Question

If the mass of an object is measured to be 53.5 ± 0.1 g and its volume is measured to be 22.30 ± 0.05 cm^3 , what is the density? Report the density uncertainty in both forms: as a percentage and as an absolute number with units.

Answer 1

`0.4 \%`

`\Delta \rho=0.0098 (g)/(cm^3)`

Step-by-step explanation:

Recall that density is defined as
`(m)/(V)` and that relative uncertainty is defined as
`(\Delta a)/(a)` where
`\Delta a` is the uncertainty in the measure and a the measure, To find the uncertainty when two physical quantities are divided, their relative uncertainties are added and then multiplied with the division result of the quantities.

`\rho=(m)/(V) \pm (m)/(V)((\Delta m)/(m)+(\Delta V)/(V))\\\rho=(53.5g)/(22.30cm^3) \pm (53.5g)/(22.30cm^3)((0.1 g)/(53.5g)+(0.05 cm^3)/(22.30cm^3))=(2.3991 \pm 0.0098 )(g)/(cm^3)}`

We have:
`\Delta \rho=0.0098 (g)/(cm^3)`

To find the percent uncertainty, we multiply the relative uncertainty by 100%.

`(\Delta \rho)/(\rho)*100 \%=(0.0098(g)/(cm^3))/(2.3991(g)/(cm^3))*100 \%\\(\Delta \rho)/(\rho)=0.0040*100 \%=0.4 \%`