JEE Challenger
More from Units and Measurements
PhyMedAdv 2018 • Paper 1

Error Calculation in Rational Function of Single Variable

Comprehension Passage

When the measurement errors associated with independent variables are given, the resulting uncertainty in any dependent quantity can be estimated. This is achieved via a series expansion, retaining only terms up to the first power of the error. For instance, consider the functional relation z=x/yz = x/y. If uncertainties in xx, yy, and zz are denoted by Δx\Delta x, Δy\Delta y, and Δz\Delta z respectively, then z±Δz=x±Δxy±Δy=xy(1±Δxx)(1±Δyy)−1.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}. Expanding (1±Δyy)−1\left(1 \pm \frac{\Delta y}{y}\right)^{-1} to first order in Δy/y\Delta y/y yields 1∓(Δy/y)1 \mp (\Delta y/y). Since fractional uncertainties from independent parameters combine additively, the absolute error in zz is given by Δz=z(Δxx+Δyy).\Delta z = z \left(\frac{\Delta x}{x} + \frac{\Delta y}{y}\right). This formulation assumes Δx/x≪1\Delta x/x \ll 1 and Δy/y≪1\Delta y/y \ll 1, allowing higher-order error terms to be neglected.

Let the quantity r=(1−a)(1+a)r = \frac{(1-a)}{(1+a)} be evaluated through the measurement of a dimensionless variable aa. If the uncertainty in determining aa is Δa\Delta a (where Δa/a≪1\Delta a/a \ll 1), what is the corresponding uncertainty Δr\Delta r in the value of rr?

Options

A

Δa(1+a)2\frac{\Delta a}{(1+a)^2}

B

2Δa(1+a)2\frac{2\Delta a}{(1+a)^2}

Correct
C

2Δa(1−a2)\frac{2\Delta a}{(1-a^2)}

D

2aΔa(1−a2)\frac{2a\Delta a}{(1-a^2)}

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To find the uncertainty Δr\Delta r in the value of r=1−a1+ar = \frac{1-a}{1+a} due to an error Δa\Delta a in the measurement of aa, we can use the method of series expansion up to the first order of the error (or differentiation with respect to aa).

Given the relation: r(a)=1−a1+ar(a) = \frac{1-a}{1+a}

Let the measured value of aa be a+Δaa + \Delta a. The corresponding value of rr is: r+Δr=1−(a+Δa)1+(a+Δa)=(1−a)−Δa(1+a)+Δar + \Delta r = \frac{1 - (a + \Delta a)}{1 + (a + \Delta a)} = \frac{(1-a) - \Delta a}{(1+a) + \Delta a}

Factoring out (1−a)(1-a) from the numerator and (1+a)(1+a) from the denominator: r+Δr=1−a1+a(1−Δa1−a)(1+Δa1+a)−1r + \Delta r = \frac{1-a}{1+a} \left(1 - \frac{\Delta a}{1-a}\right)\left(1 + \frac{\Delta a}{1+a}\right)^{-1}

Using the binomial expansion (1+x)−1≈1−x(1 + x)^{-1} \approx 1 - x for ∣x∣≪1|x| \ll 1 and retaining terms up to first order in Δa\Delta a: r+Δr≈1−a1+a(1−Δa1−a)(1−Δa1+a)r + \Delta r \approx \frac{1-a}{1+a} \left(1 - \frac{\Delta a}{1-a}\right)\left(1 - \frac{\Delta a}{1+a}\right) r+Δr≈1−a1+a[1−(11−a+11+a)Δa]r + \Delta r \approx \frac{1-a}{1+a} \left[1 - \left(\frac{1}{1-a} + \frac{1}{1+a}\right)\Delta a\right]

Simplifying the term in the parenthesis: 11−a+11+a=(1+a)+(1−a)(1−a)(1+a)=21−a2\frac{1}{1-a} + \frac{1}{1+a} = \frac{(1+a) + (1-a)}{(1-a)(1+a)} = \frac{2}{1-a^2}

Substituting this back gives: r+Δr≈1−a1+a−(1−a1+a)(2(1−a)(1+a))Δar + \Delta r \approx \frac{1-a}{1+a} - \left(\frac{1-a}{1+a}\right) \left(\frac{2}{(1-a)(1+a)}\right)\Delta a r+Δr≈r−2Δa(1+a)2r + \Delta r \approx r - \frac{2\Delta a}{(1+a)^2}

Alternatively, taking the derivative of rr with respect to aa: drda=−(1+a)−(1−a)(1)(1+a)2=−2(1+a)2\frac{dr}{da} = \frac{-(1+a) - (1-a)(1)}{(1+a)^2} = \frac{-2}{(1+a)^2}

Thus, the magnitude of the uncertainty Δr\Delta r is: Δr=∣drda∣Δa=2Δa(1+a)2\Delta r = \left|\frac{dr}{da}\right|\Delta a = \frac{2\Delta a}{(1+a)^2}

Therefore, the correct option is B.

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