The density $\rho$ of a uniform cylinder is determined by measuring its mass $m$, length $l$ and diameter $d$. The measured values of $m, l$ and $d$ are $97.42 \pm 0.02$ g, $8.35 \pm 0.05$ mm and $20.20 \pm 0.02$ mm, respectively. Calculated percentage fractional error in $\rho$ is ______.
Answer: (B) $0.82\%$
Concept. Density of a cylinder $\rho=\dfrac{m}{V}=\dfrac{4m}{\pi d^2l}$. The relative errors add with their powers, so the diameter error counts twice.
Step 1. Error formula.
$$\frac{\Delta\rho}{\rho}=\frac{\Delta m}{m}+2\frac{\Delta d}{d}+\frac{\Delta l}{l}$$
Step 2. Each term.
$$\frac{\Delta m}{m}=\frac{0.02}{97.42}=0.00021$$
$$2\frac{\Delta d}{d}=2\times\frac{0.02}{20.20}=0.00198$$
$$\frac{\Delta l}{l}=\frac{0.05}{8.35}=0.00599$$
Step 3. Add.
$$\frac{\Delta\rho}{\rho}=0.00021+0.00198+0.00599=0.00818\approx0.82\%$$
Answer. $0.82\%$ (Option B). The short length ($8.35$ mm) gives the largest share of the error.
Solution by Sreeraj P, M.Sc Physics
