If the measurement errors in all the independent quantities are known, then it is…

Physics · JEE Advanced · NTA ExamsPhysics and Measurement


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
\(z \pm \Delta z=\frac{x \pm \Delta x}{y \pm \Delta y}=\frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1}\)
The series expansion for \(\left(1 \pm \frac{\Delta y}{y}\right)^{-1},\)to first power in Δy/y, is 1 y/y). The relative errors in independent variables are always added. So the error in z will be
The above derivation makes the assumption that Δx/x << 1, Δy/y << 1 . Therefore, the higher powers of these quantities are neglected.
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 Δλ, in the determination of the decay constant
λ, in s–1, is
  1. 0.04
  2. 0.03
  3. 0.02
  4. 0.01

Answer

(C) 0.02

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