Topic: Metrology and Inspection
A wire has a mass (0.3 ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in the measurement of density is
- A. 1
- B. 2
- C. 3
- D. 4
Answer: D. 4
Explanation: Density $\rho$ is given by the formula $\rho = \frac{m}{V} = \frac{m}{\pi r² L}$. Using error propagation rules, the maximum relative percentage error is given by adding the individual relative errors: $\frac{\Delta \rho}{\rho} \% = \frac{\Delta m}{m} \% + 2\frac{\Delta r}{r} \% + \frac{\Delta L}{L} \%$. Plugging in the values gives: $(0.003/0.3) \times 100 + 2(0.005/0.5) \times 100 + (0.06/6) \times 100 = 1\% + 2(1\%) + 1\% = 4\%$.