When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass\(\left( {{\bf{lbm}}} \right)\)was employed, where\({\bf{1}}{\rm{ }}{\bf{lbm}} = {\bf{0}}.{\bf{4539}}{\rm{ }}{\bf{kg}}\). (a) If there is an uncertainty of\({\bf{0}}.{\bf{0001}}{\rm{ }}{\bf{kg}}\)in the pound-mass unit, what is its percent uncertainty? (b) Based on that percent uncertainty, what mass in pound-mass has an uncertainty of\({\bf{1}}{\rm{ }}{\bf{kg}}\)when converted to kilograms?

Percentage uncertainty: This is just \({\bf{100}}\) percent related uncertainty. Since percentage uncertainty is also a measure of the same value, it has no units

Consider the given data as below.

The mass,\(\delta m = 0.0001{\rm{ }}kg\)

The mass,\(m = 0.4539{\rm{ }}kg\)

Define the uncertainty of\(0.0001{\rm{ }}kg\)in the pound-mass unit.

\(\begin{aligned}{c}\delta m\% &= \frac{{\delta m}}{m} \times 100\% \\ &= \frac{{0.0001}}{{0.4539}} \times 100\% \\ &= 0.022\% \end{aligned}\)

Hence, the percentage uncertaintyis\(0.022\% \).

Define mass \(m\) in \(lbm\):

\(\delta m = 1{\rm{ }}kg\)

Therefore, the mass will be

\(\begin{aligned}{c}m &= \frac{{\delta m}}{{\delta m\% }} \times 100\\ &= \frac{1}{{0.022}} \times 100\end{aligned}\)

From conversion factor

\(\begin{aligned}{c}m &= 4545{\rm{ }}kg\left( {\frac{{1{\rm{ }}lbm}}{{0.4539{\rm{ }}kg}}} \right)\\ &= {10^4}{\rm{ }}{\mathop{\rm lbm}\nolimits} \end{aligned}\)

The mass is \({10^4}{\rm{ }}{\mathop{\rm lbm}\nolimits} \).