Q 11-01-010NEETNEET 2023Top questionMedium
A metal wire has mass $(0.4 \pm 0.002)$ g, radius $(0.3 \pm 0.001)$ mm and length $(5 \pm 0.02)$ cm. The maximum possible percentage error in the measurement of density will nearly be:
Answer: (C) $1.6\%$
$\rho = \dfrac{m}{\pi r^2 l}$, so
$$\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$$
$$= \frac{0.002}{0.4} + 2 \times \frac{0.001}{0.3} + \frac{0.02}{5} = 0.005 + 0.00667 + 0.004 = 0.01567$$
Percentage error $\approx 1.6\%$.
Solution by Sreeraj P, M.Sc Physics