# College Physics Problem 1.25

#### 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.

The average volume of the box is $\displaystyle Volume=l\times w\times h$ $\displaystyle Volume=\left(1.80\:cm\right)\left(2.05\:cm\right)\left(3.1\:cm\right)$ $\displaystyle Volume=11.4\:cm^3$          ☚

The percent uncertainty for each of the dimensions: $\displaystyle 1.80\pm 0.01\:cm\:\rightarrow \:\frac{0.01\:cm}{1.80\:cm}\times 100\%=0.556\%$ $\displaystyle 2.05\pm 0.02\:cm\:\rightarrow \:\frac{0.02\:cm}{2.05\:cm}\times 100\%=0.976\%$ $\displaystyle 3.1\pm 0.1\:cm\:\rightarrow \:\frac{0.1\:cm}{3.1\:cm}\times 100\%=3.226\%$

The percent uncertainty in the volume of the box is calculated by adding the percent uncertainties of the dimensions. $\displaystyle \%\:uncertainty_{volume}=0.556\%+0.976\%+3.226\%$ $\displaystyle \%\:uncertainty_{volume}=4.758\%$

The uncertainty of the volume is $\displaystyle \delta _{vol}=0.04758\times 11.4\:cm^3$ $\displaystyle \delta _{vol}=0.54\:cm^3$          ☚

Therefore, the volume is $\displaystyle Volume=11.4\pm 0.54\:cm^3$          ☚