Question #154795

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


1
Expert's answer
2021-01-13T11:37:56-0500

m=(52.5±0.1)gm=(52.5\pm0.1)g , ϵ=0.152.5100%=0.2%\epsilon=\frac{0.1}{52.5}\cdot100\%=0.2\%


h=(18.0±0.1)mmh=(18.0\pm0.1)mm , ϵ=0.118.0100%=0.6%\epsilon=\frac{0.1}{18.0}\cdot100\%=0.6\%


d=(22.0±0.1)mmd=(22.0\pm0.1)mm , ϵ=0.122.0100%=0.5%\epsilon=\frac{0.1}{22.0}\cdot100\%=0.5\%


D=mV=4mπd2h=40.05253.140.02220.018=7673(kg/m3)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)


ΔDD=(Δmm)2+(Δhh)2+(2Δdd)2\frac{\Delta D}{D}=\sqrt{(\frac{\Delta m}{m})^2+(\frac{\Delta h}{h})^2+(2\frac{\Delta d}{d})^2}


ΔDD=(0.002)2+(0.006)2+(20.005)2=0.012\frac{\Delta D}{D}=\sqrt{(0.002)^2+(0.006)^2+(2\cdot 0.005)^2}=0.012 or ϵ=1.2%\epsilon=1.2\%


ΔD=D0.012=76730.012=92(kg/m3)\Delta D=D\cdot0.012=7673\cdot0.012=92(kg/m^3)


D=(7670±90)kg/m3D=(7670\pm90)kg/m^3 , ϵ=1.2%\epsilon=1.2\%









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