Calculated Percentage Fractional Error in Density of Uniform Cylinder

The density $\rho$ of a uniform cylinder of mass $m$, length $l$, and diameter $d$ is given by the formula: $\rho = \frac{m}{V} = \frac{m}{\frac{\pi}{4} d^2 l} = \frac{4m}{\pi d^2 l}$

Taking the natural logarithm on both sides and differentiating gives the maximum relative (fractional) error in density: $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta l}{l} + 2\frac{\Delta d}{d}$

The corresponding percentage fractional error is: $\left(\frac{\Delta \rho}{\rho}\right) \times 100\% = \left( \frac{\Delta m}{m} + \frac{\Delta l}{l} + 2\frac{\Delta d}{d} \right) \times 100\%$

Given the measured values and their uncertainties:

• Mass: $m = 97.42\text{ g}$, $\Delta m = 0.02\text{ g}$
• Length: $l = 8.35\text{ mm}$, $\Delta l = 0.05\text{ mm}$
• Diameter: $d = 20.20\text{ mm}$, $\Delta d = 0.02\text{ mm}$

Calculating each individual percentage error term:

1. Percentage error in mass: $\frac{\Delta m}{m} \times 100\% = \frac{0.02}{97.42} \times 100\% \approx 0.0205\%$

2. Percentage error in length: $\frac{\Delta l}{l} \times 100\% = \frac{0.05}{8.35} \times 100\% \approx 0.5988\%$

3. Percentage error due to diameter: $2\frac{\Delta d}{d} \times 100\% = 2 \times \left(\frac{0.02}{20.20}\right) \times 100\% = \frac{0.04}{20.20} \times 100\% \approx 0.1980\%$

Summing these contributions to find the total percentage fractional error in density: $\left(\frac{\Delta \rho}{\rho}\right) \times 100\% = 0.0205\% + 0.5988\% + 0.1980\% = 0.8173\% \approx 0.82\%$

Thus, the calculated percentage fractional error in $\rho$ is $0.82\%$.

Correct Answer: B ($0.82\%$)