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 _______.

The density of a uniform cylinder is given by \(\rho = \dfrac{m}{V} = \dfrac{4m}{\pi d^2 l}\)

The percentage fractional error in density is given by:

\(\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{\Delta m}{m} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l} \right) \times 100\)

Substituting the given values:

\(\dfrac{\Delta \rho}{\rho} \times 100 = \left( \dfrac{0.02}{97.42} + 2 \times \dfrac{0.02}{20.20} + \dfrac{0.05}{8.35} \right) \times 100\)

\(\dfrac{\Delta \rho}{\rho} \times 100 = \left( 0.000205 + 0.00198 + 0.005988 \right) \times 100\)

\(\dfrac{\Delta \rho}{\rho} \times 100 = 0.008173 \times 100 = 0.8173\%\)

Rounding off to two decimal places, the percentage error is \(0.82\%\).

Answer: \(0.82\%\)