Question

The mass and diameter of a steel ball were used to determine its density the mass was measured within 1% dm3 the diameter within 3% what is the percentage error in the calculated density of the steel?

Answer 1

10%

Explanation:

density= mass/ volume`= rho= 3/4 m/(pi xx r^3)=6 m/(pi xx d^3`
the error of the misure for a result obtained by a moltiplication or a division is given by the summ of the single errors:
`%E_t= %E_m + %3 xx E_d=(1 + 3 xx 3) = 10%`

You suppose that your ball weighs 4056 kg and its weight can be 4015 kg or 4096 and its diameter is 100 cm= 1m. Well the diameter can be 97 cm or 103 cm. for example if it was 103 cm, the volume, instead of
`V= 1/6 pi xx d^3= 0,52 xx 1 m^3= 0.52 m^3` it will be
`V= 1/6 pi xx d^3= 0,52 xx 1.03^3 m^3= 0.57 m^3`
if the mass was the minimum instead of
`rho = m/V = (4056 kg)/(0.52 m^3) = 7800( kg)/(m^3)` you will have
`rho = m/V = (4015 kg)/(0.57 m^3) = 7044( kg)/(m^3)`
and the error will be `=(7800-7044)/7800 = 10%`