Question

Sol. [Key Concept: $\rho = \frac{m}{\frac{\pi}{4}d^{2}l}$, so the fractional error $\frac{\Delta\rho}{\rho} = \frac{\Delta\ m}{m} + \frac{2\Delta\ d}{d} + \frac{\Delta\ l}{l}$; convert to a percentage.]

$${m = 97.42 \pm 0.02\text{ }g }{1 = 8.35 \pm 0.05\text{ }mm }{d = 20.20 \pm 0.02\text{ }mm }{\rho = \frac{m}{\frac{\pi}{4}{\text{ }d}^{2}\mathcal{l}} }{\Rightarrow \frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + \frac{2\Delta\text{ }d}{\text{ }d} + \frac{\Delta\mathcal{l}}{\mathcal{l}} }{= \frac{0.02}{97.42} + 2 \times \frac{0.02}{20.20} + \frac{0.05}{8.35} = 0.0082 }$$% error $= 0.82\%$