Q25PE
Question
The sides of a small rectangular box are measured to be \({\bf{1}}.{\bf{80}} \pm {\bf{0}}.{\bf{01}}{\rm{ }}{\bf{cm}}\), \({\bf{2}}.{\bf{05}} \pm {\bf{0}}.{\bf{02}}{\rm{ }}{\bf{cm}}\), and \({\bf{3}}.{\bf{1}} \pm {\bf{0}}.{\bf{1}}{\rm{ }}{\bf{cm}}\) long. Calculate its volume and uncertainty in cubic centimeters
Step-by-Step Solution
VerifiedThe volume and uncertainty in cubic centimeters is \(V + \Delta V = \left( {11.439 \pm 0.544} \right){\rm{ }}c{m^3}\).
Uncertainty as a percentage is just relative uncertainty multiplied by \({\rm{100}}\). The percent uncertainty likewise lacks units since it is a ratio of comparable values.
The volume of a rectangular prism is equal to the area of the base that doubles its height. Therefore, the volume of a rectangular prism formula is given as.
\({\rm{Rectangular prism volume = }}\left( {{\rm{Length x Width x Length}}} \right){\rm{ cubic units}}\)
Given data:
Consider the given data as below.
Small Rectangle box of dimensions:
The length, \(l = 1.80 \pm 0.01{\rm{ }}cm\)
The base, \(b = 2.05 \pm 0.02{\rm{ }}cm\)
The height, \(h = 3.1 \pm 0.1{\rm{ }}cm\)
\(V = lbh\)
Substitute known values I n the above equation.
\(\begin{aligned}{\underline{\phantom{xx}}}V &= 1.80 \times 2.05 \times 3.1\\ &= 11.439{\rm{ }}c{m^3}\end{aligned}\)
Determine the uncertainty for volume as below.
\(\begin{aligned}{\underline{\phantom{xx}}}\frac{{\Delta V}}{V} &= \frac{{\Delta L}}{L} + \frac{{\Delta B}}{B} + \frac{{\Delta H}}{H}\\ &= \pm \left( {\frac{{0.01}}{{1.80}} + \frac{{0.02}}{{2.05}} + \frac{{0.1}}{{3.1}}} \right) \times 100\% \\ &= \pm 4.757\% \end{aligned}\)
Solve the above equation for uncertainty for volume as below.
\(\begin{aligned}{\underline{\phantom{xx}}}\Delta V &= \pm 4.757\% \times V\\ &= \pm 4.757\% \times 11.439\\ &= 0.544{\rm{ }}c{m^3}\end{aligned}\)
Therefore, the volume and uncertainty in cubic centimeters is \(V + \Delta V = \left( {11.439 \pm 0.544} \right){\rm{ }}c{m^3}\).