If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $$z=x/y$$. If the errors in x, y and z are Δx,Δy and Δz, respectively, then z±Δ$$z=x$$±Δxy±Δ$$y=xy1$$±Δ$$xx1$$±Δyy−$$1The$$ series expansion for $$1$$±Δyy−$$1$$, to first power in Δ$$y/y$$ , is $$1$$∓Δ$$y/y$$. The relative errors in independent variables are always added. So the error in z will be Δ$$z=z$$Δ$$xx+$$ΔyyThe above derivation makes the assumption that Δ$$x/x$$

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If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $$z=x/y$$. If the errors in x, y and z are Δx,Δy and Δz, respectively, then  z±Δ$$z=x$$±Δxy±Δ$$y=xy1$$±Δ$$xx1$$±Δyy−$$1The$$ series expansion for $$1$$±Δyy−$$1$$, to first power in Δ$$y/y$$ , is  $$1$$∓Δ$$y/y$$. The relative errors in independent variables are always added. So the error in z will be  Δ$$z=z$$Δ$$xx+$$ΔyyThe above derivation makes the assumption that  Δ$$x/x$$<<$$1$$,Δ$$y/y$$<<$$1$$  . Therefore, the higher power of these quantities are neglected.   Consider the ratio $$r=1$$−$$a1+a$$ to be determined by measuring a dimensionless quantity a. If the error in the measurement of a is ΔaΔ$$a/a$$<<$$1$$ , then what is the error Δr in determining r?In an experiment the initial number of radioactive nuclei is $$3000$$. It is found that $$1000$$ ± $$40$$ nuclei decayed in the first $$1.0$$ s. For |x| ≪ $$1$$, $$ln(1$$ $$+x)$$ $$=$$ x up to first power in x. The error ∆λ, in the determination of the decay constant  λ, in s−$$1$$ , is

Infinity Learn · Accepted Answer

$$r=1$$−$$a1+a$$ ⇒Δ$$rr=$$Δ$$1$$−$$a1$$−$$a+$$Δ$$1+a1+a$$⇒Δ$$rr=$$Δ$$a1$$−$$a+$$Δ$$a1+a$$⇒Δ$$rr=$$Δ$$a1+a+1$$−$$a1$$−$$a1+a$$∴Δ$$r=2$$Δ$$a1$$−$$a1+a1$$−$$a1+a=2$$Δ$$a(1+a)2N=N0e$$−λt⇒ln⁡$$N=ln$$⁡$$N0$$−λt⇒Δ$$NN=$$Δλt            ∴Δλ$$=402000$$×$$1=0.02$$        (N=3000-1000=2000$$ is number of nuclei left undecayed $$)

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