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Physics · Units, Dimensions & Error Analysis

JEE Advanced 2018 — Paper 1 — Question 43

Consider the ratio r=(1−a)(1+a)r = \frac{(1-a)}{(1+a)} to be determined by measuring a dimensionless quantity aa. If the error in the measurement of aa is Δa\Delta a (Δa/a≪1\Delta a / a \ll 1), then what is the error Δr\Delta r in determining rr?

  1. Option A:

    Δa(1+a)2\frac{\Delta a}{(1+a)^2}

  2. Option B:

    2Δa(1+a)2\frac{2\Delta a}{(1+a)^2}

    Correct
  3. Option C:

    2Δa(1−a2)\frac{2\Delta a}{(1-a^2)}

  4. Option D:

    2Δa(1−a2)\frac{2\Delta a}{(1-a^2)}

Answer: B

Step-by-step solution

r+Δr=1−(a+Δa)1+(a+Δa)r+\Delta r = \frac{1-(a+\Delta a)}{1+(a+\Delta a)}

Δr=1−(a+Δa)1+(a+Δa)−1−a1+a=[1−(a+Δa)](1+a)−[1+(a+Δa)](1−a)(1+a+Δa)(1+a)\Delta r = \frac{1-(a+\Delta a)}{1+(a+\Delta a)} - \frac{1-a}{1+a} = \frac{[1-(a+\Delta a)](1+a) - [1+(a+\Delta a)](1-a)}{(1+a+\Delta a)(1+a)}

∣Δr∣=∣−2Δa(1+a)2∣|\Delta r| = \left|\frac{-2\Delta a}{(1+a)^2}\right|

alternate: Δr=1−a1+a[−Δa1−a−Δa1+a]\Delta r = \frac{1-a}{1+a} \left[ \frac{-\Delta a}{1-a} - \frac{\Delta a}{1+a} \right]

Δr=2Δa(1+a)2\Delta r = \frac{2\Delta a}{(1+a)^2}

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

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