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±Δz=x±Δxy±Δy=xy1±Δxx1±Δyy−1The series expansion for 1±Δyy−1, to first power in Δy/y , is 1∓Δy/y. The relative errors in independent variables are always added. So the error in z will be Δz=zΔxx+ΔyyThe above derivation makes the assumption that Δx/x<<1,Δy/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 ΔaΔa/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
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is [OBJECT OBJECT], [OBJECT OBJECT].
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
r=1−a1+a ⇒Δrr=Δ1−a1−a+Δ1+a1+a⇒Δrr=Δa1−a+Δa1+a⇒Δrr=Δa1+a+1−a1−a1+a∴Δr=2Δa1−a1+a1−a1+a=2Δa(1+a)2N=N0e−λt⇒lnN=lnN0−λt⇒ΔNN=Δλt ∴Δλ=402000×1=0.02 (N=3000-1000=2000 is number of nuclei left undecayed )