Answer to Question #306521 in Physics for Mr. Lee

Question

The following data are the measurements of the dimensions of a box.

𝑙 = 3.0 ± 0.3 cm

ℎ = 51.1 ± 0.5 mm

𝑤 = 2.03 ± 0.01 m

How should the volume of the box be reported (central value ± uncertainty)? Write your calculations and box your final answer. 

Answer 1

The volume

$$V=lhw=3.0*0.511*203=310\:\rm cm^3$$

The relative uncertainty

$$\frac{\Delta V}{V}=\frac{\Delta l}{l}+\frac{\Delta h}{h}+\frac{\Delta w}{w}$$

$$\frac{\Delta V}{V}=\frac{0.3}{3.0}+\frac{0.5}{51.1}+\frac{0.01}{2.03}=0.11$$

$$\Delta V=310*0.11=34\rm\: cm^3$$

Finally

$$V=(310\pm34)\:\rm cm^3$$