Physics · Units, Dimensions & Error Analysis

JEE Advanced 2018 — Paper 1 — Question 44

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/yz=x / y. If the errors in x,yx, y and zz are Δx,Δy\Delta x, \Delta y and Δz\Delta z,

respectively, then

z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)−1z \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 (1±Δyy)−1\left(1 \pm \frac{\Delta y}{y}\right)^{-1}, to first power in Δy/y\Delta y / y, is 1∓(Δy/y)1 \mp(\Delta \mathrm{y} / y). The relative errors in

independent variables are always added. So the error in zz will be

Δz=z(Δxx+Δyy)\Delta z=z\left(\frac{\Delta x}{x}+\frac{\Delta y}{y}\right)

The above derivation makes the assumption that Δx/x≪1,Δy/y≪1\Delta x / x \ll 1, \Delta y / y \ll 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±401000 \pm 40 nuclei decayed in the first 1.0s. For ∣x∣≪1|x| \ll 1, ln⁡(1+x)≈x\ln(1+x) \approx x up to first power in xx. The error Δλ\Delta\lambda, in the determination of the decay constant λ\lambda, in s−1^{-1}, is

  1. Option A:

    0.04

  2. Option B:

    0.03

  3. Option C:

    0.02

    Correct
  4. Option D:

    0.01

Answer: C

Step-by-step solution

Nd=N0(1−e−λt)N_d = N_0(1-e^{-\lambda t})

1000=3000(1−e−λ⋅1)⇒e−λ=231000 = 3000 (1-e^{-\lambda \cdot 1}) \Rightarrow e^{-\lambda} = \frac{2}{3}

ΔNd=N0e−λtλΔt\Delta N_d = N_0 e^{-\lambda t} \lambda \Delta t

40=3000×23×Δλ40 = 3000 \times \frac{2}{3} \times \Delta\lambda

Δλ=0.02\Delta\lambda = 0.02

alternate :

N=N0e−λtN = N_0 e^{-\lambda t}

N+ΔN=N0e−(λ+Δλ)tN + \Delta N = N_0 e^{-(\lambda+\Delta\lambda)t}

ΔN=N0[e−(λ+Δλ)t−e−λt]\Delta N = N_0 \left[ e^{-(\lambda+\Delta\lambda)t} - e^{-\lambda t} \right]

ΔN=N0[e−λte−Δλt−e−λt]\Delta N = N_0 \left[ e^{-\lambda t} e^{-\Delta\lambda t} - e^{-\lambda t} \right]

ΔN=N0e−λt[e−Δλt−1]\Delta N = N_0 e^{-\lambda t} \left[ e^{-\Delta\lambda t} - 1 \right]

ΔN≈N0e−λt[Δλt]\Delta N \approx N_0 e^{-\lambda t} [\Delta\lambda t]

Δλ=0.02\Delta\lambda = 0.02

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Significant Figures and Error Analysis
If the measurement errors in all the independent quantities are… | JEE Advanced 2018 PYQ with Solution · DhiX AI