Problem 1-25: Solving for the uncertainty of the volume of a box with dimensional uncertainties

The sides of a small rectangular box are measured to be 1.80±0.01 cm, 2.05±0.02 cm, and 3.1±0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.

Advertisements

SOLUTION:

The average volume of the box is

\begin{align*} \text{Volume} & =l\times w\times h \\ & =\left(1.80\:\text{cm}\right)\left(2.05\:\text{cm}\right)\left(3.1\:\text{cm}\right) \\ &= 11.4\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\ \end{align*}

The percent uncertainty for each of the dimensions:

\begin{align*} 1.80\pm 0.01\:\text{cm}\:\rightarrow & \:\frac{0.01\:\text{cm}}{1.80\:\text{cm}}\times 100\%=0.556\% \\ \\ 2.05\pm 0.02\:\text{cm}\:\rightarrow & \:\frac{0.02\:\text{cm}}{2.05\:\text{cm}}\times 100\%=0.976\% \\ \\ 3.1\pm 0.1\:\text{cm}\:\rightarrow &\:\frac{0.1\:\text{cm}}{3.1\:\text{cm}}\times 100\%=3.226\% \\ \end{align*}

The percent uncertainty in the volume of the box is calculated by adding the percent uncertainties of the dimensions.

\begin{align*} \%\:\text{uncertainty}_{\text{volume}} & =0.556\%+0.976\%+3.226\% \\ & =4.758\% \end{align*}

The uncertainty of the volume is

\begin{align*} \delta _{\text{volume}} & =0.04758\times 11.4\:\text{cm}^3 \\ & =0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the volume is

\text{Volume}=11.4\pm 0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)