) 4.00 +- 0.62% (b) 4.00 +- 0.82% (c) 4.00 +- 3.2% (d) 4.00 +- 5.62%

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{d}=\frac{{\mathrm{W}}_1}{{\mathrm{W}}_1-{\mathrm{W}}_2}............\left(1\right) \\ \mathrm{d}=\frac{8}{8-6} \\ \mathrm{d}=\frac{8}{2}=4 \\ \mathrm{From}\mathrm{equation}\left(1\right), \\ \log \mathrm{d}=\log {\mathrm{W}}_1-\log \left({\mathrm{W}}_1-{\mathrm{W}}_2\right) \\ \left(\frac{∆\mathrm{d}}{\mathrm{d}}\times 100\right)=\frac{∆{\mathrm{W}}_1}{{\mathrm{W}}_1}\times 100+\frac{∆{\mathrm{W}}_1+∆{\mathrm{W}}_2}{\left({\mathrm{W}}_1-{\mathrm{W}}_2\right)}\times 100 \\ \left(\frac{∆\mathrm{d}}{\mathrm{d}}\times 100\right)=\frac{∆{\mathrm{W}}_1}{{\mathrm{W}}_1}\times 100+\frac{∆{\mathrm{W}}_1}{\left({\mathrm{W}}_1-{\mathrm{W}}_2\right)}\times 100+\frac{∆{\mathrm{W}}_2}{\left({\mathrm{W}}_1-{\mathrm{W}}_2\right)}\times 100 \\ \left(\frac{∆\mathrm{d}}{\mathrm{d}}\times 100\right)=\frac{0.05}{8}\times 100+\frac{0.05}{8-6}\times 100+\frac{0.05}{8-6}\times 100 \\ \left(\frac{∆\mathrm{d}}{\mathrm{d}}\times 100\right)=\frac{5}{8}\%+\frac{5}{2}\%+\frac{5}{2}\% \\ \left(\frac{∆\mathrm{d}}{\mathrm{d}}\times 100\right)=\frac{5+20+20}{8}\% \\ \left(\frac{∆\mathrm{d}}{\mathrm{d}}\times 100\right)=\frac{45}{8}\%=5.65\% \\ \mathrm{d}=4\pm 5.65\% \end{gathered}$$