Answer to Question #154795 in Physics for Moses

Question

During an experiment to determine the dimensions of a cylinder, a scholar used a vernier calliper and a balance and recorded the following:Mass of cylinder, m = (52.5 ± 0.1)gHeight of cylinder, h = (18.0 ± 0.1)mmDiameter of cylinder, d = (22.0 ± 0.1)mmDeduce the density of the cylinder, DDeduce the relative error in DDeduce the maximum error in DDeduce the percentage error in m Deduce the percentage error in hDeduce the percentage error in dDeduce the total percentage error in D

Answer 1

$m=(52.5\pm0.1)g$ , $\epsilon=\frac{0.1}{52.5}\cdot100\%=0.2\%$

$h=(18.0\pm0.1)mm$ , $\epsilon=\frac{0.1}{18.0}\cdot100\%=0.6\%$

$d=(22.0\pm0.1)mm$ , $\epsilon=\frac{0.1}{22.0}\cdot100\%=0.5\%$

$D=\frac{m}{V}=\frac{4m}{\pi d^2h}=\frac{4\cdot0.0525}{3.14\cdot 0.022^2\cdot0.018}=7673(kg/m^3)$

$\frac{\Delta D}{D}=\sqrt{(\frac{\Delta m}{m})^2+(\frac{\Delta h}{h})^2+(2\frac{\Delta d}{d})^2}$

$\frac{\Delta D}{D}=\sqrt{(0.002)^2+(0.006)^2+(2\cdot 0.005)^2}=0.012$ or $\epsilon=1.2\%$

$\Delta D=D\cdot0.012=7673\cdot0.012=92(kg/m^3)$

$D=(7670\pm90)kg/m^3$ , $\epsilon=1.2\%$