Question

Suppose that you determine the density of a mineral by measuring its mass (m) (4.635±0.002) g and its volume (1.13±0.05) mL. d = m/V What is the uncertainty in the computed density?

Answer 1

`\Delta \rho =0.18/mL326gm`

Step-by-step explanation:

we have error in division of 2 quantities is related as

`z=(a)/(b)\\\\(\Delta z)/(z)=(\Delta a)/(a)+(\Delta b)/(b)`

where
`\Delta a,\Delta b` are errors in quantities a,b

Thus error in density becomes

`\Delta \rho =\rho _(0)*((\Delta m)/(m)+(\Delta V)/(V))`

Applying values we get

`\Delta \rho =(4.635)/(1.13) *((0.002)/(4.635)+(0.05)/(1.13))`

thus
`\Delta \rho =0.18326gm/mL`