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

Answer:

$\Delta \rho =0.18/mL326gm$

Explanation:

we have error in division of 2 quantities is related as

$z=\frac{a}{b}\\\\\frac{\Delta z}{z}=\frac{\Delta a}{a}+\frac{\Delta b}{b}$

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

Thus error in density becomes

$\Delta \rho =\rho _{0}\times(\frac{\Delta m}{m}+\frac{\Delta V}{V})$

Applying values we get

$\Delta \rho =\frac{4.635}{1.13} \times(\frac{0.002}{4.635}+\frac{0.05}{1.13})$

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