Q.

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

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya

answer is [OBJECT OBJECT], [OBJECT OBJECT].

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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⇒ln⁡N=ln⁡N0−λt⇒ΔNN=Δλt            ∴Δλ=402000×1=0.02        (N=3000-1000=2000 is number of nuclei left undecayed )
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon